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A008951 Array read by columns: number of partitions of n into parts of 2 kinds. 14
1, 1, 1, 2, 2, 3, 4, 1, 5, 7, 2, 7, 12, 5, 11, 19, 9, 1, 15, 30, 17, 2, 22, 45, 28, 5, 30, 67, 47, 10, 42, 97, 73, 19, 1, 56, 139, 114, 33, 2, 77, 195, 170, 57, 5, 101, 272, 253, 92, 10, 135, 373, 365, 147, 20, 176, 508, 525, 227, 35, 1, 231, 684, 738, 345, 62, 2, 297 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

Fine-Riordan array S_n(m)=a(n,m) with extra row for n=0 added.

Row n of this triangle has length floor(1/2 + sqrt(2*(n+1))), n>=0. This is sequence A002024(n+1)=[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,...].

Written as triangle this becomes A103923.

a(n,m) gives also the number of partitions of n-t(m), where t(m):=A000217(m) (triangular numbers), with two kinds of parts 1,2,..m. See the column o.g.f.'s in table A103923.

In general, column m is asymptotic to exp(Pi*sqrt(2*n/3)) * 6^(m/2) * n^((m-2)/2) / (4*sqrt(3) * m! * Pi^m), equivalently to 6^(m/2) * n^(m/2) / (m! * Pi^m) * p(n), where p(n) is the partition function A000041. - Vaclav Kotesovec, Aug 28 2015

REFERENCES

H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.

LINKS

Alois P. Heinz, Columns n = 0..500, flattened

William K. Keith, Restricted k-color partitions, arXiv preprint arXiv:1408.4089, 2014

Wolfdieter Lang, First 20 rows and comments.

FORMULA

Riordan gives formula.

a(n, m) = sum over partitions of n of product(k[j], j=1..m), with k[j]=number of parts of size j (exponent of j in a given partition of n), if m>=1. If m=0 then a(n, 0)=p(n):=A000041(n) (number of partitions of n). O is counted as a part for n=0 and only for this n.

a(n, m) = sum over partitions of n of binomial(q(partition), m), with q the number of distinct parts of a given partition. m>=0.

a(n, m) = a(n-m, m-1) + a(n-m, m), n>=t(m):=m*(m+1)/2=A000217(m) (triangular numbers), else 0, with input a(n, 0) = p(n):=A000041(n).

EXAMPLE

Array begins:

m\n 0 1 2 3 4 .5 .6 .7 .8 ...

0 | 1 1 2 3 5 .7 11 15 22 ... (A000041)

1 | . 1 2 4 7 12 19 ... (A000070)

2 | . . . 1 2 .5 .9 ... (A000097)

3 | . . . . . .. .1 ... (A000098)

[1]; [1,1]; [2,2]; [3,4,1]; [5,7,2]; [7,12,5]; [11,19,9,1]...

a(3,1) = 4 because the partitions (3), (1,2) and (1^3) have q values 1,2 and 1 which sum to 4.

a(3,1) = 4 because the exponents of part 1 in the above given partitions of 3 are 0,1,3 and they sum to 4.

a(3,1) = 4 because the partitions of 3-t(1)=2 with two kinds of part 1, say 1 and 1' and one kind of part 2 are (2),(1^2), (1'^2) and (11').

MAPLE

a:= proc(n, m) option remember; `if`(n<0, 0,

      `if`(m=0, combinat[numbpart](n), a(n-m, m-1) +a(n-m, m)))

    end:

seq(seq(a(n, m), m=0..round(sqrt(2*n+2))-1), n=0..20);  # Alois P. Heinz, Nov 16 2012

MATHEMATICA

a[n_, 0] := PartitionsP[n]; a[n_, m_] /; (n >= m*(m+1)/2) := a[n, m] = a[n-m, m-1] + a[n-m, m]; a[n_, m_] = 0; Flatten[ Table[ a[n, m], {n, 0, 18}, {m, 0, Floor[1/2 + Sqrt[2*(n+1)]] - 1}]](* Jean-Fran├žois Alcover, May 02 2012, after recurrence formula *)

CROSSREFS

The first column (m=0) gives A000041(n). Columns m=1..10 are A000070 (partial sums of partition numbers), A000097, A000098, A000710, A103924-A103929.

Sequence in context: A159804 A104567 A087824 * A119473 A002122 A105689

Adjacent sequences:  A008948 A008949 A008950 * A008952 A008953 A008954

KEYWORD

nonn,tabf,nice

AUTHOR

N. J. A. Sloane.

EXTENSIONS

More terms from Robert G Bearden (nem636(AT)myrealbox.com), Apr 27 2004

Correction, comments and Riordan formulas from Wolfdieter Lang, Apr 28 2005

STATUS

approved

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Last modified February 20 12:57 EST 2019. Contains 320327 sequences. (Running on oeis4.)