A048996 Irregular triangle read by rows. Preferred multisets: numbers refining A007318 using format described in A036038.

1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 2, 2, 3, 3, 4, 1, 1, 2, 2, 1, 3, 6, 1, 4, 6, 5, 1, 1, 2, 2, 2, 3, 6, 3, 3, 4, 12, 4, 5, 10, 6, 1, 1, 2, 2, 2, 1, 3, 6, 6, 3, 3, 4, 12, 6, 12, 1, 5, 20, 10, 6, 15, 7, 1, 1, 2, 2, 2, 2, 3, 6, 6, 3, 3, 6, 1, 4, 12, 12, 12, 12, 4, 5, 20, 10, 30, 5, 6, 30, 20, 7, 21, 8, 1

Offset

0,6

Comments

This array gives in row n>=1 the multinomial numbers (call them M_0 numbers) m!/product((a_j)!,j=1..n) with the exponents of the partitions of n with number of parts m:=sum(a_j,j=1..n), given in the Abramowitz-Stegun order. See p. 831 of the given reference. See also the arrays for the M_1, M_2 and M_3 multinomial numbers A036038, A036039 and A036040 (or A080575).

For a signed version see A111786.

These M_0 multinomial numbers give the number of compositions of n >= 1 with parts corresponding to the partitions of n (in A-St order). See an n = 5 example below. The triangle with the summed entries of like number of parts m is A007318(n-1, m-1) (Pascal). - Wolfdieter Lang, Jan 29 2021

Links

Andrew Howroyd, Table of n, a(n) for n = 0..2713 (rows 0..20)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1972.

Wolfdieter Lang, First 10 rows and more.

Formula

T(n,k) = A036040(n,k) * Factorial(A036043(n,k)) / A036038(n,k) = A049019(n,k) / A036038(n,k).

If the n-th partition is P, a(n) is the multinomial coefficient of the signature of P. - Franklin T. Adams-Watters, May 30 2006

T(n,k) = A309004(A036035(n,k)). - Andrew Howroyd, Oct 19 2020

Example

Table starts:
[1]
[1]
[1, 1]
[1, 2, 1]
[1, 2, 1, 3, 1]
[1, 2, 2, 3, 3, 4, 1]
[1, 2, 2, 1, 3, 6, 1, 4, 6,  5, 1]
[1, 2, 2, 2, 3, 6, 3, 3, 4, 12, 4, 5, 10, 6, 1]
.
T(5,6) = 4 because there are four multisets using the first four digits {0,1,2,3}: 32100, 32110, 32210 and 33210
T(5,6) = 4 because there are 4 compositions of 5 that can be formed from the partition 2+1+1+1. - Geoffrey Critzer, May 19 2013
These 4 compositions 2+1+1+1, 1+2+1+1, 1+1+2+1 and 1+1+1+2 of 5 correspond to the 4 set partitions of [5] :={1,2,3,4,5}, with 4 blocks of consecutive numbers, namely {1,2},{3},{4},{5} and {1},{2,3},{4},{5} and {1},{2},{3,4},{5} and {1},{2},{3},{4,5}. - Wolfdieter Lang, May 30 2018

Maple

nmax:=9: with(combinat): for n from 1 to nmax do P(n):=sort(partition(n)): for r from 1 to numbpart(n) do B(r):=P(n)[r] od: for m from 1 to numbpart(n) do s:=0: j:=0: while s<n do j:=j+1: s:=s+B(m)[j]: x(j):=B(m)[j]: end do; jmax:=j; for r from 1 to n do q(r):=0 od: for r from 1 to n do for j from 1 to jmax do if x(j)=r then q(r):=q(r)+1 fi: od: od: A036040(n, m) := (add(q(t), t=1..n))!/(mul(q(t)!, t=1..n)); od: od: seq(seq(A036040(n, m), m=1..numbpart(n)), n=1..nmax); # Johannes W. Meijer, Jul 14 2016

Prog

(SageMath) from collections import Counter
def ASPartitions(n, k):
    Q = [p.to_list() for p in Partitions(n, length=k)]
    for q in Q: q.reverse()
    return sorted(Q)
def A048996_row(n):
    h = lambda p: product(map(factorial, Counter(p).values()))
    return [factorial(len(p))//h(p) for k in (0..n) for p in ASPartitions(n, k)]
for n in (1..10): print(A048996_row(n)) # Peter Luschny, Nov 02 2019 [corrected on notice from Sean A. Irvine, Apr 30 2022]
(PARI)
C(sig)={my(S=Set(sig)); (#sig)!/prod(k=1, #S, (#select(t->t==S[k], sig))!)}
Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
{ for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 18 2020

Crossrefs

Cf. A000670, A007318, A036035, A036038, A019538, A115621, A309004, A000079 (row sums), A000041 (row lengths).

Sequence in context: A358105 A210961 A250007 * A111786 A072811 A296559

Adjacent sequences: A048993 A048994 A048995 * A048997 A048998 A048999

Keyword

nonn,tabf

Author

Alford Arnold

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 17 2001

a(0)=1 prepended by Andrew Howroyd, Oct 19 2020

Status

approved