|
|
A116674
|
|
Triangle read by rows: T(n,k) is the number of partitions of n into odd parts and having exactly k distinct parts (n>=1, k>=1).
|
|
2
|
|
|
1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 3, 1, 5, 3, 4, 1, 2, 7, 1, 2, 8, 2, 2, 10, 3, 2, 11, 5, 2, 13, 7, 4, 12, 11, 1, 19, 11, 1, 2, 18, 17, 1, 3, 20, 21, 2, 2, 22, 27, 3, 2, 25, 32, 5, 4, 24, 41, 7, 2, 30, 46, 11, 2, 31, 56, 15, 2, 36, 62, 22, 3, 33, 80, 25, 1, 2, 39, 87, 36, 1, 4, 38, 103, 45, 2, 2, 45
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
Row n has floor(sqrt(n)) terms. Row sums yield A000009. T(n,1)=A001227(n) (n>=1). Sum(k*T(n,k),k>=1)=A038348(n-1) (n>=1).
|
|
LINKS
|
|
|
FORMULA
|
G.f.: product(1+tx^(2j-1)/(1-x^(2j-1)), j=1..infinity).
|
|
EXAMPLE
|
T(9,2)=4 because the only partitions of 9 into odd parts and having 2 distinct parts are [7,1,1],[5,1,1,1,1],[3,3,1,1,1] and [3,1,1,1,1,1,1].
Triangle starts:
1;
1;
2;
1,1;
2,1;
2,2;
2,3;
|
|
MAPLE
|
g:=product(1+t*x^(2*j-1)/(1-x^(2*j-1)), j=1..35): gser:=simplify(series(g, x=0, 34)): for n from 1 to 29 do P[n]:=coeff(gser, x^n) od: for n from 1 to 29 do seq(coeff(P[n], t, j), j=1..floor(sqrt(n))) od; # yields sequence in triangular form
# second Maple program:
with(numtheory):
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, add(b(n-i*j, i-2)*`if`(j=0, 1, x), j=0..n/i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(
b(n, iquo(n+1, 2)*2-1)):
|
|
MATHEMATICA
|
b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i<1, 0, Sum[b[n-i*j, i-2]*If[j == 0, 1, x], {j, 0, n/i}]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][b[n, Quotient[n+1, 2]*2-1]]; Table[T[n], {n, 1, 30}] // Flatten (* Jean-François Alcover, May 22 2015, after Alois P. Heinz *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|