|
%I #15 Feb 19 2015 05:22:44
%S 1,1,0,1,0,1,1,0,0,1,1,0,0,1,1,1,0,1,1,0,1,1,0,0,2,0,1,1,1,0,0,2,0,1,
%T 0,2,1,0,1,2,0,1,0,1,2,1,0,0,3,1,1,0,1,1,2,1,0,0,3,0,2,0,2,1,1,2,1,0,
%U 1,3,0,2,0,1,2,1,1,3,1,0,0,4,0,2,0,2,2,2,1,1,3,1,0,0,4,0,2,1,2,3,1,2,1,2,3
%N Triangle read by rows: T(n,k) is the number of partitions of n into odd parts such that the sum of the parts, counted without multiplicities, is equal to k (n>=1, k>=1).
%C Row sums yield A000009. T(n,n)=A000700(n). Sum(k*T(n,k),k=1..n)=A116930(n).
%H Alois P. Heinz, <a href="/A116929/b116929.txt">Rows n = 1..141, flattened</a>
%F G.f.: -1+product(1+t^(2j-1)*x^(2j-1)/(1-x^(2j-1)), j=1..infinity).
%e T(10,4) = 3 because we have [3,3,3,1], [3,3,1,1,1,1] and [3,1,1,1,1,1,1,1].
%e Triangle starts:
%e 1;
%e 1,0;
%e 1,0,1;
%e 1,0,0,1;
%e 1,0,0,1,1;
%p g:=-1+product(1+t^(2*j-1)*x^(2*j-1)/(1-x^(2*j-1)),j=1..40): gser:=simplify(series(g,x=0,20)): for n from 1 to 15 do P[n]:=sort(coeff(gser,x^n)) od: for n from 1 to 15 do seq(coeff(P[n],t^j),j=1..n) od; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
%p expand(b(n, i-2)+add(x^i*b(n-i*j, i-2), j=1..n/i))))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, iquo(n-1, 2)*2+1)):
%p seq(T(n), n=1..14); # _Alois P. Heinz_, May 14 2014
%t b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<1, 0, Expand[b[n, i-2] + Sum[x^i*b[n-i*j, i-2], {j, 1, n/i}]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, Quotient[n-1, 2]*2+1]]; Table[T[n], {n, 1, 14}] // Flatten (* _Jean-François Alcover_, Feb 19 2015, after _Alois P. Heinz_ *)
%Y Cf. A000009, A000700, A116930.
%K nonn,tabl,look
%O 1,25
%A _Emeric Deutsch_, Feb 27 2006
|
