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A367203
Product of p^(v_p(n)-1) + p^(2*floor(v_p(n)/2)) over all prime divisors p of n with valuation v_p(n).
2
1, 2, 2, 6, 2, 4, 2, 8, 12, 4, 2, 12, 2, 4, 4, 24, 2, 24, 2, 12, 4, 4, 2, 16, 30, 4, 18, 12, 2, 8, 2, 32, 4, 4, 4, 72, 2, 4, 4, 16, 2, 8, 2, 12, 24, 4, 2, 48, 56, 60, 4, 12, 2, 36, 4, 16, 4, 4, 2, 24, 2, 4, 24, 96, 4, 8, 2, 12, 4, 8, 2, 96, 2, 4, 60, 12, 4, 8, 2, 48, 108, 4, 2, 24
OFFSET
1,2
COMMENTS
If n is an odd perfect number in Euler form n = q^e * m^2, where q is prime and q == e == 1 (mod 4), then a(n) = (2*n*A048250(n))/(A007947(n)*(q+1)) and from the second formula below we deduce A367197(n)*(q+1) = 4*n.
FORMULA
a(n) is multiplicative with a(p^e) = p^(e-1) + p^(2*floor(e/2)).
a(n) = n*A048250(n)*A367197(n)/(A000203(n)*A007947(n)).
From Vaclav Kotesovec, Nov 16 2023: (Start)
Dirichlet g.f.: zeta(2*s-2) * zeta(s)^2 * Product_{p prime} ((p^s - 1)^2 * (p + 2*p^s + p^(2*s)) / p^(4*s)).
Sum_{k=1..n} a(k) ~ c * zeta(3/2)^2 * n^(3/2) / 3, where c = Product_{p prime} (1 + 1/p^5 + 2/p^(9/2) - 2/p^(7/2) - 3/p^3 + 1/p^2) = 0.834393951629126332417010554216818092635571350444778107077957309407387... (End)
EXAMPLE
a(2^3) = 2^2 + 2^(2*floor(3/2)) = 2^2 + 2^2 = 8.
MATHEMATICA
f[p_, e_] := p^(e-1) + p^(2*Floor[e/2]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 10 2023 *)
PROG
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + 2*X + p*X^2)/(1 - p^2*X^2))[n], ", ")) \\ Vaclav Kotesovec, Nov 16 2023
(SageMath)
def ss(n):
return prod(p**(valuation(n, p)-1)+p**(2*floor(valuation(n, p)/2)) for p in prime_divisors(n))
print([ss(n) for n in range(1, 41)])
(Python)
from math import prod
from sympy import factorint
def A367203(n): return prod(p**(e-1)+p**(e&-2) for p, e in factorint(n).items()) # Chai Wah Wu, Nov 17 2023
CROSSREFS
KEYWORD
nonn,mult,easy
AUTHOR
Orges Leka, Nov 10 2023
STATUS
approved