|
|
A160892
|
|
a(n) = ((2^b-1)/phi(n))*Sum_{d|n} Moebius(n/d)*d^(b-1) for b = 4.
|
|
1
|
|
|
15, 105, 195, 420, 465, 1365, 855, 1680, 1755, 3255, 1995, 5460, 2745, 5985, 6045, 6720, 4605, 12285, 5715, 13020, 11115, 13965, 8295, 21840, 11625, 19215, 15795, 23940, 13065, 42315, 14895, 26880, 25935, 32235, 26505, 49140, 21105, 40005, 35685, 52080, 25845
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
LINKS
|
|
|
FORMULA
|
Sum_{k=1..n} a(k) ~ c * n^3, where c = 5 * Product_{p prime} (1 + 1/p^2 + 1/p^3) = 5 * A330595 = 8.7446649892... .
Sum_{k>=1} 1/a(k) = (zeta(2)*zeta(3)/15) * Product_{p prime} (1 - 2/p^3 + 1/p^5) = 0.09339604419... . (End)
|
|
MATHEMATICA
|
f[p_, e_] := (p^2 + p + 1)*p^(2*e - 2); a[1] = 15; a[n_] := 15*Times @@ f @@@ FactorInteger[n]; Array[a, 40] (* Amiram Eldar, Nov 08 2022 *)
|
|
PROG
|
(PARI) a(n) = {my(f = factor(n)); 15 * prod(i = 1, #f~, (f[i, 1]^2 + f[i, 1] + 1)*f[i, 1]^(2*f[i, 2] - 2)); } \\ Amiram Eldar, Nov 08 2022
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|