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

2, 6, 66, 73786976294838206466

Offset: 1

Author

Max Alekseyev, Nov 10 2012

Keywords

• nonn
• hard
• more

Comments

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.

References

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

Links

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

Crossrefs

Intersection of A006517 and A055685.

Cf. A217468, A216822, A171959.

Formula

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