This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A320917 #26 Dec 15 2024 04:36:08
%S A320917 1,15,70,219,546,1050,2150,3315,5299,8190,13542,15330,26690,32250,
%T A320917 38220,51491,79170,79485,124166,119574,150500,203130,268710,232050,
%U A320917 330771,400350,419020,470850,684546,573300,895622,811395,947940,1187550,1173900,1160481,1826210,1862490,1868300,1809990
%N A320917 a(n) = sigma_2(n)*sigma_3(n)/sigma(n).
%C A320917 Multiplicative, because products and quotients of multiplicative functions are always multiplicative. a(p^k) for fixed k is trivially a rational function of p. The proofs below show that a(n) is always an integer, and hence a(p^k) is a polynomial in p. (Note that a(n^k) is not equal to this polynomial when n is composite.)
%C A320917 Proof from _Robert Gerbicz_ that this is always an integer:
%C A320917 First consider r=sigma_2(p^k)/sigma(p^k). Let x=p^k, then
%C A320917 r=(p^(2*k+2)-1)/(p^2-1)*(p-1)/(p^(k+1)-1)=(p^2*x^2-1)/(p^2-1)*(p-1)/(p*x-1)=(p*x+1)/(p+1)
%C A320917 Case a: k is even, then x==1 mod (p+1), so (p*x+1)==-1+1==0 mod (p+1). So here even r is an integer.
%C A320917 Case b: k is odd, then
%C A320917 sigma_3(p^k)=(p^3*x^3-1)/(p^3-1), and we must multiple this by r, and here
%C A320917 p^3*x^3-1==0 mod (p+1)
%C A320917 p^3-1==-2 mod (p+1)
%C A320917 This means that if p=2 then sigma_3(p^k) is divisible by (p+1), so we have proved the theorem.
%C A320917 If p is odd, then (p+1)/2 divides sigma_3(p^k), but we have another factor of 2 in (p*x+1), because p and x is odd.
%C A320917 It appears that a stronger result is also true: sigma_3(p^k) is divisible by (p+1) if k is odd.
%C A320917 Proof from _Giovanni Resta_ that this is always an integer:
%C A320917 We have sigma_2(p^k)*sigma_3(p^k)/sigma(p^k) =
%C A320917 ((p^(2+2*k)-1)/(p^2-1) * (p^(3+3*k)-1)/(p^3-1)) / ((p^(1+k)-1)/(p-1)) =
%C A320917 ((p^(k+1)+1)*(p^(3*k+3)-1)) / ((p+1)*(p^3-1)),
%C A320917 because p^(2+2*k)-1 is the difference of two squares and we can use x^2-1 = (x+1)*(x-1).
%C A320917 We observe that p^(3*k+3)-1 is always divisible by p^3-1 (it is indeed sigma_3(p^k)).
%C A320917 Now, if k is even, k+1 is odd, so p^(k+1)+1 is divisible by p+1 giving 1-p+p^2-p^3... +p^k as result, and we are done.
%C A320917 If k is odd, k+1 is even and in general p^(k+1)+1 is not divisible by p+1 (just as p^2+1 is not divisible by p+1).
%C A320917 However, if k is odd, then 3*k+3 is even, so p^(3*k+3)-1 beside being divisible by p^3-1 is also divisible by p+1. Since p+1 and p^3-1 have no common factors, then the ratio (p^(3*k+3)-1) / ((p+1)*(p^3-1)) is an integer and we are done.
%H A320917 Amiram Eldar, <a href="/A320917/b320917.txt">Table of n, a(n) for n = 1..10000</a>
%F A320917 a(n) = sigma_2(n)*sigma_3(n)/sigma(n).
%F A320917 Sum_{k=1..n} a(k) ~ c * n^5, where c = (Pi^6*zeta(5)/2700) * Product_{p prime} (1 - 2/p^2 + 2/p^3 - 2/p^4 + 1/p^6) = 0.1662831668... . - _Amiram Eldar_, Dec 01 2022
%t A320917 a[n_] := DivisorSigma[2, n] * DivisorSigma[3, n] / DivisorSigma[1, n]; Array[a, 40] (* _Amiram Eldar_, Aug 01 2019 *)
%t A320917 (#[[2]]#[[3]])/#[[1]]&/@Table[DivisorSigma[k,n],{n,40},{k,3}] (* _Harvey P. Dale_, Aug 14 2024 *)
%o A320917 (PARI) a(n) = sigma(n,2)*sigma(n,3)/sigma(n)
%Y A320917 Cf. A000203, A001157, A001158, A013663, A320416.
%K A320917 nonn,mult
%O A320917 1,2
%A A320917 _Franklin T. Adams-Watters_, Oct 24 2018