A286843 Even k such that k - A001065(k) = 2^m (for some m > 0).

10, 14, 22, 38, 44, 92, 110, 130, 134, 136, 152, 170, 184, 248, 250, 262, 284, 376, 410, 442, 632, 730, 752, 884, 988, 1012, 1052, 1276, 1292, 1370, 1628, 2144, 2168, 2272, 2332, 2528, 3068, 4064, 4124, 5210, 6112, 6364, 6556, 7372, 8198, 8312, 8384, 8648

Offset: 1

Author

XU Pingya, Aug 01 2017

Keywords

• nonn

Comments

Tianxin Cai conjectured that the sequence is infinite.

When p is prime, A001065(2p) = 1 + 2 + p = 3 + p. 2p - A001065(2p) = 2^m iff 2^m + 3 = p. Therefore if A057733 is infinite, Cai's conjecture is correct.

In general, for j = 2, 3, ..., if the number of primes of the form 2^m + 2^j - 1 is infinite, then Cai's conjecture is correct.

When 2^p - 1 is prime, let k = 2^p*(2^p - 1). A001065(k) = 1 + 2 + 2^2 + ... + 2^p + 2^p - 1 + 2(2^p - 1) + 2^2*(2^p - 1) + ... + 2^(p - 1)*(2^p - 1) = 2^(p + 1) - 1 + (2^p - 1)^2 = 2^(2p). k - A001065(k) = -2^p. Therefore if the number of Mersenne primes (A000668) is infinite, then there are infinitely many even k such that k - A001065(k) = -2^p.

Examples

			10 is a term as 10 - A001065(10) = 10 - 8 = 2.
22 is a term as 22 - A001065(22) = 22 - 14 = 2^3.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..893 (terms below 10^11)

Crossrefs

Cf. A000079, A001065, A005843, A057733.

Programs

Mathematica

Select[Table[2n, {n, 1, 5000}], DivisorSigma[1, 2# - DivisorSigma[1, #]] + 1 == 2(2# - DivisorSigma[1, #]) > 2 &] (* or *)
Select[2 Range[5000], IntegerQ@ Log2[2 # - DivisorSigma[1,#]] && !IntegerQ@ Log2@ # &] (* Giovanni Resta, Aug 07 2017 *)

PARI

ispower2(n) = n > 1 && n >> valuation(n, 2) == 1;
is(n) = !(n%2) && ispower2(2*n - sigma(n)); \\ Amiram Eldar, Mar 22 2024