cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A293136 Irregular triangle T(n,k) read by rows: T(n,k) is the number of strongly unimodal compositions of n (A059618) into k parts.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 0, 1, 4, 1, 0, 1, 4, 5, 0, 1, 6, 6, 2, 0, 1, 6, 10, 4, 0, 1, 8, 14, 6, 1, 0, 1, 8, 19, 14, 1, 0, 1, 10, 23, 20, 5, 0, 1, 10, 31, 30, 10, 0, 1, 12, 36, 42, 18, 2, 0, 1, 12, 44, 60, 27, 4, 0, 1, 14, 52, 76, 48, 8, 0, 1, 14, 61, 102, 68, 16, 1, 0, 1, 16, 69, 126, 101, 30, 1, 0, 1, 16, 81, 160, 138, 50, 5, 0
Offset: 0

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Author

Joerg Arndt, Oct 01 2017

Keywords

Comments

Conjecture: index k of last nonzero entry in row n of is A293137(n).

Examples

			Triangle starts:
00:  [1]
01:  [0, 1]
02:  [0, 1]
03:  [0, 1, 2]
04:  [0, 1, 2, 1]
05:  [0, 1, 4, 1]
06:  [0, 1, 4, 5]
07:  [0, 1, 6, 6, 2]
08:  [0, 1, 6, 10, 4]
09:  [0, 1, 8, 14, 6, 1]
10:  [0, 1, 8, 19, 14, 1]
11:  [0, 1, 10, 23, 20, 5]
12:  [0, 1, 10, 31, 30, 10]
13:  [0, 1, 12, 36, 42, 18, 2]
14:  [0, 1, 12, 44, 60, 27, 4]
15:  [0, 1, 14, 52, 76, 48, 8]
16:  [0, 1, 14, 61, 102, 68, 16, 1]
17:  [0, 1, 16, 69, 126, 101, 30, 1]
18:  [0, 1, 16, 81, 160, 138, 50, 5]
19:  [0, 1, 18, 90, 194, 191, 80, 10]
20:  [0, 1, 18, 102, 238, 252, 118, 22]
...
Row n=7 is [0, 1, 6, 6, 2] because in the 15 partitions of 7 there is 0 into zero parts, 1 into one part, 6 into two parts, 6 into three parts, and 2 into four parts:
[ 1]   [ 1 2 3 1 ]
[ 2]   [ 1 2 4 ]
[ 3]   [ 1 3 2 1 ]
[ 4]   [ 1 4 2 ]
[ 5]   [ 1 5 1 ]
[ 6]   [ 1 6 ]
[ 7]   [ 2 3 2 ]
[ 8]   [ 2 4 1 ]
[ 9]   [ 2 5 ]
[10]   [ 3 4 ]
[11]   [ 4 2 1 ]
[12]   [ 4 3 ]
[13]   [ 5 2 ]
[14]   [ 6 1 ]
[15]   [ 7 ]
		

Crossrefs

Cf. A059618 (row sums), A293137.
Cf. A072704 (same for weakly unimodal compositions).

Programs

  • PARI
    N=25;  x='x+O('x^N);
    T=Vec(1 + sum(n=1, N, t*x^(n) * prod(k=1, n-1, 1+t*x^k)^2));
    for(r=1,#T, print(Vecrev(T[r])) );  \\ as triangle

Formula

G.f.: 1 + Sum_{n>=1} t*x^n * ( Product_{k=1..n-1} 1 + t*x^k )^2.
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