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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.
4
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
0,2
COMMENTS
Conjecture: for all n, a(n)>0.
a(443) > 17000 if it is not 0.
LINKS
EXAMPLE
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.
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
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Vladimir Shevelev, Sep 14 2014
EXTENSIONS
More terms from Michel Marcus, Sep 15 2014
STATUS
approved