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%I #11 Feb 28 2017 12:08:11
%S 1,1,2,5,3,6,14,10,5,56,6,15,153,51,8,502,9,217,1756,25,11,7023,264,
%T 30,24363,1852,14,93629,15,6576,352782,40,3827,1377543,18,45,5200379,
%U 105812,20,20063228,21,352942,77607976,55,23,301906830,5172,185320,1166803215
%N Half the number of compositions of n into exactly two different parts with equal multiplicities.
%H Alois P. Heinz, <a href="/A242911/b242911.txt">Table of n, a(n) for n = 3..1000</a>
%F a(n) = 1/2 * Sum_{d|n} floor(d-1/2) * C(2*n/d,n/d).
%F a(p) = (p-1)/2 for odd prime p.
%F a(n) = 1/2 * (A131661(n)-A242900(n)).
%e a(6) = 5 because there are 10 compositions of 6 into exactly two different parts with equal multiplicities: [1,5], [5,1], [2,4], [4,2], [1,1,2,2], [1,2,1,2], [1,2,2,1], [2,1,1,2], [2,1,2,1], [2,2,1,1].
%p a:= n-> add(iquo(d-1, 2)*binomial(2*n/d, n/d),
%p d=numtheory[divisors](n))/2:
%p seq(a(n), n=3..60);
%t a[n_] := DivisorSum[n, Quotient[#-1, 2]*Binomial[2n/#, n/#]&]/2; Table[ a[n], {n, 3, 60}] (* _Jean-François Alcover_, Feb 28 2017, translated from Maple *)
%K nonn
%O 3,3
%A _Alois P. Heinz_, May 26 2014
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