A355885 a(n) is the smallest odd k such that k + 2^m is a de Polignac number for m = 1..n.

125, 903, 7385, 87453, 957453, 6777393, 21487809, 27035379, 1379985537, 5458529139, 15399643917, 32702289081

Offset

1,1

Comments

All terms of this sequence are composite numbers.

Note that all positive values of a(n) + 2^i - 2^j are composite from i = 1 to n.

Conjecture: this sequence is infinite and bounded, namely a(n) = K for all n >= N.

If so, then K - 2^j is a Sierpiński number for every 1 < 2^j < K, by the dual Sierpiński conjecture. Note that each positive value of K + 2^i - 2^j is composite for every i > 0.

The number K can be a (partial) solution to the open problem: are there odd (composite) numbers k such that both |(k -+ 2^m)*2^n +- 1| are composite for every pair of positive integers m,n?

By the dual Sierpiński and Riesel conjectures, these are odd numbers k such that both ||k -+ 2^m| +- 2^n| are composite for m, n > 0.

The conditional theorem: by the dual Sierpiński conjecture and by the dual Riesel conjecture; if p is an odd prime and m is a positive integer, then there exist two numbers n such that both |(p -+ 2^m)*2^n +- 1| are prime.

So if such numbers k exist, they must be composite.

Links

Table of n, a(n) for n=1..12.

Example

a(3) = 7385, because 7385 is the smallest number k such that k+2^1, k+2^2, k+2^3 are de Polignac numbers 7387, 7389, 7393.

Mathematica

depolQ[n_] := OddQ[n] && Module[{m = 2}, While[m < n && CompositeQ[n - m], m *= 2]; m > n]; a[n_] := Module[{k = 1}, While[AnyTrue[Range[1, n], !depolQ[k + 2^#] &], k++]; k]; Array[a, 5] (* Amiram Eldar, Jul 20 2022 *)

Crossrefs

Cf. A006285, A156695, A337487.

Sequence in context: A253352 A204611 A296037 * A088896 A016851 A204795

Adjacent sequences: A355882 A355883 A355884 * A355886 A355887 A355888

Keyword

nonn,hard,more

Author

Thomas Ordowski, Jul 20 2022

Extensions

More terms from Amiram Eldar, Jul 20 2022

Status

approved