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%I #59 Dec 31 2022 03:50:46
%S 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,
%T 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,
%U 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
%N a(n) = A001157(n)/A050999(n).
%C 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.
%C 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.
%C Conjecture 2: Distinct terms form A002450 without A002450(0). In other words, a(2^(n-1)) = A002450(n), for n > 0.
%C Conjecture 3: For n > 0, the list of the first 2^n - 1 terms is palindromic.
%C Conjecture 4: For n > 0, the sum of the first 2^n - 1 terms equals A006095(n+1).
%C 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
%C 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
%C a(n) is the sum of squares of powers of 2 that divide n. - _Amiram Eldar_, Nov 12 2020
%H Antti Karttunen, <a href="/A296306/b296306.txt">Table of n, a(n) for n = 1..16383</a>
%F a(n) = (4^(A007814(n) + 1) - 1)/3. - _David Radcliffe_, Dec 11 2017
%F Multiplicative with a(2^e) = (4^(e+1)-1)/3, and a(p^e) = 1 for odd primes p. - _Amiram Eldar_, Nov 12 2020
%F G.f.: Sum_{k>=0} 4^k * x^(2^k) / (1 - x^(2^k)). - _Ilya Gutkovskiy_, Dec 14 2020
%F Dirichlet g.f.: zeta(s)/(1-4/2^s). - _Amiram Eldar_, Dec 31 2022
%e A001157(4) = 21 and A050999(4) = 1, therefore a(4) = A001157(4)/A050999(4) = 21.
%t f[n_]:=DivisorSigma[2,n]/Total[Select[Divisors[n],OddQ]^2]; f/@Range[100]
%t Table[(4^(IntegerExponent[n, 2] + 1) - 1)/3, {n, 1, 100}] (* _Amiram Eldar_, Nov 12 2020 *)
%o (PARI) a(n) = sigma(n, 2)/sumdiv(n, d, d^2*(d % 2)); \\ _Michel Marcus_, Dec 11 2017
%o (Magma) [DivisorSigma(2,n)/&+[d^2:d in Divisors(n)|IsOdd(d)]:n in [1..100]]; // _Marius A. Burtea_, Jan 29 2020
%o (Python)
%o def A296306(n): return ((1<<((n&-n).bit_length()<<1))-1)//3 # _Chai Wah Wu_, Jul 16 2022
%Y Cf. A001157, A038712, A050999, A318935.
%K easy,nonn,mult
%O 1,2
%A _Ivan N. Ianakiev_, Dec 10 2017
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