A207543 Triangle read by rows, expansion of (1+y*x)/(1-2*y*x+y*(y-1)*x^2).
1, 0, 3, 0, 1, 5, 0, 0, 5, 7, 0, 0, 1, 14, 9, 0, 0, 0, 7, 30, 11, 0, 0, 0, 1, 27, 55, 13, 0, 0, 0, 0, 9, 77, 91, 15, 0, 0, 0, 0, 1, 44, 182, 140, 17, 0, 0, 0, 0, 0, 11, 156, 378, 204, 19, 0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21, 0
Offset: 0
Examples
Triangle begins : 1 0, 3 0, 1, 5 0, 0, 5, 7 0, 0, 1, 14, 9 0, 0, 0, 7, 30, 11 0, 0, 0, 1, 27, 55, 13 0, 0, 0, 0, 9, 77, 91, 15 0, 0, 0, 0, 1, 44, 182, 140, 17 0, 0, 0, 0, 0, 11, 156, 378, 204, 19 0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21 0, 0, 0, 0, 0, 0, 13, 275, 1122, 1254, 385, 23
Links
- R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.
Crossrefs
Programs
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Maple
s := (1+y*x)/(1-2*y*x+y*(y-1)*x^2): t := series(s,x,12): seq(print(seq(coeff(coeff(t,x,n),y,m),m=0..n)),n=0..11); # Peter Luschny, Aug 17 2016
Formula
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 3.
G.f.: (1+y*x)/(1-2*y*x+y*(y-1)*x^2).
Sum_{i, i>=0} T(n+i,n) = A000204(2*n+1).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A078069(n), A000007(n), A003945(n), A111566(n) for x = -1, 0, 1, 2 respectively.
Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A090131(n), A005408(n), A003945(n), A078057(n), A028859(n), A000244(n), A063782(n), A180168(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively.
From Peter Bala, Aug 17 2016: (Start)
Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:
Sum_{k = 0..p} T(p,k)*S(2*k,n) = 1/2*(2*n + 1)*(n*(n + 1))^p.
For example, for row 6 we find S(6,n) + 27*S(8,n) + 55*S(10,n) + 13*S(12,n) = 1/2*(2*n + 1)*(n*(n + 1))^6.
There appears to be a similar result for the odd power sums S(2*k + 1,n) involving A119900. (End)
Extensions
New name using a formula of the author from Peter Luschny, Aug 17 2016
Comments