A324817 a(n) = sign(A323244(n))*A001511(A323244(n)), with a(n) = 0 if A323244(n) = 0.

0, 1, 1, 2, 1, 3, 1, 2, 0, 1, 1, 2, 1, 5, 2, 2, 1, 3, 1, 2, -1, 2, 1, 2, -3, 2, 3, 2, 1, 2, 1, 2, 2, 2, -2, 2, 1, 9, -3, 2, 1, 3, 1, 2, 4, 2, 1, 2, -3, 1, 4, 2, 1, 3, -1, 2, -3, 3, 1, 2, 1, 2, 5, 2, 2, 2, 1, 2, 2, 4, 1, 2, 1, 2, 2, 2, -2, 2, 1, 2, -3, 2, 1, 2, -4, 2, -3, 2, 1, 3, -1, 2, 2, 2, -3, 2, 1, 1, -3, 2, 1, 3, 1, 2, -3

Offset

1,4

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000 (based on Hans Havermann's factorization of A156552)

Index entries for sequences related to binary expansion of n

Index entries for sequences computed from indices in prime factorization

Index entries for sequences related to sigma(n)

Formula

If A323244(n) = 0, then a(n) = 0, otherwise a(n) = sign(A323244(n)) * A001511(A323244(n)).

a(p) = 1 for all primes p.

Prog

(PARI)
A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552 by David A. Corneth
A323244(n) = ((2*A156552(n))-A323243(n)); \\ Needs also code from A323243.
A001511ext(n) = if(!n, n, sign(n)*(1+valuation(n, 2))); \\ Like A001511 but gives 0 for 0 and -A001511(-n) for negative numbers.
A324817(n) = A001511ext(A323244(n));

Crossrefs

Cf. A001511, A156552, A323243, A323244, A324724, A324725, A324731, A324732.

Sequence in context: A319299 A207031 A156248 * A106406 A123864 A035175

Adjacent sequences: A324814 A324815 A324816 * A324818 A324819 A324820

Keyword

sign

Author

Antti Karttunen, Mar 17 2019

Status

approved