A247342 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.

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

Offset: 0

Author

Vladimir Shevelev, Sep 14 2014

Keywords

• nonn
• base

Comments

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

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

Examples

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

Links

• Robert Israel, Table of n, a(n) for n = 0..442

Crossrefs

Cf. A000079, A232210, A242775, A247341.

Programs

Maple

f:= proc(n) local m,d,k,x;
    m:= 2^n;
    d:=ilog10(m);
    for k from 1 do
       x:= (10^k-1)/3*10^(d+1)+m;
       if isprime(x/2^padic:-ordp(x,2)) then return k fi
    od
end proc:
map(f, [$0..100]); # Robert Israel, Oct 30 2016

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

Extensions

More terms from Michel Marcus, Sep 15 2014