_Emeric Deutsch (deutsch(AT)duke.poly.edu), _, Feb 20 2006

editing

approved

Alois P. Heinz, <a href="/A116644/b116644.txt">Table of n, a(n) for Rows n = 0..10023614, flattened</a>

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

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

1, 1;

3, 2;

5, 2;

8, 2, 1;

10, 5;

13, 8, 1;

Alois P. Heinz, <a href="/A116644/b116644.txt">Table of n, a(n) for n = 0..10023</a>

approved

editing

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).

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

0,5

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).

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

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;

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

Cf. A000041, A116645, A116646.

nonn,tabf

Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 20 2006

approved