

A263234


Triangle read by rows: T(n,k) is the number of partitions of n having k triangular number parts (0<=k<=n).


3



1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 2, 0, 1, 1, 3, 0, 2, 0, 1, 2, 2, 4, 0, 2, 0, 1, 2, 4, 2, 4, 0, 2, 0, 1, 4, 4, 5, 2, 4, 0, 2, 0, 1, 4, 6, 5, 6, 2, 4, 0, 2, 0, 1, 5, 9, 8, 5, 6, 2, 4, 0, 2, 0, 1, 6, 10, 11, 9, 5, 6, 2, 4, 0, 2, 0, 1, 9, 13, 13, 12, 10, 5, 6, 2, 4, 0, 2, 0, 1
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OFFSET

0,8


COMMENTS

The triangular numbers are i(i+1)/2 (i=0,1,2,3,...) (A000217).
Sum of entries in row n = A000041(n) = number of partitions of n.
T(n,0) = A225044(n).
Sum_{k=0..n} k*T(n,k) = A263235(n) = total number of triangular number parts in all partitions of n.


LINKS

Alois P. Heinz, Rows n = 0..200, flattened


FORMULA

G.f.: Product_{i>0} ((1x^h(i))/((1x^i)*(1t*x^h(i))), where h(i) = i*(i+1)/2.


EXAMPLE

T(6,2) = 4 because we have [4,1,1], [3,3], [3,2,1], and [2,2,1,1] (the partitions of 6 that have 2 triangular number parts).
Triangle starts:
1;
0,1;
1,0,1;
0,2,0,1;
2,0,2,0,1;
1,3,0,2,0,1;


MAPLE

h := proc (i) options operator, arrow: (1/2)*i*(i+1) end proc: g := product((1x^h(i))/((1x^i)*(1t*x^h(i))), i = 1 .. 80): gser := simplify(series(g, x = 0, 30)): for n from 0 to 18 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 18 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form


MATHEMATICA

max = 15; h[i_] = i*(i + 1)/2; P = Product[(1  x^h[i])/((1  x^i)*(1  t*x^h[i])), {i, 1, max}] + O[x]^max;
CoefficientList[#, t]& /@ CoefficientList[P, x] // Flatten (* JeanFrançois Alcover, May 25 2018 *)


CROSSREFS

Cf. A000041, A000217, A225044, A263235.
Sequence in context: A291969 A321434 A103919 * A264394 A283310 A035445
Adjacent sequences: A263231 A263232 A263233 * A263235 A263236 A263237


KEYWORD

nonn,tabl


AUTHOR

Emeric Deutsch, Nov 12 2015


STATUS

approved



