OFFSET
1,2
COMMENTS
In general, if the function is multiplicative with a(p^e) = p^(e + m) where m > 0, then Dirichlet g.f.: Product_{primes p} (1 + p^(m+1)/(p^s - p)).
Equivalently, Dirichlet g.f.: zeta(s-m-1) * zeta(s-1) * Product_{primes p} (1 + p^(2 + m - 2*s) - p^(2 + 2*m - 2*s) - p^(1 - s)).
Sum_{k=1..n} a(k) ~ c(m) * zeta(m+1) * n^(m+2) / (m+2), where c(m) = Product_{primes p} (1 - 1/p^2 - 1/p^(m+1) + 1/p^(m+2)).
Limit_{m->oo} c(m) = 6/Pi^2 = A059956.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
Dirichlet g.f.: Product_{primes p} (1 + p^4/(p^s - p)).
Dirichlet g.f.: zeta(s-4) * zeta(s-1) * Product_{primes p} (1 + p^(5 - 2*s) - p^(8 - 2*s) - p^(1 - s)).
Sum_{k=1..n} a(k) ~ c * Pi^4 * n^5 / 450, where c = Product_{primes p} (1 - 1/p^2 - 1/p^4 + 1/p^5) = 0.5761527353856670595206110782641172754062471168028961885...
From Amiram Eldar, Sep 01 2023: (Start)
a(n) = n * A007947(n)^3 = A064549(n) * A007947(n)^2 = A361264(n) * A007947(n) = A064549(A064549(A064549(n))).
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/(p^3*(p-1))) = 1.148846213921... . (End)
MAPLE
f:= proc(n) local t;
mul(t[1]^(t[2]+3), t = ifactors(n)[2])
end proc:
map(f, [$1..50]); # Robert Israel, Mar 07 2023
MATHEMATICA
g[p_, e_] := p^(e+3); a[1] = 1; a[n_] := Times @@ g @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) for(n=1, 100, print1(direuler(p=2, n, 1 + p^4*X/(1 - p*X))[n], ", "))
(PARI) a(n) = my(f=factor(n)); for (k=1, #f~, f[k, 2] +=3); factorback(f); \\ Michel Marcus, Mar 07 2023
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Vaclav Kotesovec, Mar 06 2023
STATUS
approved