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A104471 Triangle of degree numbers for certain polynomials. 3
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, 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, 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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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).
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).
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.
REFERENCES
J. Riordan, Combinatorial Identities, Wiley, 1968.
LINKS
FORMULA
a(n, m)= sum(j*floor((n-m+1)/j), j=1..m), n>=m>=1, else 0.
EXAMPLE
[1];[2,1];[3,4,1];[4,5,4,1];[5,8,8,4,1];...
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.
CROSSREFS
See triangle A104472 where repeated numbers are omitted. a(2*n-1, n)=A024916(n).
Sequence in context: A177226 A059026 A233529 * A174828 A305309 A085643
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Mar 24 2005
STATUS
approved

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