A320917 a(n) = sigma_2(n)*sigma_3(n)/sigma(n).

1, 15, 70, 219, 546, 1050, 2150, 3315, 5299, 8190, 13542, 15330, 26690, 32250, 38220, 51491, 79170, 79485, 124166, 119574, 150500, 203130, 268710, 232050, 330771, 400350, 419020, 470850, 684546, 573300, 895622, 811395, 947940, 1187550, 1173900, 1160481, 1826210, 1862490, 1868300, 1809990

Offset: 1

Author

Franklin T. Adams-Watters, Oct 24 2018

Keywords

• nonn
• mult

Comments

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.)

Proof from Robert Gerbicz that this is always an integer:

First consider r=sigma_2(p^k)/sigma(p^k). Let x=p^k, then

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)

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.

Case b: k is odd, then

sigma_3(p^k)=(p^3*x^3-1)/(p^3-1), and we must multiple this by r, and here

p^3*x^3-1==0 mod (p+1)

p^3-1==-2 mod (p+1)

This means that if p=2 then sigma_3(p^k) is divisible by (p+1), so we have proved the theorem.

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.

It appears that a stronger result is also true: sigma_3(p^k) is divisible by (p+1) if k is odd.

Proof from Giovanni Resta that this is always an integer:

We have sigma_2(p^k)*sigma_3(p^k)/sigma(p^k) =

((p^(2+2*k)-1)/(p^2-1) * (p^(3+3*k)-1)/(p^3-1)) / ((p^(1+k)-1)/(p-1)) =

((p^(k+1)+1)*(p^(3*k+3)-1)) / ((p+1)*(p^3-1)),

because p^(2+2*k)-1 is the difference of two squares and we can use x^2-1 = (x+1)*(x-1).

We observe that p^(3*k+3)-1 is always divisible by p^3-1 (it is indeed sigma_3(p^k)).

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.

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).

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.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A001157, A001158, A013663, A320416.

Programs

Mathematica

a[n_] := DivisorSigma[2, n] * DivisorSigma[3, n] / DivisorSigma[1, n]; Array[a, 40] (* Amiram Eldar, Aug 01 2019 *)
(#[[2]]#[[3]])/#[[1]]&/@Table[DivisorSigma[k,n],{n,40},{k,3}] (* Harvey P. Dale, Aug 14 2024 *)

PARI

a(n) = sigma(n,2)*sigma(n,3)/sigma(n)

Formula

a(n) = sigma_2(n)*sigma_3(n)/sigma(n).

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