0,7

Euler transform of period 5 sequence [0,1,1,0,0,...].

Also number of partitions of n such that the number of parts is greater by one than the smallest part. - Vladeta Jovovic (vladeta(AT)Eunet.yu), Mar 04 2006

Example: a(10)=4 because we have [9,1],[6,2,2],[5,3,2], and [4,4,2]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 09 2006

Also number of partitions of n such that if the largest part is k, then there are exactly k-1 parts equal to k. Example: a(10)=4 because we have [3,3,2,2],[3,3,2,1,1],[3,3,1,1,1,1], and [2,1,1,1,1,1,1,1,1]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 09 2006

Also number of partitions of n such that if the largest part is k, then k occurs at least k+1 times. Example: a(10)=4 because we have [2,2,2,2,2],[2,2,2,2,1,1],[2,2,2,1,1,1,1], and [1,1,1,1,1,1,1,1,1,1]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 09 2006

Also number of partitions of n such that the smallest part is larger than the number of parts. Example: a(10)=4 because we have [10],[7,3],[6,4], and [5,5]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 09 2006

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 238.

G. E. Andrews, Euler's "De Partitio Numerorum", Bull. Amer. Math. Soc., 44 (No. 4, 2007), 561-573.

G. E. Andrews and R. J. Baxter, A motivated proof of the Rogers-Ramanujan identities. Amer. Math. Monthly 96 (1989), no. 5, 401-409.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 108.

R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712.

G. E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge University Press, 1999; Exercise 6(f), p. 591.

T. D. Noe, Table of n, a(n) for n=0..1000

P. Jacob and P. Mathieu, Parafermionic derivation of Andrews-type multiple-sums

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

The Rogers-Ramanujan identity is 1 + Sum_{n >= 1} t^(n*(n+1))/((1-t)*(1-t^2)*...*(1-t^n)) = Product_{n >= 1} 1/((1-t^(5*n-2))*(1-t^(5*n-3))); this is the g.f. for the sequence.

G.f.: (Product_{k>0} 1+x^(2k))*(Sum_{k>=0} x^(k^2+2k)/(Product_{i=1..k} 1-x^(4i))). - Michael Somos Oct 19 2006

a(10)=4 because we have [8,2],[7,3],[3,3,2,2], and [2,2,2,2,2].

g:=1/product((1-x^(5*j-2))*(1-x^(5*j-3)), j=1..15): gser:=series(g, x=0, 66): seq(coeff(gser, x, n), n=0..63); - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 09 2006

(PARI) a(n)=if(n<0, 0, polcoeff(sum(k=0, (sqrt(4*n+1)-1)/2, x^(k^2+k)/prod(i=1, k, (1-x^i+x*O(x^n)))), n))

Cf. A003114.

Sequence in context: A058747 A050365 A029026 this_sequence A025149 A026824 A026799

Adjacent sequences: A003103 A003104 A003105 this_sequence A003107 A003108 A003109

nonn,nice,easy

njas, Herman P. Robinson