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%I #31 Nov 26 2023 03:10:28
%S 0,1,1,1,1,2,1,1,2,2,1,2,1,2,3,1,1,3,1,2,3,2,1,2,2,2,3,2,1,4,1,1,3,2,
%T 3,3,1,2,3,2,1,4,1,2,5,2,1,2,2,3,3,2,1,4,3,2,3,2,1,4,1,2,5,1,3,4,1,2,
%U 3,4,1,3,1,2,5,2,3,4,1,2,4,2,1,4,3,2,3,2,1,6,3,2,3,2,3,2,1,3,5,3,1,4,1,2,7,2,1,4,1,4
%N Number of odd proper divisors of n. That is, the number of odd divisors of n that are less than n.
%H Antti Karttunen, <a href="/A091954/b091954.txt">Table of n, a(n) for n = 1..10000</a>
%F From _Antti Karttunen_, Oct 04 2017: (Start)
%F a(n) = Sum_{d|n, d<n} A000035(n).
%F a(n) = A001227(n) - A000035(n).
%F a(n) = A007814(A293214(n)) = A007814(A293216(n)).
%F (End)
%F G.f.: Sum_{k>=2} x^k/(1 - x^(2*k)). - _Seiichi Manyama_, Jan 23 2021
%F Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma + log(2)/2 - 1)*n, where gamma is Euler's constant (A001620). - _Amiram Eldar_, Nov 26 2023
%e The odd divisors of 15 that are less than 15 are 1, 3 and 5. Therefore there are three odd divisors of 15 that are less than 15.
%t Count[Most[Divisors[#]],_?OddQ]&/@Range[100] (* _Harvey P. Dale_, Sep 28 2012 *)
%t a[n_] := DivisorSigma[0, n/2^IntegerExponent[n, 2]] - Boole[OddQ[n]]; Array[a, 100] (* _Amiram Eldar_, Jun 11 2022 *)
%o (PARI) A091954(n) = sumdiv(n,d,(d<n)*(d%2)); \\ _Antti Karttunen_, Oct 04 2017
%o (PARI) my(N=66, x='x+O('x^N)); concat(0, Vec(sum(k=2, N, x^k/(1-x^(2*k))))) \\ _Seiichi Manyama_, Jan 23 2021
%Y Cf. A000005, A000035, A001620, A032741, A001227, A293214, A293216.
%Y Sum of the k-th powers of the odd proper divisors of n for k=0..10: this sequence (k=0), A091570 (k=1), A351647 (k=2), A352031 (k=3), A352032 (k=4), A352033 (k=5), A352034 (k=6), A352035 (k=7), A352036 (k=8), A352037 (k=9), A352038 (k=10).
%Y Sum of the k-th powers of the divisor complements of the odd proper divisors of n for k=0..10: this sequence (k=0), A352047 (k=1), A352048 (k=2), A352049 (k=3), A352050 (k=4), A352051 (k=5), A352052 (k=6), A352053 (k=7), A352054 (k=8), A352055 (k=9), A352056 (k=10).
%K easy,nonn
%O 1,6
%A _Mohammad K. Azarian_, Mar 12 2004
%E Corrected and extended by _Harvey P. Dale_, Sep 28 2012
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