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

%I A355885 #19 Oct 05 2022 04:47:25
%S A355885 125,903,7385,87453,957453,6777393,21487809,27035379,1379985537,
%T A355885 5458529139,15399643917,32702289081
%N A355885 a(n) is the smallest odd k such that k + 2^m is a de Polignac number for m = 1..n.
%C A355885 All terms of this sequence are composite numbers.
%C A355885 Note that all positive values of a(n) + 2^i - 2^j are composite from i = 1 to n.
%C A355885 Conjecture: this sequence is infinite and bounded, namely a(n) = K for all n >= N.
%C A355885 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.
%C A355885 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?
%C A355885 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.
%C A355885 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.
%C A355885 So if such numbers k exist, they must be composite.
%e A355885 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.
%t A355885 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 *)
%Y A355885 Cf. A006285, A156695, A337487.
%K A355885 nonn,hard,more
%O A355885 1,1
%A A355885 _Thomas Ordowski_, Jul 20 2022
%E A355885 More terms from _Amiram Eldar_, Jul 20 2022