A347906 -id:A347906 - cp's OEIS frontend

A130421 Numbers k such that 4^k == 2 (mod k).

1, 2, 14, 1022, 20066, 80519, 107663, 485918, 1284113, 1510313, 2531678, 3677198, 3933023, 4557713, 8277458, 8893262, 21122318, 24849833, 26358638, 39852014, 42448478, 71871113, 76712318, 80646143, 98058097, 104832833, 106694033, 131492498, 144322478, 146987033, 164360606, 168204191, 175126478, 176647378, 188997463, 196705598

Offset: 1

Jon E. Schoenfield, May 26 2007

nonn

Some terms above 10^15: 26435035805519327, 158975398896178078, 64044049390757098943, 1063423126446412987081943, 1091220919655042176978844783, 81074850280355100090334498663, 6317483763950169936179578094903, 5672799393875320397186007124651847, 170923900137537174138295268515194974, 195746953975871672436191077726091399305155458, 325665752547333314939363628501536024940097079718953, 953533053776414279913696071891872697927468471633033, 85791212788381063775490416118630897060666265030605503, 334519297382630382793758729321508383611586565722054114034741260213364710519401967713. - Max Alekseyev, Jun 18 2014

Max Alekseyev, Table of n, a(n) for n = 1..2617 (all terms below 10^15)

Cf. A006935 (odd terms times 2), A130422, A347906 (odd terms), A347908 (even terms).

Join[{1,2},Select[Range[107000000],PowerMod[4,#,#]==2&]] (* Harvey P. Dale, Jun 13 2013 *)

Terms a(28) onward from Max Alekseyev, Jun 18 2014

b-file corrected by Max Alekseyev, Oct 09 2016

A347908 Even numbers k such that 2^(2*k) == 2 (mod k).

Original entry on oeis.org

2, 14, 1022, 20066, 485918, 2531678, 3677198, 8277458, 8893262, 21122318, 26358638, 39852014, 42448478, 76712318, 131492498, 144322478, 164360606, 175126478, 176647378, 196705598, 249126626, 306789074, 317051378, 438023138, 497041538, 696970718, 996520658

Offset: 1

Jianing Song, Sep 18 2021

nonn

Numbers of the form 2*t where 2^(4*t-1) == 1 (mod t).
Even terms in A130421. Complement of A347906 in A130421.
If k > 14 is a term, then k/2 must be composite, since for odd primes p we have 2^(4*p) == 16 (mod 2*p). If k = 2*t > 2 is a term, then 2*k-1 = 4*t-1 must also be composite, since ord(2,t) | (4*k-1) and ord(2,t) <= eulerphi(t) <= t-1 < 4*t-1.
If k = 2*t > 2 is a term, then (2^(2*k) - 2)/k = (2^(4*k-1) - 1)/t is composite. See A347907 for a proof.
2*(2^t - 1) is a term if and only if 2^(t+2) == 5 (mod t) (t = 1, 3, 9, 871, 2043, 2119, 8769, ...).

			14 is a term since 14 divides 2^28 - 2.

Jianing Song, Table of n, a(n) for n = 1..1298 (contains all terms below 10^15; based on Max Alekseyev's b-file for A130421)

Cf. A130421, A347906, A347907.

isA347908(k) = if(k%4==2, k=k>>1; if(isprime(k) && k!=7, 0, Mod(2, k)^(4*k-1)==1), 0)

a(n) = A347907(n)*2.

A347907 Numbers k such that 2^(4*k-1) == 1 (mod k).

Original entry on oeis.org

1, 7, 511, 10033, 242959, 1265839, 1838599, 4138729, 4446631, 10561159, 13179319, 19926007, 21224239, 38356159, 65746249, 72161239, 82180303, 87563239, 88323689, 98352799, 124563313, 153394537, 158525689, 219011569, 248520769, 348485359, 498260329, 636381799, 638395369, 685333399, 689463889

Offset: 1

Jianing Song, Sep 18 2021

nonn

Odd numbers k such that ord(2,k) divides 4*k-1, where ord(2,k) is the multiplicative order of 2 modulo k.
Numbers k such that 2*k is in A130421.
Terms > 7 must be composite, since for odd primes p we have 2^(4*p-1) == 8 (mod p). If k > 1 is a term, then 4*k-1 must also be composite, since ord(2,k) | (4*k-1) and ord(2,k) <= eulerphi(k) <= k-1 < 4*k-1.
If k > 1 is a term, then (2^(4*k-1) - 1)/k is composite. Proof: since 4*k-1 is composite, write 4*k-1 = u*v, u >= v > 1. Proof: since 4*k-1 is composite, write 4*k-1 = u*v, u >= v > 1, then (2^(4*k-1) - 1)/k = (2^u - 1)*(2^(u*(v-1)) + ... + 2^u + 1)/k. Since k | 2^(4*k-1) - 1, there exist positive integers a,b such that a*b = k and that a | 2^u - 1 and b | 2^(u*(v-1)) + ... + 2^u + 1. Note that (2^u - 1)/a, (2^(u*(v-1)) + ... + 2^u + 1)/b >= (2^u - 1)/k >= (2^sqrt(4*k-1) - 1)/k > 1, so (2^(4*k-1) - 1)/k is the product of two integers > 1, so it is composite.
2^t - 1 is a term if and only if 2^(t+2) == 5 (mod t) (t = 1, 3, 9, 871, 2043, 2119, 8769, ...).

			7 is a term since 7 divides 2^27 - 1.

Jianing Song, Table of n, a(n) for n = 1..1298 (contains all terms below 5*10^14; based on Max Alekseyev's b-file for A130421)

Cf. A130421, A347908.

Cf. A347906 (a similar sequence).

Join[{1},Parallelize[Select[Range[69*10^7],PowerMod[2,4#-1,#]==1&]]] (* Harvey P. Dale, Apr 16 2023 *)

isA347907(k) = if(k%2 && (!isprime(k) || k==7), Mod(2, k)^(4*k-1)==1, 0)

a(n) = A347908(n)/2.

A350083 a(n) = (A006935(n) - 1) / ord(2,A006935(n)/2), where ord(k,m) is the multiplicative order of k modulo m.

Original entry on oeis.org

1, 617, 1305, 9339, 225, 5297, 6985, 1549, 174233, 46549, 93701, 66879, 431087, 593887, 1288921, 446275, 43685, 1205, 3361, 2577225, 1313, 430739, 177301, 8541, 13067, 474525, 561301, 84725, 158873, 725725, 3851, 14019, 128861, 1090301, 2529, 430667, 541673

Offset: 1

Jianing Song, Dec 12 2021

nonn

List of (2*k-1) / ord(2,k) where k ranges over the odd numbers such that 2^(2*k-1) == 1 (mod k).

			A006935(2) = 161038, so a(2) = (161038 - 1) / ord(2,161038/2) = 617.
A006935(3) = 215326, so a(3) = (215326 - 1) / ord(2,215326/2) = 1305.

Amiram Eldar, Table of n, a(n) for n = 1..1319

Cf. A006935, A347906, A350084, A300101, A174590.

list(lim) = my(v=[],d); forstep(k=1, lim, 2, if((2*k-1)%(d=znorder(Mod(2,k)))==0, v=concat(v,(2*k-1)/d))); v \\ gives a(n) for A347906(n) <= lim

a(n) = (2*A347906(n) - 1) / ord(2,A347906(n)) = (A006935(n) - 1) / A350084(n).

A350084 a(n) = ord(2,A006935(n)/2), where ord(k,m) is the multiplicative order of k modulo m.

Original entry on oeis.org

1, 261, 165, 275, 13425, 1485, 1305, 32085, 825, 3465, 2093, 3135, 495, 495, 261, 847, 9405, 552189, 198561, 261, 579261, 2475, 6237, 166725, 111111, 3393, 3565, 25245, 18585, 4437, 891891, 309455, 37125, 4833, 2301585, 14355, 11781, 3315, 915, 84975, 35259

Offset: 1

Jianing Song, Dec 12 2021

nonn

List of ord(2,k) where k ranges over the odd numbers such that 2^(2*k-1) == 1 (mod k).

			A006935(2) = 161038, so a(2) = ord(2,161038/2) = 261.
A006935(3) = 215326, so a(3) = ord(2,215326/2) = 165.

Amiram Eldar, Table of n, a(n) for n = 1..1319

Cf. A006935, A347906, A350083, A306413, A306414.

list(lim) = my(v=[],d); forstep(k=1, lim, 2, if((2*k-1)%(d=znorder(Mod(2,k)))==0, v=concat(v,d))); v \\ gives a(n) for A347906(n) <= lim

a(n) = ord(2,A347906(n)) = (A006935(n) - 1) / A350083(n).

cp's OEIS Frontend

A130421 Numbers k such that 4^k == 2 (mod k).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Extensions

A347908 Even numbers k such that 2^(2*k) == 2 (mod k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A347907 Numbers k such that 2^(4*k-1) == 1 (mod k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A350083 a(n) = (A006935(n) - 1) / ord(2,A006935(n)/2), where ord(k,m) is the multiplicative order of k modulo m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A350084 a(n) = ord(2,A006935(n)/2), where ord(k,m) is the multiplicative order of k modulo m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula