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Element 3 and set 1+(1+3^t)*(3^(t+1))
Consider the Proth space

$1+(1+3^t)*(3^{t+1})$

We will be using generating element 3

The [sub]groups generated by 3 have low order [ a lot lower than n-1 ]

Tabulating the elements of groups generated by 3 next

Using the prefix EPr to avoid clashing with existing letter conventions for groups.
Prefer G or S? Replace them in your local copy.

Conjecture: Order of 3 mod N when N of the form 1+(1+3^t)*(3^(t+1)) is 6×(1+2×t)

When t=2 we have P=1+10×27 and the set of elements modulo P is a multiplicative group.

The subgroup generated by 3 has 30 elements

The elements in EPr30 are as shown next
{ 3, 9, 27, 81, 243, 187, 19, 57, 171, 242
184, 10, 30, 90, -1, 268, 262, 244, 190, 28,
84, 252, 214, 100, 29, 87, 261, 241, 181, 1 }

If EPr30 is really a group then we need inverses so let us document those next.
- Inverse of 3 is 181 mod P
- Inverse of 9 is 241 mod P
- Inverse of 27 is 261 mod P
- Inverse of 81 is 87 mod P
- Inverse of 243 is 29 mod P
- Inverse of 187 Is 100 mod P
- Inverse of 19 is 214 mod P
- Inverse of 57 is 252 mod P
- Inverse of 171 is 84 mod P
- Inverse of 242 is 28 mod P
- Inverse of 184 is 190 mod P
- Inverse of 10 is 244 mod P
- Inverse of 30 is 262 mod P
- Inverse of 90 is 268 mod P
We could alternatively write the group EPr30 as follows:
{ 3, 9, 27, 81, (3^5), (3^6), (3^7), (3^8), (3^9), (3^10),
(3^11), -(3^12), -(3^13), -(3^14), -1,
-3, -9, -27, -81, -(3^5), -(3^6), -(3^7), -(3^8), -(3^9), -(3^10),
-(3^11), -(3^12), -(3^13), -(3^14), 1 }

From t=3 the 42 elements in EPr42 are as shown next
{ 3, 9, 27, 81, 243, 729, 2187, 2023, 1531, 55,
165, 495, 1485, 2186, 2020, 1522, 28, 84, 252, 756,
-1, 2266, 2260, 2242, 2188, 2026, 1540, 82, 246, 738,
2214, 2104, 1774, 784, 83, 249, 747, 2241, 2185, 2017,
1513, 1 }

From t=4 the 54 elements in EPr54 are as shown next
{ 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 19195,
17731, 13339, 163, 489, 1467, 4401, 13203, 19682, 19192, 17722,
13312, 82, 246, 738, 2214, 6642, -1, 19924, 19918, 19900,
19846, 19684, 19198, 17740, 13366, 244, 732, 2196, 6588, 19764,
19438, 18460, 15526, 6724, 245, 735, 2205, 6615, 19845, 19681,
19189, 17713, 13285, 1 }

From t=5 the 66 elements in EPr66 are as shown next.
{ 3, 9, 27, 81, 243, 729,
2187, 6561, 19683, 59049, 177147, 175687,
171307, 158167, 118747, 487, 1461, 4383,
13149, 39447, 118341, 177146, 175684, 171298,
158140, 118666, 244, 732, 2196, 6588,
19764, 59292, -1, 177874, 177868, 177850,
177796, 177634, 177148, 175690, 171316, 158194,
118828, 730, 2190, 6570, 19710, 59130,
177390, 176416, 173494, 164728, 138430, 59536,
721, 2193, 6579, 19737, 59211, 177633,
177145, 175681, 171289, 158113, 118585, 1 }

From t=6 the 78 elements in EPr78 are not tabulated here
From t=7 the 90 elements in EPr90 are not tabulated here
From t=8 the 102 elements in EPr102 are not tabulated here
From t=9 the 114 elements in EPr114 are not tabulated here
For t=10 the 126 elements in EPr126 are not tabulated here
For t=11 the 138 elements in EPr138 are not tabulated here
For t=12 the 150 elements in EPr150 are not tabulated here
For t=13 the 162 elements in EPr162 are not tabulated here
For t=14 the 174 elements in EPr174 are not tabulated here
For t=15 the 186 elements in EPr186 are not tabulated here
For t=16 the 198 elements in EPr198 are not tabulated here
For t=17 the 210 elements in EPr210 are not tabulated here
For t=18 the 222 elements in EPr222 are not tabulated here
For t=19 the 234 elements in EPr234 are not tabulated here
For t=20 the 246 elements in EPr246 are not tabulated here

Using sturdy element notation we might say that the groups tabulated here are
August 21, 2026
Element 3 and set 1+(-1+3^t)*(3^(t+1))
Consider the Proth space

$1+(-1+3^t)*(3^{t+1})$

We will be using generating element 3

The [sub]groups generated by 3 have low order [ a lot lower than n-1 ]

Tabulating the elements of groups generated by 3 next

Using prefix ENp to avoid clashing with existing letter conventions for groups.
Prefer G or S? Replace them in your local copy.

Conjecture: Order of 3 mod N when N of the form 1+(-1+3^t)*(3^(t+1)) is 6×(1+2×t)

When t=2 we have P=1+8×27 and the set of elements modulo P is a multiplicative group.

The subgroup generated by 3 has 30 elements

The elements in ENp30 are as shown next
{ 3, 9, 27, 81, 26, 78, 17, 51, 153, 25,
75, 8, 24, 72, -1, 214, 208, 190, 136, 191,
139, 200, 166, 64, 192, 142, 209, 193, 145, 1 }

If ENp30 is really a group then we need inverses so let us document those next.
- Inverse of 3 is 145 mod P
- Inverse of 9 is 193 mod P
- Inverse of 27 is 209 mod P
- Inverse of 81 is 142 mod P
- Inverse of 26 is 192 mod P
- Inverse of 78 is 64 mod P
- Inverse of 17 is 166 mod P
- Inverse of 51 is 200 mod P
- Inverse of 153 is 139 mod P
- Inverse of 25 is 191 mod P
- Inverse of 75 is 136 mod P
- Inverse of 8 is 190 mod P
- Inverse of 24 is 208 mod P
- Inverse of 72 is 214 mod P \\ 214==Mod((3^14),pr)^-1
We could alternatively write the group ENp30 as follows:
{ 3, 9, 27, 81, (3^5), (3^6), (3^7), (3^8), (3^9), (3^10),
(3^11), -(3^12), -(3^13), -(3^14), -1,
-3, -9, -27, -81, -(3^5), -(3^6), -(3^7), -(3^8), -(3^9), -(3^10),
-(3^11), -(3^12), -(3^13), -(3^14), 1 }

From t=3 the 42 elements in ENp42 are as shown next
{ 3, 9, 27, 81, 243, 729, 80, 240, 720, 53
159, 477, 1431, 79, 237, 711, 26, 78, 234, 702,
-1, 2104, 2098, 2080, 2026, 1864, 1378, 2027, 1867, 1387,
2054, 1948, 1630, 676, 2028, 1870, 1396, 2081, 2029, 1873,
1405, 1 }

From t=4 the 54 elements in ENp54 are as shown next
{ 3, 9, 27, 81, 243, 729, 2187, 6561,
242, 726, 2178, 6534, 161, 483, 1449, 4347,
13041, 241, 723, 2169, 6507, 80, 240, 720,
2160, 6480, -1, 19438, 19432, 19414, 19360, 19198,
18712, 17254, 12880, 19199, 18715, 17263, 12907, 19280,
18958, 17992, 15094, 6400, 19200, 18718, 17272, 12934,
19361, 19201, 18721, 17281, 12961, 1 }

From t=5 the 66 elements in ENp66 are as shown next.
{ 3, 9, 27, 81, 243, 729, 2187, 6561,
19683, 59049, 728, 2184, 6552, 19656, 58968, 485,
1455, 4365, 13095, 39285, 117855, 727, 2181, 6543,
19629, 58887, 242, 726, 2178, 6534, 19602, 58806,
-1, 176416, 176410, 176392, 176338, 176176, 175690, 174232,
169858, 156736, 117370, 175691, 174235, 169867, 156763, 117451,
175934, 174964, 172054, 163324, 137134, 58564, 175692, 174238,
169876, 156790, 117532, 176177,
175693, 174241, 169885, 156817, 117613, 1 }

From t=6 the 78 elements in ENp78 are not tabulated here
From t=7 the 90 elements in ENp90 are not tabulated here
From t=8 the 102 elements in ENp102 are not tabulated here
From t=9 the 114 elements in ENp114 are not tabulated here
For t=10 the 126 elements in ENp126 are not tabulated here
For t=11 the 138 elements in ENp138 are not tabulated here
For t=12 the 150 elements in ENp150 are not tabulated here
For t=13 the 162 elements in ENp162 are not tabulated here
For t=14 the 174 elements in ENp174 are not tabulated here
For t=15 the 186 elements in ENp186 are not tabulated here
For t=16 the 198 elements in ENp198 are not tabulated here
For t=17 the 210 elements in ENp210 are not tabulated here
For t=18 the 222 elements in ENp222 are not tabulated here
For t=19 the 234 elements in ENp234 are not tabulated here
For t=20 the 246 elements in ENp246 are not tabulated here
August 21, 2026
Element 2 and set 1+(1+2^t)*(2^(t+1))
Consider the Proth space

$1+(1+2^t)*(2^{t+1})$

We will be using generating element 2

The [sub]groups generated by 2 have low order [ a lot lower than n-1 ]

Tabulating the elements of groups generated by 2 next.

Using the prefix TPv to avoid clashing with existing letter conventions for groups.
Prefer G or S? Replace them in your local copy.

When t=2 we have P=1+5×8 and the set of elements modulo P is a multiplicative group.

The subgroup generated by 2 has 20 elements

The elements in TPv20 are as shown next
{ 2, 4, 8, 16, 32, 23, 5, 10, 20, -1
39, 37, 33, 25, 9, 18, 36, 31, 21, 1 }

If TPv20 is really a group then we need inverses so let us document those next.
- Inverse of 2 is 21 mod P
- Inverse of 4 is 31 mod P
- Inverse of 8 is 36 mod P
- Inverse of 16 is 18 mod P
- Inverse of 32 is 9 mod P
- Inverse of 23 is 25 mod P
- Inverse of 5 is 33 mod P
- Inverse of 10 is 37 mod P
- Inverse of 20 is 39 mod P
Conjecture: Order of 2 mod N when N of the form 1+(1+2^t)*(2^(t+1)) is 4×(1+2×t)

From t=3 the 28 elements in TPv28 are as shown next
{ 2, 4, 8, 16, 32, 64, 128, 111, 77, 9,
18, 36, 72, -1, 143, 141, 137, 129, 113, 81,
17, 34, 68, 136, 127, 109, 73, 1 }

We could alternatively write the group TPv28 as follows:
{ 2, 4, 8, 16, 32, 64, 15, 30, 60, 7, 14, 28, 56, -1,
-2, -4, -8, -16, -32, -64, -(2^7), -(2^8), -(2^9), -(2^10), -(2^11),
-(2^12), -(2^13), 1 }

From t=4 the 36 elements in TPv36 are as shown next
{ 2, 4, 8, 16, 32, 64, 128, 256, 512, 479,
413, 281, 17, 34, 68, 136, 272, -1, 543, 541,
537, 529, 513, 481, 417, 289, 33, 66, 132, 264,
528, 511, 477, 409, 273, 1 }

From t=5 the 44 elements in TPv44 are as shown next
{ 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
2048, 1983, 1853, 1593, 1073, 33, 66, 132, 264, 528,
1056, -1, 2111, 2109, 2105, 2097, 2081, 2049, 1985, 1857,
1601, 1089, 65, 130, 260, 520, 1040, 2080, 2047, 1981,
1849, 1585, 1057, 1 }

From t=6 the 52 elements in TPv52 are as shown next.
{ 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
2048, 4096, 8192, 8063, 7805, 7289, 6257, 4193, 65, 130,
260, 520, 1040, 2080, 4160, -1, 8319, 8317, 8313, 8305,
8289, 8257, 8193, 8065, 7809, 7297, 6273, 4225, 129, 258,
516, 1032, 2064, 4128, 8256, 8191, 8061, 7801, 7281, 6241,
4161, 1 }

From t=7 the 60 elements in TPv60 are as shown next.
{ 2, 4, 8, 16, 32, 64,
128, 256, 512, 1024, 2048, 4096,
8192, 16384, 32768, 32511, 31997, 30969,
28913, 24801, 16577, 129, 258, 516,
1032, 2064, 4128, 8256, 16512, -1,
33023, 33021, 33017, 33009, 32993, 32961,
32897, 32769, 32513, 32001, 30977, 28929,
24833, 16641, 257, 514, 1028, 2056,
4112, 8224, 16448, 32896, 32767, 32509,
31993, 30961, 28897, 24769, 16513, 1 }

From t=8 the 68 elements in TPv68 are as shown next.
{ 2, 4, 8, 16, 32, 64,
128, 256, 512, 1024, 2048, 4096,
8192, 16384, 32768, 65536, 131072, 130559,
129533, 127481, 123377, 115169, 98753, 65921,
257, 514, 1028, 2056, 4112, 8224,
16448, 32896, 65792, -1, 131583, 131581,
131577, 131569, 131553, 131521, 131457, 131329,
131073, 130561, 129537, 127489, 123393, 115201,
98817, 66049, 513, 1026, 2052, 4104,
8208, 16416, 32832, 65664, 131328, 131071,
130557, 129529, 127473, 123361, 115137, 98689, 65793, 1 }

From t=9 the 76 elements in TPv76 are not tabulated here
For t=10 the 84 elements in TPv84 are not tabulated here
For t=11 the 92 elements in TPv92 are not tabulated here
For t=12 the 100 elements in TPv100 are not tabulated here
For t=13 the 108 elements in TPv108 are not tabulated here
For t=14 the 116 elements in TPv116 are not tabulated here
For t=15 the 124 elements in TPv124 are not tabulated here
For t=16 the 132 elements in TPv132 are not tabulated here
For t=17 the 140 elements in TPv140 are not tabulated here
For t=18 the 148 elements in TPv148 are not tabulated here
For t=19 the 156 elements in TPv156 are not tabulated here
For t=20 the 164 elements in TPv164 are not tabulated here
August 21, 2026
Element 2 and set 1+(-1+2^t)*(2^(t+1))
Consider the Proth space

$1+(-1+2^t)*(2^{t+1})$

We will be using generating element 2

The [sub]groups generated by 2 have low order [ a lot lower than n-1 ]

Tabulating the elements of groups generated by 2 next.

Using the prefix TNx to avoid clashing with existing letter conventions for groups.
Prefer G or S? Replace them in your local copy.

From t=2 the elements in TNx20 are as shown next
{ 2, 4, 8, 16, 7, 14, 3, 6, 12, -1,
23, 21, 17, 9, 18, 11, 22, 19, 13, 1 }

Conjecture: Order of 2 mod N when N of the form 1+(-1+2^t)*(2^(t+1)) is 4×(1+2×t)

When t=3 we have P=1+7×16 and the set of elements modulo P is a multiplicative group.

The subgroup generated by 2 has 28 elements

The elements in TNx28 are as shown next
{ 2, 4, 8, 16, 32, 64, 15, 30, 60, 7, 14, 28, 56, -1,
111, 109, 105, 97, 81, 49, 98, 83, 53, 106, 99, 85, 57, 1 }

If TNx28 is really a group then we need inverses so let us document those next.
- Inverse of 2 is 57 mod P
- Inverse of 4 is 85 mod P
- Inverse of 8 is 99 mod P
- Inverse of 16 is 106 mod P
- Inverse of 32 is 53 mod P
- Inverse of 15 is 98 mod P
- Inverse of 30 is 49 mod P
- Inverse of 60 is 81 mod P
- Inverse of 7 is 97 mod P
- Inverse of 14 is 105 mod P
- Inverse of 28 is 109 mod P
- Inverse of 56 is 111 mod P
We could alternatively write the group TNx28 as follows:
{ 2, 4, 8, 16, 32, 64, 15, 30, 60, 7, 14, 28, 56, -1,
-2, -4, -8, -16, -32, -64, -(2^7), -(2^8), -(2^9), -(2^10), -(2^11),
-(2^12), -(2^13), 1 }

From t=4 the 36 elements in TNx36 are as shown next
{ 2, 4, 8, 16, 32, 64, 128, 256, 31,
62, 124, 248, 15, 30, 60, 120, 240, -1,
479, 477, 473, 465, 449, 417, 353, 225, 450,
419, 357, 233, 466, 451, 421, 361, 241, 1 }

From t=5 the 44 elements in TNx44 are as shown next
{ 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
63, 126, 252, 504, 1008, 31, 62, 124, 248, 496,
992, -1, 1983, 1981, 1977, 1969, 1953, 1921, 1857, 1729
1473, 961, 1922, 1859, 1733, 1481, 977, 1954, 1923, 1861,
1737, 1489, 993, 1 }

From t=6 the 52 elements in TNx52 are as shown next.
{ 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
2048, 4096, 127, 254, 508, 1016, 2032, 4064, 63, 126,
252, 504, 1008, 2016, 4032, -1, 8063, 8061, 8057, 8049,
8033, 8001, 7937, 7809, 7553, 7041, 6017, 3969, 7938, 7811,
7557, 7049, 6033, 4011, 8002, 7939, 7813, 7561, 7057, 6049,
4033, 1 }

From t=7 the 60 elements in TNx60 are as shown next.
{ 2, 4, 8, 16, 32, 64,
128, 256, 512, 1024, 2048, 4096,
8192, 16384, 255, 510, 1020, 2040,
4080, 8160, 16320, 127, 254, 508,
1016, 2032, 4064, 8128, 16256, -1,
32511, 32509, 32505, 32497, 32481, 32449,
32385, 32257, 32001, 31489, 30465, 28417,
24321, 16129, 32258, 32003, 31493, 30473,
28433, 24353, 16193, 32386, 32259, 32005,
31497, 30481. 28449, 24385, 16257, 1 }

From t=8 the 68 elements in TNx68 are as shown next.
{ 2, 4, 8, 16, 32, 64,
128, 256, 512, 1024, 2048, 4096,
8192, 16384, 32768, 65536, 511, 1022,
2044, 4088, 8176, 16352, 32704, 65408,
255, 510, 1020, 2040, 4080, 8160,
16320, 32640, 65280, -1, 130559, 130557,
130553, 130545, 130529, 130497, 130433, 130305,
130049, 129537, 128513, 126465, 122369, 114177,
97793, 65025, 130050, 129539, 128517, 126473,
122385, 114209, 97857, 65153, 130306, 130051,
129541, 128521, 126481, 122401, 114241, 97921, 65281, 1 }

From t=9 the 76 elements in TNx76 are not tabulated here
For t=10 the 84 elements in TNx84 are not tabulated here
For t=11 the 92 elements in TNx92 are not tabulated here
For t=12 the 100 elements in TNx100 are not tabulated here
For t=13 the 108 elements in TNx108 are not tabulated here
For t=14 the 116 elements in TNx116 are not tabulated here
For t=15 the 124 elements in TNx124 are not tabulated here
For t=16 the 132 elements in TNx132 are not tabulated here
For t=17 the 140 elements in TNx140 are not tabulated here
For t=18 the 148 elements in TNx148 are not tabulated here
For t=19 the 156 elements in TNx156 are not tabulated here
For t=20 the 164 elements in TNx164 are not tabulated here

Using sturdy element notation we might say the results tabulated in this post are
August 20, 2026
Element 1+3^t and set 1+(1+3^t)*(3^(t+1))
Consider the Proth space

$1+(1+3^t)*(3^{t+1})$

We will be using generating element 1+3^t

The [sub]groups generated by 1+3^t have low order [ a lot lower than n-1 ]

Tabulating the elements of groups generated by 1+3^t next

Using the prefix EPu to avoid clashing with existing letter conventions for groups.
Prefer G or S? Replace them in your local copy.

When t=2 we have P=1+10×27 and the set of elements modulo P is a multiplicative group.

Our generator 1+3^t is 10 and the subset it generates has 5 elements.

The elements in EPu5 are { 10, 100, 187, 244, 1 }

If EPu5 is really a group then we need inverses so let us document those next.
- Inverse of 10 is 244 mod P
- Inverse of 100 is 187 mod P
For 28 from t= 3 the elements in EPu42 are as shown next.
{ 28, 784, 1531, 2026, 3, 84, 83, 55, 1540, 9,
252, 249, 165, 82, 27, 756, 747, 495, 246, 81,
-1, 2241, 1485, 738, 243, 2266, 2185, 2186, 2214, 729,
2260, 2017, 2020, 2104, 2187, 2242, 1513, 1522, 1774, 2023,
2188, 1 }

The seventh to last element is always -27

The thirteenth to last element is always 729 alternatively written as 3^6

Every twelfth element after 3^6 from the end of the subgroup is a power of 3.

In order we would write them as follows

$3^6, 3^{12}, 3^{18}, 3^{24}, 3^{30}, 3^{36}, 3^{42}, 3^{48}, 3^{54}, 3^{60}$

Writing them longhand we might say
- thirteenth from last is 3^6
- twentyfifth from last is 3^12
- …
- fortyninth from last is 3^24
- …
In calculation terms we can jump to any portion of our group and be at most 11 elements away from the element we require.

This can be a tremendous labour saver as instead of repeated powering to obtain an element of the subgroup, we can jump near our required element and then conduct a much smaller number of repeated powering to gain our result.

For 82 from t=4 the 27 elements in EPu27 are shown next.
{ 82, 6274, 13339, 17740, 9, 738, 735, 489, 244,
81, 6642, 6615, 4401, 2196, 729, 19924, 19681, 19682,
19764, 6561, 19900, 17713, 17722, 18460, 19195, 19684, 1 }

Conjecture: The number of elements in the group generated by 1+3^t working modulo P where P is 1+(1+3^t)*(3^(t+1)) must be a multiple of 1+2×t

For 244 from t=5 the 22 elements in EPu22 are not tabulated here
For 730 from t=6 the 39 elements in EPu39 are not tabulated here
For 2188 from t=7 the 90 elements in EPu90 are not tabulated here
For 6562 from t=8 the 17 elements in EPu17 are not tabulated here
For 19682 from t=9 the 114 elements in EPu114 are not tabulated here
For 59048 from t=10 the 63 elements in EPu63 are not tabulated here
For 177146 from t=11 the 46 elements in EPu46 are not tabulated here
For 531440 from t=12 the 75 elements in EPu75 are not tabulated here
For 1594322 from t=13 the elements in EPu162 are not tabulated here
For 4782968 from t=14 the elements in EPu29 are not tabulated here
For 14348906 from t=15 the elements in EPu186 are not tabulated here
For 43046720 from t=16 the elements in EPu99 are not tabulated here
For 129140162 from t=17 the elements in EPu70 are not tabulated here
For 387420488 from t=18 the elements in EPu111 are not tabulated here
For 1162261466 from t=19 the elements in EPu234 are not tabulated here
For 3486784400 from t=20 the elements in EPu41 are not tabulated here
August 20, 2026
Element -1+3^t and set 1+(-1+3^t)*(3^(t+1))
Consider the Proth space

$1+(-1+3^t)*(3^{t+1})$

We will be using generating element −1+3^t

The [sub]groups generated by −1+3^t have low order [ a lot lower than n-1 ]

Tabulating the elements of groups generated by −1+3^t next.

Using the ENt to avoid clashing with existing letter conventions for groups. Prefer G or S? Replace them in your local copy.

For 8 from t=2 elements ENt5 are { 8, 64, 78, 190, 1 }

For 26 from t= 3 the elements in ENt42 are as shown next.
{ 26, 676, 720, 1864, 3, 78, 2028, 53, 1378, 9
234, 1870, 159, 2027, 27, 702, 1396, 477, 1867, 81
2106, 2081, 1431, 1387, 243, 2104, 2029, 79, 2054, 729
2098, 1873, 237, 1948, 80, 2080, 1405, 711, 1630, 240
2026, 1 }

When t=4 we have P=1+80×243 and the set of elements modulo P is a multiplicative group.

Our generator −1+3^t is 80 and the subset it generates has 27 elements.

The elements in ENt27 are as shown next
{ 80, 6400, 6534, 17254, 9, 720, 18718, 483, 19199, 81,
6480, 12934, 4347, 17263, 729, 19438, 19201, 241, 19280, 6561,
19414, 17281, 2169, 17992, 726, 19198, 1 }

If ENt27 is really a group then we need inverses so let us document those next.
- Inverse of 80 is 19198 mod P
- Inverse of 6400 is 726 mod P
- Inverse of 6534 is 17992 mod P
- Inverse of 17254 is 2169 mod P
- Inverse of 9 is 17281 mod P
- Inverse of 720 is 19414 mod P
- Inverse of 18718 is 6561 mod P
- Inverse of 483 is 19280 mod P
- Inverse of 19199 is 241 mod P
- Inverse of 81 is 19201 mod P
- Inverse of 6480 is 19438 mod P
- Inverse of 12934 is 729 mod P
- Inverse of 4347 is 17263 mod P
For 242 from t=5 the 22 elements in ENt22 are as shown next
{ 242, 58564, 58968, 156736, 27, 6534, 169876, 4365,
174235, 729, 176418, 176177, 117855, 117451, 19683, 176392,
169885, 6543, 172054, 2184, 175690, 1 }

The seventh to last element is always -27

The thirteenth to last element is always 729 alternatively written as 3^6

For 728 from t=6 the 39 elements in ENt39 are as shown next
{ 728, 529984, 531198, 1414990, 81, 58968,
1533142, 39339, 1572463, 6561, 1592134, 1589953,
2185, 1590680, 531441, 1591894, 1415233, 176985,
1474120, 59022, 1572454, 9, 6552, 1585582,
4371, 1589951, 729, 530712, 1061182, 354051, 1415071, 59049,
1592110, 1572481, 19665, 1579024, 6558, 1589950, 1 }

Every twelfth element after 3^6 from the end of the subgroup is a power of 3.

In order we would write them as follows

$3^6, 3^{12}, 3^{18}, 3^{24}, 3^{30}, 3^{36}, 3^{42}, 3^{48}, 3^{54}, 3^{60}$

Writing them longhand we might say
- thirteenth from last is 3^6
- twentyfifth from last is 3^12
- …
- fortyninth from last is 3^24
- …
In calculation terms we can jump to any portion of our group and be at most 11 elements away from the element we require.

This can be a tremendous labour saver as instead of repeated powering to obtain an element of the subgroup, we can jump near our required element and then conduct a much smaller number of repeated powering to gain our result.

For 2186 from t=7 the 90 elements in ENt90 are as shown next.
{ 2186,4778596,4782240,12748024,243,531198,
13811068,354213,14165227,59049,14342338,14322673,
19677,14329228,6560,14340160,9561565,4781511,
11154430,1594080,13810906,81,177066,14165254,
118071,14283307,19683,14342344,14335789,6559,
14337974,4782969,14341618,12748753,1593837,13279708
531360,14165200,27,59022,14283316,39357,
14322667,6561,-1,14340161,9563751,9560107,
1594323,14342104,13811149,531279,13988134,177120,
14283298,9,19674,14322670,13119,14335787,
2187,4780782,9560836,3187917,12748267,531441,
14342266,14165281,177093,14224276,59040,14322664,
3,6558,14335788,4373,9559378,729,
1593594,12748510,1062639,13810987,177147,14342320,
14283325,59031,14302990,19680,14335786,1 }

The sixth element can be obtained as 531198==Mod(-27,14342347)^-1 alternatively

The twelfth element can be obtained as 14322673==Mod(729,14342347)^-1 alternatively

For 6560 from t=8 the 68 elements in ENt17 are as shown next.
{ 6560, 43033600, 43044534, 114771574, 729, 4782240,
124337998, 3188403, 127526239, 531441, 129120454, 128943361,
177129, 129002392, 59046, 129100798, 1 }

Conjecture: The number of elements in the group generated by −1+3^t working modulo P where P is 1+(-1+3^t)*(3^(t+1)) must be a multiple of 1+2×t

For 19682 from t=9 the 114 elements in ENt114 are not tabulated here
For 59048 from t=10 the 63 elements in ENt63 are not tabulated here
For 177146 from t=11 the 46 elements in ENt46 are not tabulated here
For 531440 from t=12 the 75 elements in ENt75 are not tabulated here
For 1594322 from t=13 the elements in ENt162 are not tabulated here
For 4782968 from t=14 the elements in ENt29 are not tabulated here
For 14348906 from t=15 the elements in ENt186 are not tabulated here
For 43046720 from t=16 the elements in ENt99 are not tabulated here
For 129140162 from t=17 the elements in ENt70 are not tabulated here
For 387420488 from t=18 the elements in ENt111 are not tabulated here
For 1162261466 from t=19 the elements in ENt234 are not tabulated here
For 3486784400 from t=20 the elements in ENt41 are not tabulated here
August 20, 2026
Element 1+2^t and set 1+(1+2^t)*(2^(t+1))
Consider the Proth space

$1+(1+2^t)*(2^{t+1})$

We will be using generating element 1 + 2^t

The [sub]groups generated by 1+2^t have low order [ a lot lower than n-1 ]

A related Conjecture is given next.

Conjecture: Order of 1+2^t mod P when P is prime and P of the form 1+(1+2^t)*(2^(t+1)) divides Lcm(2+4×t−(t+1),2+4×t)

Tabulating the elements of groups generated by 1+2^t next.

Using the TPy to avoid clashing with existing letter conventions for groups. Prefer G or S? Replace them in your local copy.

For 5 from t=2 the 20 elements in TPy20 are as shown next
{ 5,25,2,10,9,4,20,18,8,40,36,16,39,31,32,37,21,23,33,1}
which when put in numeric order are
1,2,4,5,8,9,10,16,18,20,21,23,25,31,32,33,36,37,39,40

Elements that do not appear are as shown next
3,6,7,11,12,13,14,15,17,19,22,24,26,27,28,29,30,34,35,38

When t=3 we have P=1+9×16 and the set of elements modulo P is a multiplicative group.

Our generator 1+2^t is 9 and the subset it generates has 14 elements.

The elements in TPy14 are
{ 9,81,4,36,34,16,144,136,64,141,109,111,129,1 }

If TPy14 is really a group then we need inverses so let us document those next.
- Inverse of 9 is 129 mod P
- Inverse of 81 is 111 mod P
- Inverse of 4 is 109 mod P
- Inverse of 36 is 141 mod P
- Inverse of 34 is 64 mod P
- Inverse of 16 is 136 mod P
For 17 from t=4 the 36 elements in TPy36 are as shown next
{ 17,289,8,136,132,64,543,511,512,529,
273,281,417,4,68,66,32,-1,528,256,
537,409,413,481,2,34,33,16,272,264,
128,541,477,479,513,1 }

For 33 from t=5 the 11 elements in TPy11 are as shown next
{ 33,1089,16,528,520,256,2109,1981,1983,2049,1 }

For 65 from t=6 the 52 elements in TPy52 are as shown next
{ 65,4225,32,2080,2064,1024,8313,7801,7805,8065,
2,130,129,64,4160,4128,2048,8305,7281,7289,
7809,4,260,258,128,8320,8256,4096,8289,6241,
6257,7297,8,520,516,256,8319,8191,8192,8257,
4161,4193,6273,16,1040,1032,512,8317,8061,8063,
8193,1 }

The fifth to last element is always -4

Every eighth element after 16 from the end of the subgroup is a power of 2.

In order we would write them as follows

$2^4, 2^8, 2^{12}, 2^{16}, 2^{24}, 2^{28}, 2^{32}, 2^{36}, 2^{40}$

Writing them longhand we might say
eight from last is 16
sixteenth from last is 2^8
twentyfourth from last is 2^12
…
eightieth from last is 2^40

In calculation terms we can jump to any portion of our group and be at most 7 elements away from the element we require.

This can be a tremendous labour saver as instead of repeated powering to obtain an element of the subgroup, we can jump near our required element and then conduct a much smaller number of repeated powering to gain our result.

For 129 from t=7 the 30 elements in TPy30 are as shown next.
{ 129, 16641, 64, 8256, 8224, 4096, 33009, 30961, 30969, 32001,
4, 516, 514, 256, -1, 32896, 16384, 32961, 24769, 24801,
28929, 16, 2064, 2056, 1024, 33021, 32509, 32511, 32769, 1 }

The fourth element can be obtained as
8256==Mod(-4,33025)^-1 alternatively

The eighth element can be obtained as
30961==Mod(16,33025)^-1 alternatively

For 257 from t=8 the 68 elements in TPy68 are as shown next.
{ 257, 66049, 128, 32896, 32832, 16384, 131553, 123361,
123377, 127489, 8, 2056, 2052, 1024, 131583, 131071,
131072, 131329, 65793, 65921, 98817, 64, 16448, 16416,
8192, 131569, 127473, 127481, 129537, 4, 1028, 1026,
512, -1, 131328, 65536, 131457, 98689, 98753, 115201,
32, 8224, 8208, 4096, 131577, 129529, 129533, 130561, 2, 514
513, 256, 65792, 65664, 32768, 131521, 115137, 115169, 123393, 16,
4112, 4104, 2048, 131581, 130557, 130559, 131073, 1 }

Conjecture: The number of elements in the group generated by 1+2^t working modulo P where P is 1+(1+2^t)*(2^(t+1)) must be a multiple of 1+2×t

For 513 from t=9 the 19 elements in TPy19 are as shown next
{ 513, 263169, 256, 131328, 131200, 65536, 525249, 492481,
492513, 408929, 16, 8208, 8200, 4096, 525309, 523261,
523263, 524289, 1 }

For 1025 from t=10 the 84 elements in TPy84 are not tabulated here
For 2049 from t=11 the 46 elements in TPy46 are not tabulated here
For 4097 from t=12 the elements in TPy100 are not tabulated here
For 8193 from t=13 the 27 elements in TPy27 are not tabulated here
For 16385 from t=14 the elements in TPy116 are not tabulated here
For 32769 from t=15 the elements in TPy62 are not tabulated here
For 65537 from t=16 the elements in TPy132 are not tabulated here
For 131073 from t=17 the elements in TPy35 are not tabulated here
For 262145 from t=18 the elements in TPy148 are not tabulated here
For 524289 from t=19 the elements in TPy78 are not tabulated here
For 1048577 from t=20 elements in TPy164 are not tabulated here

Using sturdy element notation we might say the results tabulated in this post are
August 20, 2026
Does widening a definition weaken it?

Why do we need another definition?

We have the definition for generator being an element of a group.

But a sturdy element is instead an element of a set.

We could widen the definition for generator to be an element of a set.

Would that create problems?

Should we say super generator instead of sturdy element?

But this might require or imply a generator could be an element of a set.

There is nothing wrong with describing the results of using a sturdy element modulo enclosing set as being a cyclic group.

However to call those results a cyclic subgroup implies a parent group and that is not always the case.

We could widen the definition of subgroup to include a parent set also.

Would that create problems?

A new definition “super element” avoids having to widen existing definitions.

Perhaps because a debate about widening generator or subgroup has not yet happened shows that these elements and their behaviour have not yet been studied more widely.

“They are just generators”
But they do not belong to a group [in all cases]

“They are super generators”
We need to be careful we do not imply group membership [in all cases] by reusing the term generator

“They generate cyclic subgroups”
We need to be careful we do not imply group membership [in all cases] by the current definition of a subgroup

But wouldn’t having a sturdy element require a new notation?

If we reuse the angle brackets that we use for generator typically then we could risk confusion.

How about double angle brackets

<<-1+2^t>> for the sturdy element -1+2^t

<<1+2^t>> for the sturdy element 1+2^t

<<-1+3^t>> for the sturdy element -1+3^t

<<1+3^t>> for the sturdy element 1+3^t

Each of those sturdy elements has a context set or enclosing set so should be quoted in more complete form as shown next.

August 15, 2026
Defining a sturdy element

The set of elements modulo 35 is a set

Working modulo 35 does not give you a group because we can find zero divisors and those are not invertible.

Multiples of 5 and 7 are zero divisors because they are divisors of 35.

There is no inverse for the elements 5, 7, 10, 14, 15, 20, 21, 25, 28,30 in that set

Listing the elements not on that list gives us a subset consisting of 24 elements as follows
{ 1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34 }

Does the zero divisor 5 generate a group?

Powering of 5 just gives us a series of non-invertible results
{ 5, 25, 20, 30, 10, 15 }

Our generated results do not include the identity element 1 or similar and as we noted there are no inverses.

Does the element 6 generate a group?

We obtain a cyclic group having 2 elements { 6, 1 }

Does the zero divisor 7 generate a group?

Successive powering of 7 just gives us a series of non-invertible results { 7, 14, 28, 21 }

Our generated results do not include the identity element 1 or similar and as we noted there are no inverses.

Does the element 3 generate a multiplicative group?

We obtain a cyclic group having 12 elements
{ 3, 9, 27, 11, 33, 29, 17, 16, 13, 4, 12, 1 }

Does the element 11 generate a multiplicative group?

We obtain a cyclic group having 3 elements { 11, 16, 1 }

Next we attempt to define what a super element is

(i) A sturdy element requires context
[ a set in which it’s properties are special ]

(ii) A sturdy element always generates a group [or stronger] from that enclosing context.

(iii) A sturdy element generates a multiplicative group whose order is

$<= \frac{(n−1) }{2}$

(iv a) A sturdy element shares the context set with an element that generates a maximal group (n-1) elements (depending on the value of our source variable [t])

(iv b) A sturdy element shares the context set with an element that generates a near maximal group (n-1-zero divisors based adjustment) elements (depending on the value of our source variable [t])

For the set

$1+(-1+2^t)*(2^{t+1})$

the element -1 + 2^t is a sturdy element

For the set

$1+(-1+2^t)*(2^{t+1})$

the element 392+t is not a sturdy element

Setting t=5 we obtain 397 for 392+t

Working modulo 1985 we obtain the following from successive powering of 397
{ 397, 794, 1588, 1191, 397, … }

That subset cannot be a group because it contains a zero divisor in 397

That series of elements is not a group so our element 392+t fails property (ii) and is therefore not a sturdy element.

For the set

$1+(1+2^t)*(2^{t+1})$

the element 1 + 2^t is a sturdy element.

For the set

$1+(1+2^t)*(2^{t+1})$

the element 105+t is not a sturdy element.

Setting t=4 we obtain 109 for 105+t

Working modulo 545 we obtain the following from successive powering of 109 { 109, 436, 109 }

That subset cannot be a group as it contains a zero divisor in 109

That series of elements is not a group so our element 105+t fails property (ii) and is therefore not a sturdy element.

For the set

$1+(-1+3^t)*(3^{t+1})$

the element -1 + 3^t is a sturdy element

For the set

$1+(-1+3^t)*(3^{t+1})$

the element 38+t is not a sturdy element

Setting t=5 we obtain 43 for 38+t

Working modulo 2107 we obtain the following from successive powering of 43
{ 43, 1849, 1548, 1247, 946, 645, 344, 43, … }

That subset cannot be a group as it contains a zero divisor in 43

That series of elements is not a group so our element 38+t fails property (ii) and is therefore not a sturdy element.

For the set

$1+(1+3^t)*(3^{t+1})$

the element 1 + 3^t is a sturdy element

For the set

$1+(1+3^t)*(3^{t+1})$

the element 25406+t is not a sturdy element

Setting t=5 we obtain 25411 for 25406+t

Working modulo 177877 we obtain the following from successive powering of 25411 { 25411 }

That subset cannot be a group as it contains a zero divisor in 25411

That series of elements is not a group so our element 24506+t fails property (ii) and is therefore not a sturdy element.

Author: Gary Wright 2026

This post appears as the first chapter in a draft book and is available as a pdf

August 14, 2026
Abundant relations in 1+(-1+2^t)*(2^(t+1))

The number space

$1+(-1+2^t)*(2^{t+1})$

has an abundance of relations

In this post I provide links to pages about those relations

Relation ‘lowercase b’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase c’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase d’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase f’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase g’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase h’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase i’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase n’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase p’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase q’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase r’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase s’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase t’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase u’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase w’ in group 1+(-1+2^t)*(2^(t+1))

Relation ‘lowercase z’ in group 1+(-1+2^t)*(2^(t+1))

July 27, 2026