A300909 Sum of 4th powers dividing n.
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 82, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17
Offset: 1
Examples
a(16) = 17 because 16 has 5 divisors {1, 2, 4, 8, 16} among which 2 divisors {1, 16} are 4th powers and 1 + 16 = 17.
L.g.f.: L(x) = x + x^2/2 + x^3/3 + x^4/4 + x^5/5 + x^6/6 + x^7/7 + x^8/8 + x^9/9 + x^10/10 + x^11/11 + x^12/12 + x^13/13 + x^14/14 + x^15/15 + 17*x^16/16 + x^17/17 + ...
exp(L(x)) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + x^15 + 2*x^16 + 2*x^17 + ... + A046042(n)*x^n + ...
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
- A. Dixit, B. Maji, A. Vatwani, Voronoi summation formula for the generalized divisor function sigma_z^k(n), arXiv:2303.09937 [math.NT], 2023, sigma(z=4,k=4,n).
Programs
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Maple
N:= 1000: # for a(1)..a(N) V:= Vector(N,1): for m from 2 to floor(N^(1/4)) do R:= [seq(i,i=m^4 .. N, m^4)]; V[R]:= map(`+`,V[R],m^4) od: convert(V,list); # Robert Israel, Mar 15 2018
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Mathematica
Table[DivisorSum[n, # &, IntegerQ[#^(1/4)] &], {n, 112}] nmax = 112; Rest[CoefficientList[Series[Sum[k^4 x^k^4/(1 - x^k^4), {k, 1, 10}], {x, 0, nmax}], x]] nmax = 112; Rest[CoefficientList[Series[-Log[Product[(1 - x^k^4), {k, 1, 10}]], {x, 0, nmax}], x] Range[0, nmax]] f[p_, e_] := (p^(4*(1 + Floor[e/4])) - 1)/(p^4 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 01 2020 *) -
PARI
a(n) = sumdiv(n, d, d*ispower(d, 4)); \\ Michel Marcus, Mar 15 2018
Formula
G.f.: Sum_{k>=1} k^4*x^(k^4)/(1 - x^(k^4)).
L.g.f.: -log(Product_{k>=1} (1 - x^(k^4))) = Sum_{n>=1} a(n)*x^n/n.
D.g.f.: zeta(s)*zeta(4s-4). - Robert Israel, Mar 15 2018
Sum_{k=1..n} a(k) ~ zeta(5/4)*n^(5/4)/5 - n/2. - Vaclav Kotesovec, Dec 01 2020
Comments