1,4

Also number of partitions of n such that the difference between the two largest distinct parts is 2 (it is assumed that 0 is a part in each partition). Example: a(6)=3 because we have [4,2],[3,1,1,1] and [2,2,2]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 08 2006

G.f.: Sum_{m>0} (x^(2*m) / Product_{i>m} (1-x^i)}. More generally, g.f. for number of partitions of n such that the least part occurs exactly k times is Sum_{m>0} (x^(k*m)/Product_{i>m} (1-x^i)}.

G.f.=sum(x^(2k-2)*(1-x^(k-1))/product(1-x^j, j=1..k), k=1..infinity). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 08 2006

a(6)=3 because we have [4,1,1],[3,3] and [2,2,1,1].

g:=sum(x^(2*k)/product(1-x^j, j=k+1..80), k=1..70): gser:=series(g, x=0, 55): seq(coeff(gser, x, n), n=1..51); - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 08 2006

(* do first *) Needs["DiscreteMath`Combinatorica`"] (* then *) f[n_] := Block[{p = Partitions[n], l = PartitionsP[n], c = 0, k = 1}, While[k < l + 1, q = PadLeft[ p[[k]], 3]; If[ q[[1]] != q[[3]] && q[[2]] == q[[3]], c++ ]; k++ ]; c]; Table[ f[n], {n, 51}] (from Robert G. Wilson v Jul 23 2004)

(PARI) {q=sum(m=1, 100, x^(2*m)/prod(i=m+1, 100, 1-x^i, 1+O(x^60)), 1+O(x^60)); for(n=1, 51, print1(polcoeff(q, n), ", "))} - Klaus Brockhaus, Jul 21 2004

Cf. A002865, A097091, A097092, A097093.

Cf. A117989.

Sequence in context: A132888 A124774 A056610 this_sequence A108949 A167704 A052959

Adjacent sequences: A096370 A096371 A096372 this_sequence A096374 A096375 A096376

easy,nonn

Vladeta Jovovic (vladeta(AT)eunet.rs), Jul 19 2004

Edited and extended by Robert G. Wilson v (rgwv(AT)rgwv.com) and Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jul 21 2004