

A178866


Basic Multinomial Coefficients


7



1, 1, 1, 3, 1, 10, 1, 15, 15, 10, 1, 105, 35, 21, 1, 280, 210, 105, 56, 35, 28, 1, 1260, 1260, 378, 280, 126, 84, 36, 1, 6300, 3150, 2520, 2100, 1575, 945, 630, 210, 126, 120, 45, 1, 34650, 17325, 15400, 6930, 6930, 5775, 4620, 4620, 990, 462, 330, 165, 55, 1
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OFFSET

1,4


COMMENTS

All multinomial coefficients (MC's) are equal, but some are more equal than others. These are the basic multinomial coefficients (BMC's). The ordinary multinomial coefficients can be generated with the basic multinomial coefficients; see A178867.
A number n can be partitioned in A000041(n) different ways. The seven partitions of n=5 are e.g. [5] = [1+4] = [2+3] = [1+1+3] = [1+2+2] = [1+1+1+2] = [1+1+1+1+1]. We observe that the kth partition of n will consist of a certain number of 1s (i.e., "singles"), a certain number of 2s (i.e., "pairs"), a certain number of 3s (i.e., "triples"), a certain number of 4s (i.e., "4tuples") and so on. We denote with qt the number of ttuples in the kth partition of n. We observe that for the third partition of n=5 there is one pair (q2=1) and one triple (q3=1).
The multinomial coefficients are defined by M3[n,k] = n!/product((t!)^qt*(qt)!, t = 1..n), see Abramowitz and Stegun. For the third partition M3[5,3] = 10, so there are 10 different ways of distributing one pair (B1, B1) and one triple (B2, B2, B2) over five positions.
We define the BMC's as the multinomial coefficients M3[n,k] for which there are no singles (q1=0) in the kth partition of n for n>=2. Furthermore we define a(1) = 1.
The number of a(n) terms in a triangle row lead to sequence A002865(n) (n>=2). The row sums lead to sequence A000296(n) (n>=2).


LINKS

Table of n, a(n) for n=1..56.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 24, pp. 822832.
Kevin Brown, The Collector's Problem
J. W. Meijer and M. Nepveu, Euler's ship on the Pentagonal Sea, Acta Nova, Volume 4, No.1, December 2008. pp. 176187.


FORMULA

n = sum(qt*t, t=1..n)
M3[n,k] = n!/product((t!)^qt*(qt)!, t = 1..n)


EXAMPLE

The first few triangle rows are (P = Pair; T = Triple; 4T = 4 Tuple; etc..):
n = 1: 1;
n = 2: 1 (1*P);
n = 3: 1 (1*T);
n = 4: 3 (2*P), 1 (1*4T);
n = 5: 10 (1*P+1*T), 1 (1*5T);
n = 6: 15 (3*P), 15 (1*4T+1*P), 10 (2*T), 1 (1*6T);
n = 7: 105 (1*T+2*P), 35 (1*4T+1*T), 21 (1*5T+1*P), 1 (1*7T);


MAPLE

with(combinat): nmax:=11; A178866(1):=1: T:=1: for n from 1 to nmax do y(n):=numbpart(n): P(n):=sort(partition(n)): k:=0: for r from 1 to y(n) do if P(n)[r, 1]>1 then k:=k+1; B(k):=P(n)[r]: fi; od: A002865(n):=k; for k from 1 to A002865(n) do s:=0: j:=1: while s<n do s:=s+B(k)[j]: x(j+1):=B(k)[j]: j:=j+1; end do; jmax:=j; for r from 1 to n do q(r):=0 od: for r from 2 to n do for j from 1 to jmax do if x(j)=r then q(r):=q(r)+1 fi: od: od: M3[n, k]:= n!/(product((t!)^q(t)*q(t)!, t=1..n)): od: a:=sort([seq(M3[n, k], k=1..A002865(n))], `>`): for k from 1 to A002865(n) do M3[n, k]:=a[k] od: for k from 1 to A002865(n) do T:=T+1: A178866(T):= M3[n, k]: od: od: seq(A178866(n), n=1..T);


CROSSREFS

Cf. A036040 (version 1), A080575 (version 2) and A178867 (version 3).
Sequence in context: A226646 A127613 A211360 * A019427 A008299 A016478
Adjacent sequences: A178863 A178864 A178865 * A178867 A178868 A178869


KEYWORD

easy,nonn,tabf


AUTHOR

Johannes W. Meijer and Manuel Nepveu (Manuel.Nepveu(AT)tno.nl), Jun 21 2010, Jun 24 2010


STATUS

approved



