A216822 Numbers n such that 2^n == 2 (mod n*(n+1)).

1, 5, 13, 29, 37, 61, 73, 157, 181, 193, 277, 313, 397, 421, 457, 541, 561, 613, 661, 673, 733, 757, 877, 997, 1093, 1153, 1201, 1213, 1237, 1289, 1321, 1381, 1453, 1621, 1657, 1753, 1873, 1905, 1933, 1993, 2017, 2137, 2341, 2473, 2557, 2593, 2797, 2857, 2917

Offset: 1

Author

V. Raman, Sep 17 2012

Keywords

• easy
• nonn

Comments

a(17) = 561 is the first composite number in the sequence. - Charles R Greathouse IV, Sep 19 2012

Intersection of { A015919(n) } and { A192109(n)-1 }. - Max Alekseyev, Apr 22 2013

Links

• V. Raman and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 2000 terms from V. Raman)
• Mersenne Forum, Prime Conjecture

Crossrefs

Cf. A069051 (prime n such that 2^n == 2 (mod n*(n-1))).

Cf. A217466 (prime terms of the sequence).

Cf. A217465 (composite terms of the sequence)

Programs

Mathematica

Select[Range[1, 10000], Mod[2^# - 2, # (# + 1)] == 0 &] (* T. D. Noe, Sep 19 2012 *)
Join[{1},Select[Range[3000],PowerMod[2,#,#(#+1)]==2&]] (* Harvey P. Dale, Oct 05 2022 *)

PARI

is(n)=Mod(2,n*(n+1))^n==2; \\ Charles R Greathouse IV, Sep 19 2012

Python

A216822_list = [n for n in range(1,10**6) if n == 1 or pow(2,n,n*(n+1)) == 2] # Chai Wah Wu, Mar 25 2021

Extensions

a(1)=1 prepended by Max Alekseyev, Dec 29 2017