

A116599


Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts equal to 2 (n>=0, 0<=k<=floor(n/2)).


2



1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 2, 1, 6, 3, 1, 1, 8, 4, 2, 1, 11, 6, 3, 1, 1, 15, 8, 4, 2, 1, 20, 11, 6, 3, 1, 1, 26, 15, 8, 4, 2, 1, 35, 20, 11, 6, 3, 1, 1, 45, 26, 15, 8, 4, 2, 1, 58, 35, 20, 11, 6, 3, 1, 1, 75, 45, 26, 15, 8, 4, 2, 1, 96, 58, 35, 20, 11, 6, 3, 1, 1, 121, 75, 45, 26, 15, 8, 4, 2, 1
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OFFSET

0,5


COMMENTS

Row n has 1 + floor(n/2) terms.
Row sums are the partition numbers (A000041).


LINKS

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened


FORMULA

T(n,0) = A027336(n), Sum_{k=0..floor(n/2)} k*T(n,k) = A024786(n).
Column k has g.f.: x^(2*k)/[(1x)*Product_{j>=0} ((1x^j))] (k=0,1,2,...).
G.f.: 1/[(1x)*(1t*x^2)*Product_{j>=3}( (1x^j) )].
T(n,k) = p(n2*k)  p(n2*k2) for k<=(n2)/2;
T(n, floor(n/2))=1 (follows at once from the g.f.).


EXAMPLE

T(6,1)=3 because we have [4,2], [3,2,1] and [2,1,1,1,1].
Triangle starts:
1;
1;
1,1;
2,1;
3,1,1;
4,2,1;
6,3,1,1;
8,4,2,1;


MAPLE

with(combinat): T:=proc(n, k) if k=floor(n/2) then 1 elif k<=(n2)/2 then numbpart(n2*k)numbpart(n2*k2) fi end: for n from 0 to 18 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form


MATHEMATICA

nn = 20; p = Product[1/(1  x^i), {i, 3, nn}]; f[list_] := Select[list, # > 0 &]; Map[f, CoefficientList[Series[p /(1  x)/(1  y x^2), {x, 0, nn}], {x, y}]] // Flatten (* Geoffrey Critzer, Jan 22 2012 *)


CROSSREFS

Cf. A000041, A027336, A024786.
Sequence in context: A174066 A089178 A187489 * A138121 A138151 A207378
Adjacent sequences: A116596 A116597 A116598 * A116600 A116601 A116602


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Feb 18 2006


EXTENSIONS

Keyword tabl changed to tabf by Michel Marcus, Apr 09 2013


STATUS

approved



