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A104471 Triangle of degree numbers for certain polynomials. 3

%I #11 Aug 29 2019 17:31:23

%S 1,2,1,3,4,1,4,5,4,1,5,8,8,4,1,6,9,11,8,4,1,7,12,12,15,8,4,1,8,13,18,

%T 16,15,8,4,1,9,16,19,22,21,15,8,4,1,10,17,22,23,27,21,15,8,4,1,11,20,

%U 26,30,28,33,21,15,8,4,1,12,21,29,34,35,34,33,21,15,8,4,1,13,24,30,37,39,41

%N Triangle of degree numbers for certain polynomials.

%C This is the rectangular array ay(N,m):=sum(j*floor(N/j),j=1..m), m>=1, N>=1, written as a triangle a(n,m):=ay(n-m+1,m).

%C If A(j,x) is a polynomial of exact degree N(j):=floor(N/j), with some N>=1 and j=1,2,..., then F(m,x):=product(A(j,x^j),j=1..m) has degree A(N,m). This product appears in Fine's identity (on the lhs) if a finite product with m factors is taken. See the Riordan reference p. 182 eq.(20).

%C This choice of the degree N(j) guarantees that in F(m,x) all coefficients of x^n for n=0,...,N are correctly given. Due to Fine's identity (on the rhs) the coefficient of x^n of F(m,x) is given by the sum over all partitions of n with number of parts m of the product(a(j,k(j)),j=1..m), where a(j,p) is [x^p]A(j,x) and k(j)>=0 is the exponent of j in the considered partition of n into m parts. If n<m then k(j)=0 for all j=n+1,..,m.

%D J. Riordan, Combinatorial Identities, Wiley, 1968.

%H W. Lang: <a href="/A104471/a104471.txt">First 10 rows and array ay(N,m).</a>

%F a(n, m)= sum(j*floor((n-m+1)/j), j=1..m), n>=m>=1, else 0.

%e [1];[2,1];[3,4,1];[4,5,4,1];[5,8,8,4,1];...

%e a(5,3)=ay(3,3)=8 because for N=3 and m=3 one has N(1)=3, N(2)=1 and N(3)=1 and (a(1,0)+ a(1,1)*x^1+ ...+a(1,3)*x^3)*(a(2,0)+a(2,1)*x^2)*(a(3,0)+a(3,1)*x^3) has exact degree 3+2+3=8.

%Y See triangle A104472 where repeated numbers are omitted. a(2*n-1, n)=A024916(n).

%K nonn,easy,tabl

%O 1,2

%A _Wolfdieter Lang_, Mar 24 2005

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