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Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the odd part (A000265) of concatenation b_k 2^n is prime, or a(n)=0 if there is no such prime.
4

%I #18 Oct 31 2016 04:10:47

%S 1,2,1,1,1,1,4,3,2,1,3,1,1,6,1,1,1,3,1,15,29,5,1,2,3,6,1,6,20,6,3,50,

%T 3,22,8,5,5,1,84,8,7,36,3,6,7,20,6,6,8,1,6,3,2,38,1,5,3,2,5,16,1,12,

%U 13,7,1,4,16,5,32,1,6,13,4,150,7,29,17,9,12,34

%N Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the odd part (A000265) of concatenation b_k 2^n is prime, or a(n)=0 if there is no such prime.

%C Conjecture: for all n, a(n)>0.

%C a(443) > 17000 if it is not 0.

%H Robert Israel, <a href="/A247342/b247342.txt">Table of n, a(n) for n = 0..442</a>

%e 2^0=1 and already 31 is prime. So a(0)=1;

%e 2^1=2, but odd part of 32 is 1 (nonprime); then consider odd part of 332. It is 83 that is prime. So a(1)=2.

%p f:= proc(n) local m,d,k,x;

%p m:= 2^n;

%p d:=ilog10(m);

%p for k from 1 do

%p x:= (10^k-1)/3*10^(d+1)+m;

%p if isprime(x/2^padic:-ordp(x,2)) then return k fi

%p od

%p end proc:

%p map(f, [$0..100]); # _Robert Israel_, Oct 30 2016

%o (PARI) a(n) = {k = 0; while (! ((val = eval(concat(Str((10^k-1)/3), Str(2^n)))) && isprime(val/2^valuation(val, 2))), k++); k;} \\ _Michel Marcus_, Sep 15 2014

%Y Cf. A000079, A232210, A242775, A247341.

%K nonn,base

%O 0,2

%A _Vladimir Shevelev_, Sep 14 2014

%E More terms from _Michel Marcus_, Sep 15 2014