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A178866 Basic Multinomial Coefficients.

%I A178866 #13 Jul 22 2025 08:16:26
%S A178866 1,1,1,3,1,10,1,15,15,10,1,105,35,21,1,280,210,105,56,35,28,1,1260,
%T A178866 1260,378,280,126,84,36,1,6300,3150,2520,2100,1575,945,630,210,126,
%U A178866 120,45,1,34650,17325,15400,6930,6930,5775,4620,4620,990,462,330,165,55,1
%N A178866 Basic Multinomial Coefficients.
%C A178866 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.
%C A178866 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 k-th 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., "4-tuples") and so on. We denote with qt the number of t-tuples in the k-th partition of n. We observe that for the third partition of n=5 there is one pair (q2=1) and one triple (q3=1).
%C A178866 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.
%C A178866 We define the BMC's as the multinomial coefficients M3[n,k] for which there are no singles (q1=0) in the k-th partition of n for n>=2. Furthermore we define a(1) = 1.
%C A178866 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).
%H A178866 M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 24, pp. 822-832.
%H A178866 Kevin Brown, <a href="http://www.mathpages.com/home/kmath047.htm">The Collector's Problem</a>
%H A178866 J. W. Meijer and M. Nepveu, <a href="https://www.ucbcba.edu.bo/wp-content/uploads/PDF/Acta-Nova/v4.n1.Meijer.pdf">Euler's ship on the Pentagonal Sea</a>, Acta Nova, Volume 4, No.1, December 2008. pp. 176-187.
%F A178866 n = sum(qt*t, t=1..n)
%F A178866 M3[n,k] = n!/product((t!)^qt*(qt)!, t = 1..n)
%e A178866 The first few triangle rows are (P = Pair; T = Triple; 4-T = 4 Tuple; etc..):
%e A178866 n = 1: 1;
%e A178866 n = 2: 1 (1*P);
%e A178866 n = 3: 1 (1*T);
%e A178866 n = 4: 3 (2*P), 1 (1*4-T);
%e A178866 n = 5: 10 (1*P+1*T), 1 (1*5-T);
%e A178866 n = 6: 15 (3*P), 15 (1*4-T+1*P), 10 (2*T), 1 (1*6-T);
%e A178866 n = 7: 105 (1*T+2*P), 35 (1*4-T+1*T), 21 (1*5-T+1*P), 1 (1*7-T);
%p A178866 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);
%Y A178866 Cf. A036040 (version 1), A080575 (version 2) and A178867 (version 3).
%K A178866 easy,nonn,tabf
%O A178866 1,4
%A A178866 _Johannes W. Meijer_ and Manuel Nepveu (Manuel.Nepveu(AT)tno.nl), Jun 21 2010, Jun 24 2010