



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
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OFFSET

1,2


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 kth powers of the divisors of n is divisible by the sum of the kth powers of the odd divisors of n.
Conjecture 2: Distinct terms form A002450 without A002450(0). In other words, a(2^(n1)) = 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).


LINKS

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


FORMULA

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


EXAMPLE

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


MATHEMATICA

f[n_]:=DivisorSigma[2, n]/Total[Select[Divisors[n], OddQ]^2]; f/@Range[100]


PROG

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


CROSSREFS

Cf. A001157, A050999.
Sequence in context: A144879 A049411 A070729 * A101693 A063476 A126325
Adjacent sequences: A296303 A296304 A296305 * A296307 A296308 A296309


KEYWORD

easy,nonn,mult


AUTHOR

Ivan N. Ianakiev, Dec 10 2017


STATUS

approved



