{"id":132,"date":"2026-08-20T15:39:47","date_gmt":"2026-08-20T15:39:47","guid":{"rendered":"https:\/\/mathrelated.co.uk\/?p=132"},"modified":"2026-09-03T18:37:57","modified_gmt":"2026-09-03T18:37:57","slug":"super-element-12t-and-set-112t2t1","status":"publish","type":"post","link":"https:\/\/mathrelated.co.uk\/index.php\/2026\/08\/20\/super-element-12t-and-set-112t2t1\/","title":{"rendered":"Element 1+2^t and set 1+(1+2^t)*(2^(t+1))"},"content":{"rendered":"\n

Consider the Proth space<\/p>\n\n\n\n

1<\/mn>+<\/mo>(<\/mo>1<\/mn>+<\/mo>2<\/mn>t<\/mi><\/msup>)<\/mo>\u2217<\/mo>(<\/mo>2<\/mn>t<\/mi>+<\/mo>1<\/mn><\/mrow><\/msup>)<\/mo><\/mrow>1+(1+2^t)*(2^{t+1})<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

We will be using generating element 1 + 2^t<\/p>\n\n\n\n

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

A related Conjecture is given next.
<\/p>\n\n\n\n

Conjecture<\/strong>: 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\u00d7t\u2212(t+1),2+4\u00d7t)
<\/p>\n\n\n\n

Tabulating the elements of groups generated by 1+2^t next.<\/p>\n\n\n\n

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

For 5 from t=2 the 20 elements in TPy20<\/strong> 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
<\/p>\n\n\n\n

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<\/p>\n\n\n\n

When t=3 we have P=1+9\u00d716 and the set of elements modulo P is a multiplicative group.
<\/p>\n\n\n\n

Our generator 1+2^t is 9 and the subset it generates has 14 elements.
<\/p>\n\n\n\n

The elements in TPy14<\/strong> are
{ 9,81,4,36,34,16,144,136,64,141,109,111,129,1 }
<\/p>\n\n\n\n

If TPy14<\/strong> is really a group then we need inverses so let us document those next.<\/p>\n\n\n\n

    \n
  • Inverse of 9 is 129 mod P<\/li>\n\n\n\n
  • Inverse of 81 is 111 mod P<\/li>\n\n\n\n
  • Inverse of 4 is 109 mod P<\/li>\n\n\n\n
  • Inverse of 36 is 141 mod P<\/li>\n\n\n\n
  • Inverse of 34 is 64 mod P<\/li>\n\n\n\n
  • Inverse of 16 is 136 mod P<\/li>\n<\/ul>\n\n\n\n

    For 17 from t=4 the 36 elements in TPy36<\/strong> 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<\/strong>,272,264,
    128,541<\/strong>,477,479,513,1 }
    <\/p>\n\n\n\n

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

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

    The fifth to last element is always -4<\/p>\n\n\n\n

    Every eighth element after 16 from the end of the subgroup is a power of 2.<\/p>\n\n\n\n

    In order we would write them as follows<\/p>\n\n\n\n

    2<\/mn>4<\/mn><\/msup>,<\/mo>2<\/mn>8<\/mn><\/msup>,<\/mo>2<\/mn>12<\/mn><\/msup>,<\/mo>2<\/mn>16<\/mn><\/msup>,<\/mo>2<\/mn>24<\/mn><\/msup>,<\/mo>2<\/mn>28<\/mn><\/msup>,<\/mo>2<\/mn>32<\/mn><\/msup>,<\/mo>2<\/mn>36<\/mn><\/msup>,<\/mo>2<\/mn>40<\/mn><\/msup><\/mrow>2^4, 2^8, 2^{12}, 2^{16}, 2^{24}, 2^{28}, 2^{32}, 2^{36}, 2^{40}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

    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<\/p>\n\n\n\n

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

    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.<\/p>\n\n\n\n

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

    The fourth element can be obtained as
    8256==Mod(-4,33025)^-1 alternatively
    <\/p>\n\n\n\n

    The eighth element can be obtained as
    30961==Mod(16,33025)^-1 alternatively<\/p>\n\n\n\n

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

    Conjecture<\/strong>: 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\u00d7t<\/p>\n\n\n\n

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

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


    Using sturdy element notation we might say the results tabulated in this post are<\/p>\n\n\n\n

    \"Super<\/figure>\n\n\n\n

    <\/p>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    Consider the Proth space 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\u00d7t\u2212(t+1),2+4\u00d7t) Tabulating the elements […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5,6],"tags":[7,8],"class_list":["post-132","post","type-post","status-publish","format-standard","hentry","category-mathematics","category-numbertheory","tag-cyclic","tag-group"],"_links":{"self":[{"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/posts\/132","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/comments?post=132"}],"version-history":[{"count":9,"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/posts\/132\/revisions"}],"predecessor-version":[{"id":304,"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/posts\/132\/revisions\/304"}],"wp:attachment":[{"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/media?parent=132"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/categories?post=132"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathrelated.co.uk\/index.php\/wp-json\/wp\/v2\/tags?post=132"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}