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Total number of parts in all self-conjugate partitions of n. Also, sum of largest parts of all self-conjugate partitions of n.
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%I #22 Dec 23 2019 23:31:58

%S 0,1,0,2,2,3,3,4,7,8,9,10,15,16,18,23,30,32,35,42,51,59,63,73,89,100,

%T 106,125,145,160,174,198,229,255,274,310,355,388,420,472,534,582,631,

%U 701,784,859,928,1021,1144,1243,1338,1475,1630,1767,1909,2089,2299

%N Total number of parts in all self-conjugate partitions of n. Also, sum of largest parts of all self-conjugate partitions of n.

%H T. D. Noe, <a href="/A067619/b067619.txt">Table of n, a(n) for n = 0..1000</a>

%H Arnold Knopfmacher and Neville Robbins, <a href="http://www.plouffe.fr/OEIS/citations/robbinspart.pdf">Identities for the total number of parts in partitions of integers</a>, Util. Math. 67 (2005), 9-18.

%F G.f.: A(q) = Sum_{n >= 1} n*q^(2*n-1)*(1+q)*(1+q^3)*...*(1+q^(2*n-3)).

%F From _Peter Bala_, Aug 20 2017: (Start)

%F Let F(q) = Product_{i >= 1} (1 + q^(2*i-1)). Then A(q) = Sum_{n >= 0} ( F(q) - Product_{i = 1..n} (1 + q^(2*i-1)) ).

%F It follows that the above sum A(q) satisfies -A(q-1) = 1 + q + 3*q^2 + 12*q^3 + 61*q^4 + ..., the g.f. for A158691, row-Fishburn matrices of size n. (End)

%t CoefficientList[Series[Sum[n*q^(2n-1)*Product[1+q^k, {k, 1, 2n-3, 2}], {n, 1, 30}], {q, 0, 60}], q]

%Y Cf. A000700, A000701, A006128, A015723, A046682, A079499, A158691.

%K easy,nonn

%O 0,4

%A _Naohiro Nomoto_, Feb 01 2002

%E Edited by _Dean Hickerson_, Feb 11 2002