A294893 Number of divisors d of n such that Stern polynomial B(d,x) is irreducible.

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 3, 2, 2, 1, 2, 3, 2, 2, 2, 1, 3, 1, 2, 2, 1, 3, 3, 1, 2, 2, 3, 1, 2, 1, 2, 3, 2, 3, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 2, 1, 3, 3, 2, 2, 2, 3, 2, 1, 2, 2, 3, 1, 3, 1, 2, 3

Offset: 1

Author

Antti Karttunen, Nov 10 2017

Keywords

• nonn

Comments

Number of terms > 1 of A186891 that divide n.

Examples

			For n=25, with divisors [1, 5, 25], both B(5,x) and B(25,x) are irreducible, thus a(25)=2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..22001

Crossrefs

Cf. A186891, A283991, A294891, A294892, A294894.

Cf. also A294883.

Differs from A001221 for the first time at n=25.

Programs

PARI

ps(n) = if(n<2, n, if(n%2, ps(n\2)+ps(n\2+1), 'x*ps(n\2)));
A283991(n) = polisirreducible(ps(n));
A294893(n) = sumdiv(n,d,A283991(d));

Formula

a(n) = Sum_{d|n} A283991(d).

a(n) + A294894(n) = A000005(n).

a(n) = A294891(n) + A283991(n).