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Define g(k) = min(n: n >= 0, 2^n + k prime). Then a(n) = min(odd k: g(k) = n).
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%I #27 Apr 19 2018 09:58:28

%S 1,3,7,23,31,47,199,83,61,257,139,953,991,647,1735,383,511,1337,1069,

%T 713,271,1937,3223,5213,751,8477,4339,353,1501,287,829,1553,2371,1811,

%U 11185,3023,7381,7937,6439,1433,13975,2897,4183

%N Define g(k) = min(n: n >= 0, 2^n + k prime). Then a(n) = min(odd k: g(k) = n).

%C Previous name: a(n) = min(k : A067760((k-1)/2)) = n.

%C a(n) is the first odd number k for which 2^m + k is the first prime value, as m ranges from 0 to n, or 0 if no such k exists. Thus it is the first k for which A067760((k-1)/2) = n, and therefore also the first k for which you need to test primality of exactly n values to show that it is not a dual SierpiƄski number.

%C In the name, g(n) = A067760(n) except for n=1. - _Michel Marcus_, Apr 07 2018

%H Hugo van der Sanden, <a href="/A260350/b260350.txt">Table of n, a(n) for n = 0..1087</a>

%F For n>=2, a(n) = (min(k : A067760((k-1)/2)) = n). - _Michel Marcus_, Apr 07 2018

%e 2^i + 7 is composite for i < 2 (with values 8, 9) but prime for i = 2 (11); the smaller odd numbers 1, 3 and 5 each yield a prime for smaller i, so a(2) = 7.

%o (PARI) g(k) = {my(j=0); while (!isprime(2^j+k), j++); j;}

%o a(n) = {my(k = 1); while(g(k) != n, k+=2); k;} \\ _Michel Marcus_, Apr 07 2018

%Y Cf. A067760, A076336, A260349.

%K nonn

%O 0,2

%A _Hugo van der Sanden_, Jul 23 2015

%E New name from _Hugo van der Sanden_ and _Michel Marcus_, Apr 07 2018