a(n,k) tabf head (staircase) for A048996 M0 (or M_0) multinomial numbers for partitions of n in Abramowitz-Stegun (A-St) order. M0([a_1,...,a_n]) = sum(a_j,j=1..n)!/product((a_j)!,j=1..n) = m!/product((a_j)!,j=1..n). The row number is n and m is the number of parts of a partition of n. k numbers the partitions in the A-ST order. n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 1 1 2 1 1 3 1 2 1 4 1 2 1 3 1 5 1 2 2 3 3 4 1 6 1 2 2 1 3 6 1 4 6 5 1 7 1 2 2 2 3 6 3 3 4 12 4 5 10 6 1 8 1 2 2 2 1 3 6 6 3 3 4 12 6 12 1 5 20 10 6 15 7 1 9 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 10 1 2 2 2 2 1 3 6 6 6 3 6 3 3 4 12 12 6 12 24 4 4 6 5 20 20 30 30 20 1 6 30 15 60 15 7 42 35 8 28 9 1 . . . n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 The sequence of row lengths is A000041: [1, 2, 3, 5, 7, 11, 15, 22, 30, 42,...] (partition numbers). One could add the row for n=0 with a 1, if the part 0 is considered for n=0, and only for this n. For the ordering of this tabf array a(n,k) see Abramowitz-Stegun ref. pp. 831-2. E.g. a(4,4) refers to the fourth partition of n=4 in this ordering, namely (1^2,3^1)=[1,1,3], whence a(4,4)=3 because (2+1)!/(2!*1!)=3. Or a(6,5) = 3 for the partition (1^2,4^1)=[1,1,4] of n=6 from the same computation. a(7,10) = 12 from the 10th partition of n=7 which his (1^2,2^1,3^1)=[1,1,2,3] with (2+1+1)!/(2!*1!*1!) = 12. ####################################### Changed May 03 2007: a(n,m) into a(n,k). Added May 03 2007: The coefficients a(n,k) of row n appear in the calculation of the (1+sum(f[j]*x^j))^p as coefficients of x^n as follows: [x^n]((1+sum(f[j]*x^j))^p) = sum(sum(binomial(p,m)*M0(n,a_1,..,a_n)*product(f[j]^a_j,j=1..n),(a_1,..,a_n) from Pa(n,m)),m=1..min{n,p}) with m:=sum(a_j,j=1..n) and Pa(n,m) the set of partitions of n with m parts written in exponential form (1^a_1,...,n^a_n). If a_j=0 then j is not recorded. M0(n,a_1,...,a_n):=m!/product(a_j,j=1..n) are the numbers given as a(n,k) above if k is the k-th partition in A-St order. Example: n=4, p=2: partitions m=1: (4^1), m=2: (1^1,3^1) and (2^2). [x^4]((1+sum(f[j]*x^j))^2) = binomial(2,1)*(1!/1!)*f_4 +binomial(2,2)((2!/(1!*1!))*f_1*f_3 + (2!/2!)*(f_2)^2) = 2*f_4 + 2*f_1*f_3 + (f_2)^2. ######################################## Added Oct 12 2007: These M0 (or M_0) numbers for partitions (in short M0 partition numbers) are identical with the M_1 (or M1) partition numbers for the exponents of the partitions of n, read as partitons of m (the part number). See A036038 for the M_1 number array. The M0 numbers enter in the calculation of [x^n] A(x)^m, with an o.g.f. A(x) (ordinary convolutions). For A(x)=1/(1-x) the M0 numbers appear directly, and the sum over all M0 numbers for fixed part number m is binomial(n-1,m-1) (Pascal triangle). ########################################### e.o.f.#######################################################################################