A217468 -id:A217468 - cp's OEIS frontend

A297412 Numbers k such that A019320(k) is in A217468.

43, 114, 163, 258, 326, 379, 487, 758, 762, 883, 974, 978, 1459, 1766, 2274, 2647, 2918, 2922, 3079, 3943, 5294, 5298, 5419, 6158, 7886, 8754, 9199, 10838, 11827, 14407, 15882, 16759, 18398, 18474, 18523, 23654, 23658, 24967, 26407, 28814, 32514, 33518, 37046, 37339, 39367

Offset: 1

Max Alekseyev, Dec 29 2017

nonn

Set difference of A297413 and A072226.

is_A297412(n) = my(m=polcyclo(n, 2)); Mod(2, m*(m-1))^m==2 && !ispseudoprime(m);

A069051 Primes p such that p-1 divides 2^p-2.

Original entry on oeis.org

2, 3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367, 42463, 71443, 77659, 95923, 99079, 113779, 117307, 143263, 174763, 175447, 184843, 265483, 304039, 308827

Offset: 1

Benoit Cloitre, Apr 03 2002

nonn

These are the prime values of n such that 2^n == 2 (mod n*(n-1)). - V. Raman, Sep 17 2012
These are the prime values p such that n^(2^(p-1)) is congruent to n or -n (mod p) for all n in Z/pZ, the commutative ring associated with each term. This results follows from Fermat's little theorem. - Philip A. Hoskins, Feb 08 2013
A prime p is in this sequence iff p-1 belongs to A014741. For p>2, this is equivalent to (p-1)/2 belonging to A014945. - Max Alekseyev, Aug 31 2016
From Thomas Ordowski, Nov 20 2018: (Start)
Conjecture: if n-1 divides 2^n-2, then (2^n-2)/(n-1) is squarefree.
Numbers n such that b^n == b (mod (n-1)*n) for every integer b are 2, 3, 7, and 43; i.e., only prime numbers of the form A014117(k) + 1. (End)
These are primes p such that p^2 divides b^(2^p-2) - 1 for every b coprime to p. - Thomas Ordowski, Jul 01 2024

Charles R Greathouse IV, Table of n, a(n) for n = 1..3314 (First 553 terms from V. Raman)
Mersenne Forum, Prime Conjecture

Cf. A216822, A217468, A211203, A211349, A330382.

a(n)-1 form subsequence of A014741; (a(n)-1)/2 for n>1 forms a subsequence of A014945.

Filtered([1..350000],p->IsPrime(p) and (2^p-2) mod (p-1)=0); # Muniru A Asiru, Dec 03 2018

[p : p in PrimesUpTo(310000) | IsZero((2^p-2) mod (p-1))]; // Vincenzo Librandi, Dec 03 2018

Select[Prime[Range[10000]], Mod[2^# - 2, # - 1] == 0 &] (* T. D. Noe, Sep 19 2012 *)
Join[{2,3},Select[Prime[Range[30000]],PowerMod[2,#,#-1]==2&]] (* Harvey P. Dale, Apr 17 2022 *)

isA069051(p)=Mod(2,p-1)^p==2 && isprime(p); \\ Charles R Greathouse IV, Sep 19 2012

from sympy import prime
for n in range(1,350000):
    if (2**prime(n)-2) % (prime(n)-1)==0:
        print(prime(n)) # Stefano Spezia, Dec 07 2018

a(1) added by Charles R Greathouse IV, Sep 19 2012

A217465 Composite integers k such that 2^k == 2 (mod k*(k+1)).

Original entry on oeis.org

561, 1905, 4033, 4681, 5461, 6601, 8481, 11305, 13741, 13981, 16705, 23377, 30121, 31417, 41041, 49141, 52633, 57421, 88357, 88561, 101101, 107185, 121465, 130561, 162193, 196021, 196093, 204001, 208465, 219781, 266305, 276013, 278545, 282133, 285541, 314821, 334153, 341497, 390937, 399001

Offset: 1

V. Raman, Oct 04 2012

nonn

Terms A019320(k) belongs to this sequence for k in A297415. - Max Alekseyev, Dec 29 2017

Amiram Eldar, Table of n, a(n) for n = 1..12766 (terms below 10^12; terms 1..100 from Harvey P. Dale)
Mersenne Forum, Prime Conjecture, 2012.

Subsequence of A216822.
Contains A303531 as a subsequence.
Cf. A019320, A217466, A217468, A297415, A298758.

Select[Range[400000],!PrimeQ[#]&&PowerMod[2,#,#(#+1)]==2&] (* Harvey P. Dale, Oct 12 2012 *)

for(n=1,10000,if((2^n)%(n*(n+1))==2&&isprime(n)==0,printf(n",")))

forcomposite(n=4,10^6, if(Mod(2,n*(n+1))^n==2, print1(n", "))) \\ Charles R Greathouse IV, Aug 29 2024

from sympy import isprime
A217465_list = [n for n in range(1,10**6) if pow(2,n,n*(n+1)) == 2 and not isprime(n)] # Chai Wah Wu, Mar 25 2021

A219037 Numbers k such that k divides 2^k + 2 and (k-1) divides 2^k + 1.

Original entry on oeis.org

2, 6, 66, 73786976294838206466

Offset: 1

Max Alekseyev, Nov 10 2012

Also, numbers k such that 2^k == k-2 (mod k*(k-1)).
The sequence is infinite: if m is in this sequence, then so is 2^m + 2.
No other terms below 10^20.

W. Sierpinski, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #18.

Kin Y. Li et al., Solution to Problem 323, Mathematical Excalibur 14(2), 2009, p. 3.

Intersection of A006517 and A055685.

Cf. A217468, A216822, A171959.

Conjecture: a(n+1) = 2^a(n) + 2 for all n.

A330382 Composite numbers k such that k-1 divides 2^k-2.

Original entry on oeis.org

55, 295, 343, 1027, 1135, 1315, 1807, 2059, 2395, 3403, 4375, 5335, 6175, 6499, 7183, 7939, 9235, 10207, 12643, 13123, 14155, 16003, 16255, 19495, 21547, 23815, 27595, 27703, 30619, 35479, 37927, 43219, 45487, 48007, 48763, 50275, 55567, 58483, 64387, 64639, 74899

Offset: 1

Amiram Eldar and Thomas Ordowski, Dec 12 2019

nonn

If k is in this sequence, then 2^k-1 is also a term, so this sequence is infinite.
Also 2^p-1 is in this sequence for such prime p in A069051 that 2^p-1 is composite.
Theorem: if k-1 | 2^k-2, then m-1 | 2^m-2, where m = 2^k-1.
Conjecture: k-1 | 2^k-2 for k = (2^n-1)^3 if and only if n(n-1) | 2^n-2 for n > 2.
It seems that A007013(n)^3 for n > 1 and A007013(n) for n > 4 are in this sequence.
These are the composites k for which M - 1 divides 2^M - 2 where M = 2^k - 1. - Thomas Ordowski, Jul 01 2024

Amiram Eldar, Table of n, a(n) for n = 1..5000

Cf. A007013, A014741, A054723, A069051, A217468.

A217468 is a subsequence.

Select[Range[75000], CompositeQ[#] && Divisible[PowerMod[2, #, # - 1] - 2, # - 1] &]

forcomposite(k=1,75000,if(!((2^k-2)%(k-1)),print1(k,", "))) \\ Hugo Pfoertner, Dec 12 2019

Composites of A014741(n) + 1. - Thomas Ordowski, Jul 01 2024

A374841 Numbers k such that 2^(2^k-2) == 1 (mod k^2).

Original entry on oeis.org

3, 7, 19, 43, 73, 127, 163, 337, 341, 379, 487, 601, 881, 883, 937, 1387, 1459, 1801, 2593, 2647, 2857, 3079, 3529, 3673, 3943, 4057, 4201, 4681, 5419, 5461, 5881, 6121, 6481, 6529, 6553, 6571, 6841, 7481, 7993, 8233, 8911, 9001, 9199, 9241, 9721, 10261, 10657, 11161, 11827, 12241

Offset: 1

Thomas Ordowski, Jul 22 2024

nonn

If p is an odd prime and 2^(2^p-2) == 1 (mod p), then 2^(2^p-2) == 1 (mod p^2).
If 2^(k-1) == 1 (mod k) and 2^(2^k-2) == 1 (mod k), then 2^(2^k-2) == 1 (mod k^2).
Composite terms that are not Fermat pseudoprimes to base 2 are 66709, 951481, ...
Note that 66709 = 19*3511 and 951481 = 271*3511, where 3511 is a Wieferich prime.

			3 is a term, because 3^2 divides 2^(2^3-2) - 1 = 2^6 - 1 = 63.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..250 from Michel Marcus)

Cf. A001220, A001567, A069051 (> 2 is a subsequence), A217468 (subsequence).

Select[Range[12500], PowerMod[2, 2^# - 2, #^2] == 1 &] (* Amiram Eldar, Jul 22 2024 *)

isok(k) = Mod(2, k^2)^(2^k-2) == 1; \\ Michel Marcus, Jan 05 2025

More terms from Amiram Eldar Jul 22 2024

A297413 Numbers k such that 2^m == 2 (mod m*(m-1)), where m=A019320(k).

Original entry on oeis.org

2, 3, 6, 7, 14, 19, 38, 42, 43, 86, 114, 127, 163, 254, 258, 326, 379, 487, 758, 762, 883, 974, 978, 1459, 1766, 2274, 2647, 2918, 2922, 3079, 3943, 5294, 5298, 5419, 6158, 7886, 8754, 9199, 10838, 11827, 14407, 15882, 16759, 18398, 18474, 18523, 23654, 23658, 24967, 26407, 28814, 32514, 33518, 37046, 37339, 39367

Offset: 1

Max Alekseyev, Dec 29 2017

nonn

Also, numbers k such that A019320(k) belongs to A069051 or A217468.

Cf. A069051, A217468, A297414.

Contains A297412 as a subsequence.

is_A297413(k) = my(m=polcyclo(k,2)); Mod(2,m*(m-1))^m==2;

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A297412 Numbers k such that A019320(k) is in A217468.

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A069051 Primes p such that p-1 divides 2^p-2.

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A217465 Composite integers k such that 2^k == 2 (mod k*(k+1)).

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A219037 Numbers k such that k divides 2^k + 2 and (k-1) divides 2^k + 1.

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A330382 Composite numbers k such that k-1 divides 2^k-2.

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A374841 Numbers k such that 2^(2^k-2) == 1 (mod k^2).

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A297413 Numbers k such that 2^m == 2 (mod m*(m-1)), where m=A019320(k).

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