A328959 a(n) = sigma_0(n) - 2 - (omega(n) - 1) * nu(n), where sigma_0 = A000005, nu = A001221, omega = A001222.

-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0

Offset

1,72

Comments

Conjecture: All terms are nonnegative except for a(1) = -1.

Links

Antti Karttunen, Table of n, a(n) for n = 1..100000

Index entries for sequences computed from exponents in factorization of n

Formula

a(n) = A000005(n) - A307408(n). - Antti Karttunen, Nov 17 2019

Example

a(72) = sigma_0(72) - 2 - (omega(72) - 1) * nu(72) = 12 - 2 - (5 - 1) * 2 = 2.

Mathematica

Table[DivisorSigma[0, n]-2-(PrimeOmega[n]-1)*PrimeNu[n], {n, 100}]

Prog

(PARI)
A307408(n) = 2+((bigomega(n)-1)*omega(n));
A328959(n) = (numdiv(n) - A307408(n)); \\ Antti Karttunen, Nov 17 2019

Crossrefs

The positions of positive terms are conjectured to be A320632.

Positions of first appearances are A328963.

omega(n) * nu(n) is A113901(n).

(omega(n) - 1) * nu(n) is A307409.

sigma_0(n) - omega(n) * nu(n) is A328958(n).

Cf. A000005, A001221, A001222, A112798, A124010, A307408, A323023, A328956, A328960, A328961, A328962, A328965.

Sequence in context: A185778 A071164 A027345 * A086080 A369054 A281477

Adjacent sequences: A328956 A328957 A328958 * A328960 A328961 A328962

Keyword

sign

Author

Gus Wiseman, Nov 02 2019. The idea for this sequence came from Mats Granvik.

Status

approved