A296306 a(n) = A001157(n)/A050999(n).

1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 341, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 1365, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 341, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 5461, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 341, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 1365

Offset: 1

Author

Ivan N. Ianakiev, Dec 10 2017

Keywords

• easy
• nonn
• mult

Comments

a(n) is the sum of the second powers of the divisors of n divided by the sum of the second powers of the odd divisors of n.

Conjecture 1: For any nonnegative integer k and positive integer n, the sum of the k-th powers of the divisors of n is divisible by the sum of the k-th powers of the odd divisors of n.

Conjecture 2: Distinct terms form A002450 without A002450(0). In other words, a(2^(n-1)) = A002450(n), for n > 0.

Conjecture 3: For n > 0, the list of the first 2^n - 1 terms is palindromic.

Conjecture 4: For n > 0, the sum of the first 2^n - 1 terms equals A006095(n+1).

To prove Conjecture 1 all one needs to do is to prove that the sum of the k-th powers of the divisors of n divided by the sum of the k-th powers of the odd divisors of n equals: a) A001511(n), for k = 0, and b) ((2^k)^A001511(n) - 1)/(2^k - 1), for k > 0. - Ivan N. Ianakiev, Jan 29 2020

Conjecture 1 indeed follows from multiplicativity of sigma_k, in particular sigma_k(2^j (2m+1)) = sigma_k(2^j) sigma_k(2m+1). Conjecture 3 follows from Radcliffe's formula, since A007814 has this property. - M. F. Hasler, Jan 31 2020

a(n) is the sum of squares of powers of 2 that divide n. - Amiram Eldar, Nov 12 2020

Examples

			A001157(4) = 21 and A050999(4) = 1, therefore a(4) = A001157(4)/A050999(4) = 21.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16383

Crossrefs

Cf. A001157, A038712, A050999, A318935.

Programs

Magma

[DivisorSigma(2,n)/&+[d^2:d in Divisors(n)|IsOdd(d)]:n in [1..100]]; // Marius A. Burtea, Jan 29 2020

Mathematica

f[n_]:=DivisorSigma[2,n]/Total[Select[Divisors[n],OddQ]^2]; f/@Range[100]
Table[(4^(IntegerExponent[n, 2] + 1) - 1)/3, {n, 1, 100}] (* Amiram Eldar, Nov 12 2020 *)

PARI

a(n) = sigma(n, 2)/sumdiv(n, d, d^2*(d % 2)); \\ Michel Marcus, Dec 11 2017

Python

def A296306(n): return ((1<<((n&-n).bit_length()<<1))-1)//3 # Chai Wah Wu, Jul 16 2022

Formula

a(n) = (4^(A007814(n) + 1) - 1)/3. - David Radcliffe, Dec 11 2017

Multiplicative with a(2^e) = (4^(e+1)-1)/3, and a(p^e) = 1 for odd primes p. - Amiram Eldar, Nov 12 2020

G.f.: Sum_{k>=0} 4^k * x^(2^k) / (1 - x^(2^k)). - Ilya Gutkovskiy, Dec 14 2020

Dirichlet g.f.: zeta(s)/(1-4/2^s). - Amiram Eldar, Dec 31 2022