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A116644 Triangle read by rows: T(n,k) is the number of partitions of n having exactly k doubletons (n>=0, k>=0). By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses). 4
1, 1, 1, 1, 3, 3, 2, 5, 2, 8, 2, 1, 10, 5, 13, 8, 1, 20, 9, 1, 26, 12, 4, 33, 21, 2, 46, 25, 5, 1, 58, 37, 6, 75, 48, 11, 1, 101, 59, 16, 125, 84, 19, 3, 157, 115, 23, 2, 206, 135, 39, 5, 253, 187, 46, 4, 317, 238, 63, 8, 1, 403, 292, 90, 7, 494, 382, 108, 17, 1, 608, 490, 139, 18 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

Apparently, rows n with p(p+1)<=n<(p+1)(p+2) have at most p+1 terms. Row sums are the partition numbers (A000041). T(n,0)=A116645(n). Sum(k*T(n,k),k>=0)=A116646(n).

LINKS

Alois P. Heinz, Rows n = 0..614, flattened

FORMULA

G.f.: G(t,x) = product(1+x^j+tx^(2j)+x^(3j)/(1-x^j), j=1..infinity).

EXAMPLE

T(6,2) = 1 because [2,2,1,1] is the only partition of 6 with 2 doubletons.

Triangle starts:

1;

1;

1,  1;

3;

3,  2;

5,  2;

8,  2, 1;

10, 5;

13, 8, 1;

MAPLE

g:=product(1+x^j+t*x^(2*j)+x^(3*j)/(1-x^j), j=1..35): gser:=simplify(series(g, x=0, 35)): P[0]:=1: for n from 1 to 24 do P[n]:=coeff(gser, x^n) od: for n from 0 to 24 do seq(coeff(P[n], t, j), j=0..degree(P[n])) od; # sequence given in triangular form

CROSSREFS

Cf. A000041, A116645, A116646.

Sequence in context: A238238 A117937 A110897 * A166462 A328177 A320776

Adjacent sequences:  A116641 A116642 A116643 * A116645 A116646 A116647

KEYWORD

nonn,tabf

AUTHOR

Emeric Deutsch, Feb 20 2006

STATUS

approved

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Last modified December 7 05:14 EST 2019. Contains 329839 sequences. (Running on oeis4.)