Mathrelated

Category: abstractalgebra

  • Mathematical discovery

    There are many ways that Mathematical discovery can happen.

    Doing an Undergraduate degree or Masters or PhD are ways that discoveries can be made.

    Working in Algebra for many years and asking “what if” is another way.

    My discovery, depending on your point of view, is one or more of the following:

    • Discovered a strong probable prime or ‘is prime’ test for a particular set
    • “fixed the b” so same base used all the time for a prime test of Proth numbers
    • Discovered a set in which a probable prime test behaves strongly.

    Thought it worth describing the process that has allowed that to happen and will talk subjectively about that next.
    ( Do jump ahead to the Conjectures at the bottom of this post if you are interested more in that )

    Have spent many years working in the set

    1+(−1+2t)∗(2t+1)1+(-1+2^t)*(2^{t+1})

    Documented lots of algebraic observations, and through them, was able to come up with an order conjecture involving a least common multiple.

    Nothing too exciting so far.

    Through thinking about order of two elements in particular, began to see the importance of thinking lengthwise about things with decreasing and increasing values of t.

    This is in contrast to thinking about isomorphism primarily, where we tend to view order as an important separator and then move laterally between [groups].


    Went so far as to propose a new definition “sturdy element” to facilitate this thinking.

    Tabulated [a lot] of group examples using those elements.
    ( See other posts on this site )

    Adjusted the set definition slightly from

    1+(−1+2t)∗(2t+1)1+(-1+2^t)*(2^{t+1})

    to nearby sets

    This was a key step

    Broke away from considering only sets we can describe as Proth numbers and considered other sets.

    Still thinking about order and patterns and came up with some new conjectures.

    To try and add a few chapters to my draft book, returned to sets of Proth numbers but considered slightly different powers.

    This was another key step.

    During this time was always asking “what if” and “suppose” type questions of what I was seeing.


    This supposing and enquiring was another key step.


    Spotted something interesting when working with groups in set

    1+(−1+2t)∗(2t+3)1+(-1+2^t)*(2^{t+3})

    Factored the order of a couple of groups to see if there was anything to see.


    Made a supposition and tested it for a couple of examples.


    Found that the pattern did not apply in all cases.


    Asked the question did it only apply to primes.


    There was the discovery [ see conjecture next ]

    Created a script to test things out.


    Used a computer algebra package to run the script to see it work with larger examples.


    The largest prime found [9769 digits] using this probable prime test as a filter is next.

    1+(−1+216224)×(216227)  is  Prime1+(-1+2^{16224})×(2^{16227}) ~ ~ is ~ ~ Prime

    Looked for a further set with something that looked algebraic

    The largest prime found [386 digits] using this probable prime test as a filter is next.

    1+(1+2639)×(2642)  is  Prime1+(1+2^{639})×(2^{642}) ~ ~ is ~ ~ Prime

    The early build up to my discovery involved tabulating over 100 groups.

    This tabulation is documented in the early chapters of my (draft) book.

  • Strong probable prime test using element 2 and set 1+(1+2^t)×(2^(t+3))

    Consider the Proth space

    1+(1+2t)∗(2t+3)1+(1+2^t)*(2^{t+3})

    Using the above conjecture we have established the following

    1+(1+2639)×(2642)  is  Prime1+(1+2^{639})×(2^{642}) ~ ~ is ~ ~ Prime

    By tabulating order of groups next we provide examples where order follows the conjecture and other examples where the Proth number used is composite.

    We will be using generating element 2

    When t=3 we have P=1+9×64 and the set of elements modulo P is a multiplicative group.


    The subgroup generated by 2 has 144 elements

    Order is (2^4)×(1+2^3) so we are a [probable] prime.

    The elements in TPc144 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 447,
    317, 57, 114, 228, 456, 335, 93, 186, 372, 167,
    334, 91, 182, 364, 151, 302, 27, 54, 108, 216,
    432, 287, 574, 571, 565, 553, 529, 481, 385, 193,
    386, 195, 390, 203, 406, 235, 470, 363, 149, 298,
    19, 38, 76, 152, 304, 31, 62, 124, 248, 496,
    415, 253, 506, 435, 293, 9, 18, 36, 72, 144,
    288, -1, 575, 573, 569, 561, 545, 513, 449, 321,
    65, 130, 260, 520, 463, 349, 121, 242, 484, 391,
    205, 410, 243, 486, 395, 213, 426, 275, 550, 523,
    469, 361, 145, 290, 3, 6, 12, 24, 48, 96,
    192, 384, 191, 382, 187, 374, 171, 342, 107, 214,
    428, 279, 558, 539, 501, 425, 273, 546, 515, 453,
    329, 81, 162, 324, 71, 142, 284, 568, 559, 541,
    505, 433, 289, 1 }

    Next we look at 2177=7*311 and see order is 465 so does not follow the rule established in the conjecture as the Proth number is composite.

    For 2177 from t=4 the 465 elements in TPc465 are not tabulated here
    For 8449 from t=5 the 840 elements in TPc840 are not tabulated here
    For 33281 from t=6 the 7953 elements in TPc7953 are not tabulated here
    For 132097 from t=7 the 6972 elements in TPc6972 are not tabulated here
    For 526337 from t=8 the 17688 elements in TPc17688 are not tabulated here
    For 2101249 from t=9 the 525312 elements in TPc525312 are not tabulated here
    For 8396801 from t=10 the 298680 elements in TPc298680 are not tabulated here

    A script based on the conjecture is shown next

    Running that script for t to 500 gives some small examples that fit with the conjecture

    A question that applies to all such searches once t becomes large is whether the primes exist.

    This is an open question not answered here.

    Do adjust the value for starter to search from the starting t value you require and run the script in Pari/GP or adapt it for your favoured computer algebra package.

  • Element 2 and set 1+(-1+2^t)×(2^(t+3))

    Consider the Proth space

    1+(−1+2t)∗(2t+3)1+(-1+2^t)*(2^{t+3})

    Using the above conjecture we have established the following

    1+(−1+216224)×(216227)  is  Prime1+(-1+2^{16224})×(2^{16227}) ~ ~ is ~ ~ Prime

    By tabulating order of groups next we provide examples where order follows the conjecture and other examples where the Proth number used is composite.

    We will be using generating element 2

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

    The subgroup generated by 2 has 8 elements

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

    The elements in TNz8 are { 2, 4, 8, -1, 15, 13, 9, 1 }
    If TNz8 is really a group then we need inverses so let us document those next.

    • Inverse of 2 is 9 mod P
    • Inverse of 4 is 13 mod P
    • Inverse of 8 is 15 mod P

    Later we look at 1921=17*113 and see order is 56 so does not follow the rule established in the conjecture as the Proth number is composite.

    For 97 from t=2 the 48 elements in TNz48 are as shown next.
    { 2, 4, 8, 16, 32, 64, 31, 62, 27, 54,
    11, 22, 44, 88, 79, 61, 25, 50, 3, 6,
    12, 24, 48, -1, 95, 93, 89, 81, 65, 33,
    66, 35, 70, 43, 86, 75, 53, 9, 18, 36,
    72, 47, 94, 91, 85, 73, 49, 1 }

    For 449 from t=3 the 224 elements in TNz224 are as shown next.
    { 2, 4, 8, 16, 32, 64, 128, 256, 63, 126,
    252, 55, 110, 220, 440, 431, 413, 377, 305, 161,
    322, 195, 390, 331, 213, 426, 403, 357, 265, 81,
    162, 324, 199, 398, 347, 245, 41, 82, 164, 328,
    207, 414, 379, 309, 169, 338, 227, 5, 10, 20,
    40, 80, 160, 320, 191, 382, 315, 181, 362, 275,
    101, 202, 404, 359, 269, 89, 178, 356, 263, 77,
    154, 308, 167, 334, 219, 438, 427, 405, 361, 273,
    97, 194, 388, 327, 205, 410, 371, 293, 137, 274,
    99, 198, 396, 343, 237, 25, 50, 100, 200, 400,
    351, 253, 57, 114, 228, 7, 14, 28, 56, 112,
    224, -1, 447, 445, 441, 433, 417, 385, 321, 193,
    386, 323, 197, 394, 339, 229, 9, 18, 36, 72,
    144, 288, 127, 254, 59, 118, 236, 23, 46, 92,
    184, 368, 287, 125, 250, 51, 102, 204, 408, 367,
    285, 121, 242, 35, 70, 140, 280, 111, 222, 444,
    439, 429, 409, 369, 289, 129, 258, 67, 134, 268,
    87, 174, 348, 247, 45, 90, 180, 360, 271, 93,
    186, 372, 295, 141, 282, 115, 230, 11, 22, 44,
    88, 176, 352, 255, 61, 122, 244, 39, 78, 156,
    312, 175, 350, 251, 53, 106, 212, 424, 399, 349,
    249, 49, 98, 196, 392, 335, 221, 442, 435, 421,
    393, 337, 225, 1 }

    For 1921 from t=4 the 56 elements in TNz56 are as shown next.
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
    127, 254, 508, 1016, 111, 222, 444, 888, 1776, 1631,
    1341, 761, 1522, 1123, 325, 650, 1300, 679, 1358, 795,
    1590, 1259, 597, 1194, 467, 934, 1868, 1815, 1709, 1497,
    1073, 225, 450, 900, 1800, 1679, 1437, 953, 1906, 1891,
    1861, 1801, 1681, 1441, 961, 1 }

    For 7937 from t=5 the 3968 elements in TNz3968 are not tabulated here
    For 32257 from t=6 the 16128 elements in TNz16128 are not tabulated here
    For 130049 from t=7 the 10603 elements in TNz10603 are not tabulated here
    For 522241 from t=8 the 14457 elements in TNz14457 are not tabulated here
    For 2093057 from t=9 the 20520 elements in TNz20520 are not tabulated here
    For 8380417 from t=10 the 4190208 elements in TNz4190208 are not tabulated here
    For 33538049 from t=11 the 16769024 elements in TNz16769024 are not tabulated here
    For 134184961 from t=12 the 246480 elements in TNz246480 are not tabulated here

    Prime testing of some of the larger examples found using the conjecture is shown next

    1+(−1+216224)×(216227)  can be written as  (232451)−(216227)+11+(-1+2^{16224})×(2^{16227}) ~ ~ can ~ be ~ written ~ as ~~ (2^{32451})-(2^{16227})+1

    Further work on this set of numbers has produced the following improved conjecture

  • Sure it’s structure preserving

    An isomorphism is said to be structure preserving


    In fact relationships between elements are maintained even if the elements are renamed or reordered


    But what if a finite multiplicative cyclic group has a secondary feature?

    Allowing both relabelling and reordering means this secondary feature is no longer accessible or usable.

    Does this prevent the mapping being structure preserving?

    A truly structure preserving mapping would allow the use of the secondary feature even after reordering and relabelling.

    Best illustrated with examples.

    Using a generating element of 65 and working modulo 8321 we obtain the 52 element multiplicative group 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 }

    There is a secondary feature in that group that allows us to jump to a portion of our group and be at most 7 elements from the element we require.


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

    In order we would write them as follows

    24,28,212,216,220,224,228,232,236,2402^4, 2^8, 2^{12}, 2^{16}, 2^{20}, 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
    thirtysecond from last is 2^16
    fortieth from last is 2^20
    fortyeighth from last is 2^24

    The fourth item is in bold in the group listing above and represents 2^24 mod 8321

    Using sturdy element notation we might say the group we tabulated above beginning 65 is

    Sturdy element notation for 1+2^t for set 1+(1+2^t)*(2^(t+1)) for t=6

    Current theory tells us that there exists an isomorphism between this finite abelian cyclic group and the additive group Z52.


    Such an isomorphism would use both relabelling and reordering.

    The problem then becomes how to access the secondary feature?
    We are now [after isomorphism] the group Z52 whose elements are:
    { 1,2,3,4,5,6,7,8,9,10,11,12,13,
    14,15,16,17,18,19,20,21,22,23,24,25,26,
    27,28,29,30,31,32,33,34,35,36,37,38,39,
    40,41,42,43,44,45,46,47,48,49,50,51,52 }

    By allowing reordering without being specific about tracking each element and how it maps individually, we have no way of locating an element of the secondary feature once in Z52


    This prevents us from jumping to a section of the group and navigating a maximum of seven elements from that jump destination.

    That secondary feature is part of our original multiplicative cyclic group.

    We are unable to use that secondary feature.


    Therefore we are not truly structure preserving.

    Next a further example.

    Using a generating element of 257 and working modulo 131585 we obtain the 68 element multiplicative group 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 }

    Using sturdy element notation we might say the group we tabulated above beginning 257 is

    Sturdy element notation for 1+2^t for set 1+(1+2^t)*(2^(t+1)) for t=8

    Labelling that 68 element multiplicative cyclic group as TPy68.


    With a jump or secondary feature similar to our previous example

    eight from last is 16
    sixteenth from last is 2^8
    twentyfourth from last is 2^12
    thirtysecond from last is 2^16
    fortieth from last is 2^20
    fortyeighth from last is 2^24
    fiftysixth from last is 2^28
    sixtyfourth from last is 2^32

    The fourth item is in bold in the group listing above and represents 2^32 mod 131585.


    Current theory tells us that there exists an isomorphism between this finite abelian cyclic group and the additive group Z68.

    Such an isomorphism would use both relabelling and reordering.

    The problem then becomes how to access the secondary feature?

    We are now [after isomorphism] the group Z68 whose elements are
    { 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,
    18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,
    35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,
    52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68 }

    By allowing reordering without being specific about tracking each element and how it maps individually, we have no way of locating an element of the secondary feature once in Z68


    This prevents us from jumping to a section of the group and navigating a maximum of seven elements from that jump destination.

    That secondary feature is part of our original multiplicative cyclic group TPy68.

    We are unable to use that secondary feature.

    Therefore we are not truly structure preserving.

    If the jump feature is an essential part of the multiplication operation, then you cannot have a structure preserving map.


    Do we make a special case for this type of group, or do we modify or qualify current theory?


    A question that might be hard to answer is do we in fact have the jump feature in the Z group and just need to add something to the labelling to allow it to operate [again]?


    Want to try and highlight in bold the elements in the Z group?
    Would that be enough?

  • Conjectures 1

    Assorted conjectures 1

    Conjecture: Factors of N when t is prime and N of the form 1+(-1+2^t)*(2^(t+1)) and 1+2*t is prime must be either 5 or of the form 1+4*k*(1+2*t) where k is a positive integer

    Example for t=11 pr=8384513 we have the factors 277 and 30269 where 1+2*t is 23

    277=1+4*3 *23

    30269=1+4*7*47*23

    8384513==277*30269

    Conjecture: Order of b mod N when N of the form

    1+(−1+bt)×bt1+(−1+b^t) ×b^t

    is 6×t where b and t are positive integers >=2

    Conjecture: Order of b mod N when N of the form

    1+(1+bt)×bt1+(1+b^t) ×b^t

    is 3×t where b and t are positive integers >=2

    Conjecture: There are an infinite number of cyclic groups having finite order.

    Posts on this site provide an abundant source of cyclic groups.

    1+(−1+216224)×(216227)  is  Prime1+(-1+2^{16224})×(2^{16227}) ~ ~ is ~ ~ Prime

    1+(1+2639)×(2642)  is  Prime1+(1+2^{639})×(2^{642}) ~ ~ is ~ ~ Prime

  • Element 3 and set 1+(1+3^t)×3^t

    Consider the space

    1+(1+3t)∗(3t)1+(1+3^t)*(3^{t})

    We will be using generating element 3

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

    Tabulating the elements of groups generated by 3 next

    Using prefix NPf 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) is 3×t


    When t=3 we have P=1+28×27 and the set of elements modulo P is a multiplicative group.


    The subgroup generated by 3 has 9 elements


    The elements in NPf9 are as shown next
    { 3, 9, 27, 81, 243, 729, 673, 505, 1 }


    If NPf9 is really a group then we need inverses so let us document those next.

    • Inverse of 3 is 505 mod P
    • Inverse of 9 is 673 mod P
    • Inverse of 27 is 729 mod P
    • Inverse of 81 is 243 mod P

    For 6643 from t=4 the 12 elements in NPf12 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 6397, 5905, 4429, 1 }


    For 59293 from t=5 the 15 elements in NPf15 are as shown next
    { 3, 9, 27, 81, 243,
    729, 2187, 6561, 19683, 59049,
    58561, 57097, 52705, 39529, 1 }


    For 532171 from t=6 the 18 elements in NPf18 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 19683, 59049, 177147, 531441,
    529981, 525601, 512461, 473041, 354781, 1 }

    For 4785157 from t=7 the 21 elements in NPf21 are as shown next
    { 3, 9, 27, 81, 243, 729, 2187,
    6561, 19683, 59049, 177147, 531441, 1594323, 4782969,
    4778593, 4765465, 4726081, 4607929, 4253473, 3190105, 1 }


    For 43053283 from t=8 the 24 elements in NPf24 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 19683, 59049, 177147, 531441,
    1594323, 4782969, 14348907, 43046721, 43033597, 42994225,
    42876109, 42521761, 41458717, 38269585, 28702189, 1 }


    For 387440173 from t=9 the 27 elements in NPf27 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 19683, 59049, 177147, 531441,
    1594323, 4782969, 14348907, 43046721, 129140163,
    387420489, 387381121, 387263017, 386908705, 385845769,
    382656961, 373090537, 344391265, 258293449, 1 }

    For 3486843451 from t=10 the 30 elements in NPf30 are as shown next
    { 3, 9, 27, 81, 243,
    729, 2187, 6561, 19683, 59049,
    177147, 531441, 1594323, 4782969, 14348907,
    43046721, 129140163, 387420489, 1162261467, 3486784401,
    3486666301, 3486312001, 3485249101, 3482060401, 3472494301,
    3443796001, 3357701101, 3099416401, 2324562301, 1 }

    For 31381236757 from t=11 the 33 elements in NPf33 are as shown next
    { 3, 9, 27, 81, 243,
    729, 2187, 6561, 19683, 59049,
    177147, 531441, 1594323, 4782969, 14348907,
    43046721, 129140163, 387420489, 1162261467, 3486784401,
    10460353203, 31381059609, 31380705313, 31379642425, 31376453761, 31366887769, 31338189793, 31252095865,
    30993814081, 30218968729, 27894432673, 20920824505, 1 }


    For 282430067923 from t=12 the 36 elements in NPf36 are as shown next
    { 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049,
    177147, 531441, 1594323, 4782969, 14348907,
    43046721, 129140163, 387420489, 1162261467, 3486784401,
    10460353203, 31381059609, 94143178827, 282429536481,
    282428473597, 282425284945, 282415718989, 282387021121,
    282300927517, 282042646705, 281267804269, 278943276961,
    271969695037, 251048949265, 188286711949, 1 }

    For t=13 the 39 elements in NPf39 are not tabulated here
    For t=14 the 42 elements in NPf42 are not tabulated here
    For t=15 the 45 elements in NPf45 are not tabulated here
    For t=16 the 48 elements in NPf48 are not tabulated here
    For t=17 the 51 elements in NPf51 are not tabulated here
    For t=18 the 54 elements in NPf54 are not tabulated here
    For t=19 the 57 elements in NPf57 are not tabulated here
    For t=20 the 60 elements in NPf60 are not tabulated here


    Using sturdy element notation we might say the results tabulated in this post are

    Sturdy element notation for 3 for set 1+(1+3^t)*(3^t) for t=3,..,12

  • Element 3 and set 1+(-1+3^t)×3^t

    Consider the space

    1+(−1+3t)∗(3t)1+(-1+3^t)*(3^{t})

    We will be using generating element 3

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

    Tabulating the elements of groups generated by 3 next

    Using prefix NNe 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) is 6×t

    When t=2 we have P=1+8×9 and the set of elements modulo P is a multiplicative group.


    The subgroup generated by 3 has 12 elements

    The elements in NNe12 are { 3, 9, 27, 8, 24, -1, 70, 64, 46, 65, 49, 1 }

    If NNe12 is really a group then we need inverses so let us document those next.

    • Inverse of 3 is 49 mod P
    • Inverse of 9 is 65 mod P
    • Inverse of 27 is 46 mod P
    • Inverse of 8 is 64 mod P
    • Inverse of 24 is 70 mod P

    We could alternatively write the group NNe12 as follows:
    { 3, 9, 27, (3^4), (3^5), -1,
    -3, -9, -27, -(3^4), -(3^5), 1, }

    For 703 from t=3 the 18 elements in NNe18 are as shown next
    { 3, 9, 27, 81, 243, 26, 78, 234, -1
    700, 694, 676, 622, 460, 677, 625, 469, 1 }

    For 6481 from t=4 the 24 elements in NNe24 are as shown next
    { 3, 9, 27, 81, 243, 729, 2187, 80, 240, 720,
    2160, -1, 6478, 6472, 6454, 6400, 6238, 5752, 4294, 6401,
    6241, 5761, 4321, 1 }

    For 58807 from t=5 the 30 elements in NNe30 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 19683, 242, 726, 2178,
    6534, 19602, -1, 58804, 58798, 58780,
    58726, 58564, 58078, 56620, 52246, 39124,
    58565, 58081, 56629, 52273, 39205, 1 }

    For 530713 from t=6 the 36 elements in NNe36 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 19683, 59049, 177147, 728,
    2184, 6552, 19656, 58968, 176904, -1,
    530710, 530704, 530686, 530632, 530470, 529984,
    528526, 524152, 511030, 471664, 353566, 529985,
    528529, 524161, 511057, 471745, 353809, 1 }

    For 4780783 from t=7 the 42 elements in NNe42 are as shown next
    { 3, 9, 27, 81, 243, 729,
    2187, 6561, 19683, 59049, 177147, 531441,
    1594323, 2186, 6558, 19674, 59022, 177066,
    531198, 1593594, -1, 4780780, 4780774, 4780756,
    4780702, 4780540, 4780054, 4778596, 4774222, 4761100,
    4721734, 4603636, 4249342, 3186460, 4778597, 4774225,
    4761109, 4721761, 4603717, 4249585, 3187189, 1 }

    For t=8 the 48 elements in NNe48 are not tabulated here
    For t=9 the 54 elements in NNe54 are not tabulated here
    For t=10 the 60 elements in NNe60 are not tabulated here
    For t=11 the 66 elements in NNe66 are not tabulated here
    For t=12 the 72 elements in NNe72 are not tabulated here
    For t=13 the 78 elements in NNe78 are not tabulated here
    For t=14 the 84 elements in NNe84 are not tabulated here
    For t=15 the 90 elements in NNe90 are not tabulated here
    For t=16 the 96 elements in NNe96 are not tabulated here
    For t=17 the 102 elements in NNe102 are not tabulated here
    For t=18 the 108 elements in NNe108 are not tabulated here
    For t=19 the 114 elements in NNe114 are not tabulated here
    For t=20 the 120 elements in NNe120 are not tabulated here

    Using sturdy element notation we might say the results tabulated in this post are

    Sturdy element notation for 3 for set 1+(-1+3^t)*(3^t) for t=2,..,7

  • Element 2 and set 1+(1+2^t)*(2^t)

    Consider the space

    1+(1+2t)∗(2t)1+(1+2^t)*(2^{t})

    We will be using generating element 2

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

    Tabulating the elements of groups generated by 2 next

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

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

    When t=3 we have P=1+9×8 and the set of elements modulo P is a multiplicative group.


    The subgroup generated by 2 has 9 elements

    The elements in NPb9 are as shown next
    { 2, 4, 8, 16, 32, 64, 55, 37, 1 }

    If NPb9 is really a group then we need inverses so let us document those next.

    • Inverse of 2 is 37 mod P
    • Inverse of 4 is 55 mod P
    • Inverse of 8 is 64 mod P
    • Inverse of 16 is 32 mod P

    From 273 for t=4 the 12 elements in NPb12 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 239, 205, 137, 1 }


    From 1057 for t=5 the 15 elements in NPb15 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
    991, 925, 793, 529, 1 }

    From 4161 for t=6 the 18 elements in NPb18 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
    2048, 4096, 4031, 3901, 3641, 3121, 2081, 1 }

    From 16513 for t=7 the 21 elements in NPb21 are as shown next
    { 2, 4, 8, 16, 32, 64, 128,
    256, 512, 1024, 2048, 4096, 8192, 16384,
    16255, 15997, 15481, 14449, 12385, 8257, 1 }

    From 65793 for t=8 the 24 elements in NPb24 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256,
    512, 1024, 2048, 4096, 8192, 16384, 32768, 65536,
    65279, 64765, 63737, 61681, 57569, 49345, 32897, 1 }

    For t=9 the 27 elements in NPb27 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 512,
    1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144,
    261631, 260605, 258553, 254449, 264241, 229825, 196993,
    131329, 1 }

    For t=10 the 30 elements in NPb30 are as shown next
    { 2, 4, 8, 16, 32, 64, 128,
    256, 512, 1024, 2048, 4096, 8192,
    16384, 32768, 65536, 131072, 262144,
    524288, 1048576, 1047551, 1045501, 1041401, 1033201,
    1016801, 984001, 918401, 787201, 524801, 1 }

    For t=11 the 33 elements in NPb33 are not tabulated here

    For t=12 the 36 elements in NPb36 are as shown next
    { 2, 4, 8, 16, 32, 64, 128,
    256, 512, 1024, 2048, 4096, 8192,
    16384, 32768, 65536, 131072, 262144,
    524288, 1048576, 2097152, 4194304, 8388608, 16777216,
    16773119, 16764925, 16748537, 16715761, 16650209, 16519105,
    16256897, 15732481, 14683649, 12585985, 8390657, 1 }

    For t=13 the 39 elements in NPb39 are not tabulated here
    For t=14 the 42 elements in NPb42 are not tabulated here
    For t=15 the 45 elements in NPb45 are not tabulated here
    For t=16 the 48 elements in NPb48 are not tabulated here
    For t=17 the 51 elements in NPb51 are not tabulated here
    For t=18 the 54 elements in NPb54 are not tabulated here
    For t=19 the 57 elements in NPb57 are not tabulated here


    For t=20 the 60 elements in NPb60 are as shown next

    { 2, 4, 8, 16, 32,
    64, 128, 256, 512, 1024,
    2048, 4096, 8192, 16384, 32768,
    65536, 131072, 262144, 524288, 1048576,
    2097152, 4194304, 8388608, 16777216, 33554432,
    67108864, 134217728, 268435456, 536870912, 1073741824,
    2147483648, 4294967296, 8589934592,
    17179869184, 34359738368, 68719476736,
    137438953472, 274877906944, 549755813888,
    1099511627776, 1099510579199, 1099508482045,
    1099504287737, 1099495899121, 1099479121889,
    1099445567425, 1099378458497, 1099244240641, 1098975804929, 1098438933505, 1097365190657,
    1095217704961, 1090922733569, 1082332790785,
    1065152905217, 1030793134081, 962073591809,
    824634507265, 549756338177, 1 }

    Using sturdy element notation we might say the results tabulated in this post are

    Sturdy element notation for 2 for set 1+(1+2^t)*(2^t) for t=3,4,5,6,7,8,9,10,12,20

  • Element 2 and set 1+(-1+2^t)*(2^t)

    Consider the space

    1+(−1+2t)∗(2t)1+(-1+2^t)*(2^{t})

    We will be using generating element 2

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

    Tabulating the elements of groups generated by 2 next

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

    From t=3 the elements in NNa18 are
    { 2, 4, 8, 16, 32, 7, 14, 28, -1,
    55, 53, 49, 41, 25, 50, 43, 29, 1 }

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

    When t=4 we have P=1+15×16 and the set of elements modulo P is a multiplicative group.


    The subgroup generated by 2 has 24 elements

    The elements in NNa24 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 15, 30, 60, 120, -1
    239, 237, 233, 225, 209, 177, 113, 226, 211, 181, 121, 1 }

    If NNa24 is really a group then we need inverses so let us document those next.

    • Inverse of 2 is 121 mod P
    • Inverse of 4 is 181 mod P
    • Inverse of 8 is 211 mod P
    • Inverse of 16 is 226 mod P
    • Inverse of 32 is 113 mod P
    • Inverse of 64 is 177 mod P
    • Inverse of 128 is 209 mod P
    • Inverse of 15 is 225 mod P
    • Inverse of 30 is 233 mod P
    • Inverse of 60 is 237 mod P
    • Inverse of 120 is 239 mod P

    We could alternatively write the group NNa24 as follows:
    { 2, 4, 8, 16, 32, 64, (2^7), (2^8), (2^9), (2^10), (2^11), -1,
    -2, -4, -8, -16, -32, -64, -(2^7), -(2^8), -(2^9), -(2^10), -(2^11), 1 }

    From t=5 the 30 elements in NNa30 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 31,
    62, 124, 248, 496, -1, 991, 989, 985, 977, 961,
    929, 865, 737, 481, 962, 931, 869, 745, 497, 1 }

    From t=6 the 36 elements in NNa36 are as shown next
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
    2048, 63, 126, 252, 504, 1008, 2016, -1, 4031, 4029,
    4025, 4017, 4001, 3969, 3905, 3777, 3521, 3009, 1985, 3970,
    2907, 3781, 3529, 3025, 2017, 1 }

    From t=7 the 42 elements in NNa42 are as shown next.
    { 2, 4, 8, 16, 32, 64, 128,
    256, 512, 1024, 2048, 4096, 8192, 127,
    254, 508, 1016, 2032, 4064, 8128, -1,
    16255, 16253, 16249, 16241, 16225, 16193, 16129,
    16001, 15745, 15233, 14209, 12161, 8065, 16130,
    16003, 15749, 15241, 14225, 12193, 8129, 1 }

    From t=8 the 48 elements in NNa48 are as shown next.
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
    2048, 4096, 8192, 16384, 32768, 255, 510, 1020, 2040, 4080,
    8160, 16320, 32640, -1, 65279, 65277, 65273, 65265, 65249, 65217,
    65153, 65025, 64769, 64257, 63233, 57089, 48897, 32513, 65026,
    64771, 64261, 63241, 61201, 57121, 48961, 32641, 1 }

    From t=9 the 54 elements in NNa54 are as shown next.
    { 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
    2048, 4096, 8192, 16384, 32768, 65536, 131072, 511, 1022, 2044,
    4088, 8176, 16352, 32704, 65408, 130816, -1, 261631,
    261629, 261625, 261617, 261601, 261569, 261505, 261377, 261121,
    260609, 259585, 257537, 253441, 245249, 228865,
    196097, 130561, 261122, 260611, 259589, 257545,
    253457, 245281, 228929, 196225, 130817, 1 }

    From t=10 the 60 elements in NNa60 are as shown next.
    { 2, 4, 8, 16, 32, 64, 128, 256,
    512, 1024, 2048, 4096, 8192, 16384, 32768, 65536,
    131072, 262144, 524288, 1023, 2046, 4092, 8184, 16368,
    32736, 65472, 130944, 261888, 523776, -1, 1047551, 1047549,
    1047545, 1047537, 1047521, 1047489, 1047425, 1047297, 1047041, 1046529, 1045505, 1043457, 1039361, 1031169, 1014785,
    982017, 916481, 785409, 523265, 1046530, 1045507,
    1043461, 1039369, 1031185, 1014817, 982081,
    916609, 785665, 523777, 1 }


    For t=11 the 66 elements in NNa66 are not tabulated here
    For t=12 the 72 elements in NNa72 are not tabulated here
    For t=13 the 78 elements in NNa78 are not tabulated here
    For t=14 the 84 elements in NNa84 are not tabulated here
    For t=15 the 90 elements in NNa90 are not tabulated here
    For t=16 the 96 elements in NNa96 are not tabulated here
    For t=17 the 102 elements in NNa102 are not tabulated here
    For t=18 the 108 elements in NNa108 are not tabulated here
    For t=19 the 114 elements in NNa114 are not tabulated here
    For t=20 the 120 elements in NNa120 are not tabulated here


    Using sturdy element notation we might say the results tabulated in this post are

    Sturdy element notation for 2 for set 1+(-1+2^t)*(2^t) for t=3,..,10

  • Group referencing becomes tricky when large

    Providing P is prime, we can refer to a [multiplicative] subgroup easily and in keeping with convention.


    Let G be the multiplicative group modulo 390001

    Let S be the [multiplicative] subgroup generated by 5.

    S has 24 elements as shown next
    { 5, 25, 125, 625, 3125, 15625, 78125, 624,
    3120, 15600, 78000, -1, 389996, 389976, 389876, 389376,
    386876, 374376, 311876, 389377, 386881, 374401, 312001, 1 }

    There are two problems potentially as we look at larger / other examples

    1. The first part G can involve some very large numbers – what happens when you reach 12 digits or more?
    2. The second part where we talk of subgroups* might be problematic when the modulo is not prime.

    *When talking of multiplicative subgroups we are usually doing this in the context of a multiplicative main group.
    However when the modulo is composite, there is no group under multiplication from which to subgroup from.

    Using sturdy element notation we might refer to the above group instead as

    sturdy element notation for element 5 for set 1+(-1+5^t)*(5^t) for t=4

    It really is a matter of preference as to which you consider easier.

    For this next 24 element example, we are hitting problem 1 (getting large) and problem 2 (composite N) so will not use a G and S way of navigating.

    The 24 element multiplicative group
    { 5, 25, 125, 625, 3125,
    15625, 78125, 390625, 1953125, 9765625,
    48828125, 244140625, 1220703125, 6103515625,
    30517578125, 152587890625, 152586328121, 152578515601,
    152539453001, 152344140001, 151367575001,
    146484750001, 122070625001, 1 }

    can only properly be described (in my opinion) using sturdy element notation as shown next.

    sturdy element notation for element 5 for set 1+(1+5^t)*(5^t) for t=8

    Where we to try to talk in modulo terms we would be saying modulo 152588281251 for any attempt to describe a main group G.
    But we do not have a multiplicative group G?