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 A117955 Number of partitions of n into exactly 2 types of odd parts. 5
 0, 0, 0, 1, 1, 2, 3, 5, 4, 7, 8, 10, 11, 13, 12, 19, 18, 20, 22, 25, 24, 30, 31, 36, 33, 39, 38, 45, 45, 48, 51, 57, 54, 60, 56, 69, 67, 72, 72, 79, 78, 84, 84, 90, 87, 97, 97, 112, 99, 107, 112, 117, 115, 126, 118, 131, 134, 137, 136, 152, 143, 149, 149, 163, 152, 174, 164 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,6 REFERENCES N. BENYAHIA TANI, S. BOUROUBI, O. KIHEL, An effective approach for integer partitions using exactly two distinct sizes of parts, Bulletin du Laboratoire, 03 (2015) 18 - 27; Availaible on line at http://www.liforce.usthb.dz. D Christopher, T Nadu, Partitions with Fixed Number of Sizes, Journal of Integer Sequences, 15 (2015), #15.11.5. LINKS Alois P. Heinz, Table of n, a(n) for n = 1..10000 N. Benyahia Tani, Sadek Bouroubi, Enumeration of the Partitions of an Integer into Parts of a Specified Number of Different Sizes and Especially Two Sizes, J. Int. Seq. 14 (2011) # 11.3.6 FORMULA G.f.=sum(sum(x^(2i+2j-2)/[(1-x^(2i-1))(1-x^(2j-1))], j=1..i-1), i=1..infinity). G.f. for number of partitions of n into exactly m types of odd parts is obtained if we substitute x(i) with -Sum_{k>0}(x^(2*n-1)/(x^(2*n-1)-1))^i in the cycle index Z(S(m); x(1),x(2),..,x(m)) of the symmetric group S(m) of degree m. - Vladeta Jovovic, Sep 20 2007 EXAMPLE a(8)=5 because we have [7,1],[5,3],[5,1,1,1],[3,3,1,1] and [3,3,1,1]. MAPLE g:=sum(sum(x^(2*i+2*j-2)/(1-x^(2*i-1))/(1-x^(2*j-1)), j=1..i-1), i=1..40): gser:=series(g, x=0, 75): seq(coeff(gser, x^n), n=1..72); CROSSREFS Cf. A002133. Sequence in context: A127515 A256996 A099424 * A074049 A193973 A245057 Adjacent sequences:  A117952 A117953 A117954 * A117956 A117957 A117958 KEYWORD nonn AUTHOR Emeric Deutsch, Apr 05 2006 STATUS approved

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Last modified September 26 10:43 EDT 2017. Contains 292518 sequences.