

A355885


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


0



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



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



KEYWORD

nonn,hard,more


AUTHOR



EXTENSIONS



STATUS

approved



