A208513 Triangle of coefficients of polynomials u(n,x) jointly generated with A111125; see the Formula section.
1, 1, 1, 1, 4, 1, 1, 9, 6, 1, 1, 16, 20, 8, 1, 1, 25, 50, 35, 10, 1, 1, 36, 105, 112, 54, 12, 1, 1, 49, 196, 294, 210, 77, 14, 1, 1, 64, 336, 672, 660, 352, 104, 16, 1, 1, 81, 540, 1386, 1782, 1287, 546, 135, 18, 1, 1, 100, 825, 2640, 4290, 4004, 2275, 800, 170, 20, 1
Offset: 1
Examples
First five rows: 1; 1, 1; 1, 4, 1; 1, 9, 6, 1; 1, 16, 20, 8, 1; First five polynomials u(n,x): u(1,x) = 1; u(2,x) = 1 + x; u(3,x) = 1 + 4*x + x^2; u(4,x) = 1 + 9*x + 6*x^2 + x^3; u(5,x) = 1 + 16*x + 20*x^2 + 8*x^3 + x^4;
Links
- G. C. Greubel, Rows n = 1..50 of the triangle, flattened
- Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials
Programs
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Magma
A208513:= func< n,k | k eq 1 select 1 else (2*(n-1)/(n+k-2))*Binomial(n+k-2, 2*k-2) >; [A208513(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 02 2022
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Mathematica
(* First program *) u[1, x_]:=1; v[1, x_]:=1; z=16; u[n_, x_]:= u[n-1, x] + x*v[n-1, x]; v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x] + 1; Table[Expand[u[n, x]], {n, 1, z/2}] Table[Expand[v[n, x]], {n, 1, z/2}] cu = Table[CoefficientList[u[n, x], x], {n,z}]; TableForm[cu] Flatten[%] (* A208513 *) Table[Expand[v[n, x]], {n,z}] cv = Table[CoefficientList[v[n, x], x], {n,z}]; TableForm[cv] Flatten[%] (* A111125 *) (* Second program *) T[n_, k_]:= If[k==1, 1, ((n-1)/(k-1))*Binomial[n+k-3, 2*k-3]]; Table[T[n, k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Feb 02 2022 *) -
Sage
def A208513(n,k): return 1 if (k==1) else ((n-1)/(k-1))*binomial(n+k-3, 2*k-3) flatten([[A208513(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 02 2022
Formula
Coefficients of u(n, x) from the mixed recurrence relations:
u(n,x) = u(n-1,x) + x*v(n-1,x),
v(n,x) = u(n-1,x) + (x+1)*v(n-1,x) + 1,
where u(1,x) = 1, u(2,x) = 1+x, v(1,x) = 1, v(2,x) = 3+x.
From Peter Bala, May 01 2012: (Start)
Working with an offset of 0: T(n,0) = 1; T(n,k) = (n/k)*binomial(n+k-1,2*k-1) = (n/k)*A078812(n,k) for k > 0. Cf. A156308.
O.g.f.: ((1-t)^2 + t^2*x)/((1-t)*((1-t)^2-t*x)) = 1 + (1+x)*t + (1+4*x+x^2)*t^2 + ....
u(n+1,x) = -1 + (b(2*n,x) + 1)/b(n,x), where b(n,x) = Sum_{k = 0..n} binomial(n+k, 2*k)*x^k are the Morgan-Voyce polynomials of A085478.
This triangle is formed from the even numbered rows of A211956 with a factor of 2^(k-1) removed from the k-th column entries.
(End)
T(n, k) = (2*(n-1)/(n+k-2))*binomial(n+k-2, 2*k-2). - G. C. Greubel, Feb 02 2022
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