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A366889
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Dirichlet inverse of the highest power of two that divides the sum of divisors of n.
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3
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1, -1, -4, 0, -2, 4, -8, 0, 15, 2, -4, 0, -2, 8, 8, 0, -2, -15, -4, 0, 32, 4, -8, 0, 3, 2, -64, 0, -2, -8, -32, 0, 16, 2, 16, 0, -2, 4, 8, 0, -2, -32, -4, 0, -30, 8, -16, 0, 63, -3, 8, 0, -2, 64, 8, 0, 16, 2, -4, 0, -2, 32, -120, 0, 4, -16, -4, 0, 32, -16, -8, 0, -2, 2, -12, 0, 32, -8, -16, 0, 272, 2, -4, 0, 4, 4, 8, 0
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OFFSET
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1,3
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COMMENTS
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LINKS
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FORMULA
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a(1) = 1, and for n > 1, a(n) = -Sum_{d|n, d<n} A082903(n/d) * a(d).
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PROG
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(PARI)
A082903(n) = (2^valuation(sigma(n), 2));
memoA366889 = Map();
A366889(n) = if(1==n, 1, my(v); if(mapisdefined(memoA366889, n, &v), v, v = -sumdiv(n, d, if(d<n, A082903(n/d)*A366889(d), 0)); mapput(memoA366889, n, v); (v)));
(Python)
from functools import lru_cache
from sympy import divisor_sigma, divisors
@lru_cache(maxsize=None)
def A366889(n): return 1 if n==1 else -sum((1<<(~(m:=int(divisor_sigma(d))) & m-1).bit_length())*A366889(n//d) for d in divisors(n, generator=True) if d>1) # Chai Wah Wu, Jan 03 2024
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CROSSREFS
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KEYWORD
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sign,mult
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AUTHOR
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STATUS
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approved
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