A340362 a(n) = A005940(n) - A324106(n), where A324106(n) is multiplicative with a(p^e) = A005940(p^e).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 24, 0, 0, -32, 0, -12, 0, 0, 0, -12, 0, 0, 16, 0, 0, 140, 0, 0, 0, 0, 0, 114, 0, 0, 0, 150, 0, 280, 0, 0, 48, 0, 0, 180, 0, -108, -64, 0, 0, -70, -24, 0, 0, 0, 0, 18, 0, 140, -24, 0, 0, 0, 0, 0, 32, 330, 0, -60, 0, 0, 280, 300, 0, 632, 0, 300

Offset: 1

Author

Antti Karttunen, Jan 06 2021

Keywords

• sign

Comments

It is conjectured that A070776 gives the positions of all zeros after the initial a(1) = 0. If that holds, then for all i, j: a(i) = a(j) => A340363(i) = A340363(j).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A005940, A070776, A340364, A340365, A340366.

Cf. also A305800, A340363.

Programs

PARI

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940
A324106(n) = { my(f=factor(n)); prod(i=1, #f~, A005940(f[i,1]^f[i,2])); };
A340362(n) = (A005940(n)-A324106(n));