A081858 -id:A081858 - cp's OEIS frontend

A231917 Numbers n such that 16n^2 + 10n + 1 divides 2^n - 1.

0, 11, 75, 156, 179, 215, 216, 239, 371, 431, 504, 551, 564, 624, 651, 711, 936, 999, 1040, 1076, 1296, 1304, 1416, 1680, 1884, 1911, 2079, 2324, 2615, 2696, 3176, 3224, 3236, 3500, 3696, 3780, 3879, 4040, 4044, 4215, 4340, 4368, 4431, 4599, 4604, 4859, 5019

Offset: 1

Arkadiusz Wesolowski, Nov 15 2013

nonn

The prime numbers of this sequence are in A231916.

Chris Caldwell, The Prime Glossary, Mersenne number

Subsequence of A081858. Supersequence of A231916.

Cf. A000225, A028993, A231918.

Prepend[Select[Range[5019], PowerMod[2, #, 16*#^2 + 10*# + 1] == 1 &], 0]

A247094 Integers of the form (2^k + 1)/(2k + 1).

Original entry on oeis.org

1, 2, 3, 5, 27, 565, 7085, 48771, 1266205, 9099507, 17602325, 128207979, 26494256091, 11147523830125, 84179432287299, 165269711096165, 281629680514649643, 4246732448623781667, 126774939137440139965, 1925041114036033717685, 14833445639443302757131

Offset: 1

Juri-Stepan Gerasimov, Nov 18 2014

nonn

a(A103579(n)) is a subsequence.

Numbers n such that 2n + 1 divides 2^n + 1: 0, 1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, ...

			1 is in this sequence because (2^1 + 1)/(2*1 + 1) = 1,
2 is in this sequence because (2^0 + 1)/(2*0 + 1) = 2,
3 is in this sequence because (2^5 + 1)/(2*5 + 1) = 3.

Colin Barker, Table of n, a(n) for n = 1..400

Cf. A081856, A081858, A103579, A224486, A247132.

s=[]; for(k=0, 100, t=(2^k + 1)/(2*k + 1); if(type(t)=="t_INT", s=concat(s, t))); s=vecsort(s,,8) \\ Colin Barker, Nov 18 2014

a(19) corrected by Colin Barker, Nov 18 2014

A233089 Numbers n such that 2n-1 and 2n+1 divide 2^n-1.

Original entry on oeis.org

8, 128, 228, 648, 3240, 5976, 13160, 23760, 23940, 24840, 32768, 37224, 78540, 82800, 139248, 166716, 238368, 278520, 280368, 288360, 516528, 633420, 664668, 731808, 734448, 1145520, 1211100, 1377240, 1425816, 1484568, 1627640, 2055060, 2131080, 2292780

Offset: 1

Michel Marcus, Dec 04 2013

nonn

Intersection of A081856 and A081858.

Numbers n such that 4n^2-1 divides 2^n-1. - Charles R Greathouse IV, Dec 04 2013

			2^8-1=255 is divisible by 2*8-1=15 and by 2*8+1=17.

Charles R Greathouse IV, Table of n, a(n) for n = 1..5000

Select[Range[23*10^5],PowerMod[2,#,2#+{1,-1}]=={1,1}&] (* Harvey P. Dale, Dec 19 2014 *)

isok(n) = !((2^n-1) % (2*n-1)) && !((2^n-1) % (2*n+1));

is(n)=Mod(2,4*n^2-1)^n==1 \\ Charles R Greathouse IV, Dec 04 2013

A247132 Numbers k such that 2k - 1 divides 2^k + 1.

Original entry on oeis.org

1, 194997, 1463649, 1957025, 4657005, 6464145, 17214725, 70930629, 76938345, 319359365, 336837501, 429872625, 486213525, 1343289717, 1831683645, 2163016845, 2430979425, 2950546137, 3463374005, 5031564525, 5608791441, 8993704797, 9596401485, 12556945401, 13492461125, 14559291285, 18429009725

Offset: 1

Juri-Stepan Gerasimov, Nov 20 2014

nonn

			1 is in this sequence because (2^1 + 1)/(2*1 - 1) = 3.

Cf. A081856, A081858, A247094.

[n: n in [1..2000000] | Denominator((2^n + 1)/(2*n - 1)) eq 1];

Select[Range[200000], IntegerQ[(2^# + 1) / (2 # - 1)] &] (* Vincenzo Librandi, Nov 20 2014 *)

is(n)=Mod(2,2*n-1)^n==-1 \\ Charles R Greathouse IV, Nov 20 2014

a(5)-a(27) from Charles R Greathouse IV, Nov 20 2014

A250203 Numbers n such that the Phi_n(2) is the product of exactly two primes and is divisible by 2n+1.

Original entry on oeis.org

11, 20, 23, 35, 39, 48, 83, 96, 131, 231, 303, 375, 384, 519, 771, 848, 1400, 1983, 2280, 2640, 2715, 3359, 6144, 7736, 7911, 11079, 13224, 16664, 24263, 36168, 130439, 406583

Offset: 1

Eric Chen, Mar 13 2015

Here Phi_n is the n-th cyclotomic polynomial.
Is this sequence infinite?
Phi_n(2)/(2n+1) is only a probable prime for n > 16664.
a(33) > 2000000.
Subsequence of A005097 (2 * a(n) + 1 are all primes)
Subsequence of A081858.
2 * a(n) + 1 are in A115591.
Primes in this sequence are listed in A239638.
A085021(a(n)) = 2.
All a(n) are congruent to 0 or 3 (mod 4). (A014601)
All a(n) are congruent to 0 or 2 (mod 3). (A007494)
Except the term 20, all even numbers in this sequence are divisible by 8.

			Phi_11(2) = 23 * 89 and 23 = 2 * 11 + 1, so 11 is in this sequence.
Phi_35(2) = 71 * 122921 and 71 = 2 * 35 + 1, so 35 is in this sequence.
Phi_48(2) = 97 * 673 and 97 = 2 * 48 + 1, so 48 is in this sequence.

Eric Chen, Gord Palameta, Factorization of Phi_n(2) for n up to 1280
Will Edgington, Factorization of completely factored Phi_n(2) [from Internet Archive Wayback Machine]
Henri Lifchitz and Renaud Lifchitz, PRP records. Search for (2^a-1)/b
Samuel Wagstaff, The Cunningham project

Cf. A239638, A085724, A072226, A005384, A005097, A081858, A014601, A014664, A001917, A115591.

Select[Range[10000], PrimeQ[2*# + 1] && PowerMod[2, #, 2*# + 1] == 1 &&
PrimeQ[Cyclotomic[#, 2]/(2*#+1)] &]

isok(n) = if (((x=polcyclo(n, 2)) % (2*n+1) == 0) && (omega(x) == 2), print1(n, ", ")); \\ Michel Marcus, Mar 13 2015

A343679 Lucasian pseudoprimes: composite numbers k such that 2^(k-1) == k+1 (mod k(2k+1)).

Original entry on oeis.org

150851, 452051, 1325843, 1441091, 4974971, 5016191, 15139199, 19020191, 44695211, 101276579, 119378351, 128665319, 152814531, 187155383, 203789951, 223782263, 307367171, 387833531, 392534231, 470579831, 505473263, 546748931, 626717471, 639969891, 885510239, 974471243, 1147357559

Offset: 1

Thomas Ordowski, Apr 26 2021

nonn

These are pseudoprimes k == 3 (mod 4) such that 2k+1 is prime.
Proof. Let q = 2k+1 be prime, where k == 3 (mod 4) is a pseudoprime. We have q == 7 (mod 8), so 2 is a square mod q, which gives 2^((q-1)/2) == 1 (mod q), by Euler's criterion. Thus, 2^k == 1 (mod q), which implies 2^(k-1) == (q+1)/2 (mod q), so that 2^(k-1) == k+1 (mod q). The conclusion that 2^(k-1) == k+1 (mod kq) follows from the assumption that k is a pseudoprime and from the Chinese remainder theorem. - Carl Pomerance (in a letter to the author), Apr 14 2021
Note that if p is a Lucasian prime, i.e., p == 3 (mod 4) with 2p+1 prime; then (2^p-1)/(2p+1) == 1 (mod p), hence 2^p-2p-2 == 0 (mod p(2p+1)), so 2^(p-1) == p+1 (mod p(2p+1)).

Cf. A001567, A002515, A081858.

Select[Range[10^7], CompositeQ[#] && PowerMod[2, #-1, #*(2*#+1)] == #+1 &] (* Amiram Eldar, Apr 26 2021 *)

More terms from Amiram Eldar, Apr 26 2021

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.

Original entry on oeis.org

1, 3, 2, 2, 3, 2, 2, 7, 4, 2, 7, 2, 2, 3, 2, 6, 6, 5, 2, 6, 4, 6, 7, 2, 2, 12, 2, 5, 6, 2, 2, 21, 10, 2, 6, 2, 8, 7, 5, 2, 3, 2, 21, 6, 8, 15, 18, 5, 4, 6, 2, 2, 17, 2, 6, 6, 8, 5, 19, 9, 2, 12, 2, 18, 18, 2, 14, 7, 4, 2, 6, 4, 10, 7, 2, 10, 12, 15, 6, 6, 4, 2, 16, 2, 2, 19, 2, 5, 6, 2, 2, 6, 10, 9, 21, 2, 4, 32, 2, 2, 6

Offset: 0

Juri-Stepan Gerasimov, Jun 25 2025

nonn

			1 is the term because 2*0 + 1 = 1 is divisor of (2^1 + 2*0 + 1)^2 - 1 = 3^2 - 1 = 8.

Cf. A003462 (numbers m > 0 such that a(m) = 3), A005384 (primes p such that a(p) = 2), A005408 (odd numbers), A076481 (primes q such that a(q) = 3), A081858 (numbers k numbers k >= 0 such that 2k + 1 divides 2^k - 1), A102781 (numbers k such that 2k + 1 divides (2^k + 2*k + 1)^2 - 1), A224486 (numbers k such that 2k + 1 divides 2^k + 1).

[#[k: k in [1..2*n+1] | ((2^k+2*n+1)^2 - 1) mod (2*n + 1) eq 0]: n in [0..100]];

a[n_]:=Length[Select[Range[2n+1],Divisible[(2^#+2n+1)^2-1,2n+1] &]]; Array[a,101,0] (* Stefano Spezia, Jun 25 2025 *)

a(n) = sum(k=1, 2*n+1, !Mod((2^k + 2*n + 1)^2 - 1, 2*n + 1)); \\ Michel Marcus, Jun 25 2025

cp's OEIS Frontend

A231917 Numbers n such that 16*n^2 + 10*n + 1 divides 2^n - 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A247094 Integers of the form (2^k + 1)/(2k + 1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions

A233089 Numbers n such that 2n-1 and 2n+1 divide 2^n-1.

Views

Author

Keywords

Comments

Examples

Links

Programs

Mathematica

PARI

PARI

A247132 Numbers k such that 2k - 1 divides 2^k + 1.

Views

Author

Keywords

Examples

Crossrefs

Programs

Magma

Mathematica

PARI

Extensions

A250203 Numbers n such that the Phi_n(2) is the product of exactly two primes and is divisible by 2n+1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A343679 Lucasian pseudoprimes: composite numbers k such that 2^(k-1) == k+1 (mod k(2k+1)).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Extensions

A385326 The number of positive k <= 2*n + 1 such that 2*n + 1 divides (2^k + 2*n + 1)^2 - 1.

Views

Author

Keywords

Examples

Crossrefs

Programs

Magma

Mathematica

PARI

A231917 Numbers n such that 16n^2 + 10n + 1 divides 2^n - 1.

A385326 The number of positive k <= 2n + 1 such that 2n + 1 divides (2^k + 2*n + 1)^2 - 1.