A128520 q-inverse of x-x^2-x^3. Coefficients of a solution to the functional equation x = f(x) - f(x) f(q x) - f(x) f(q x) f(q^2 x).

1, 1, 1, 2, 1, 4, 2, 3, 1, 6, 8, 8, 7, 3, 5, 1, 8, 18, 21, 28, 17, 23, 14, 11, 5, 8, 1, 10, 32, 50, 73, 73, 77, 79, 58, 57, 44, 39, 22, 18, 8, 13, 1, 12, 50, 103, 166, 221, 238, 289, 256, 269, 226, 231, 176, 166, 113, 119, 74, 62, 36, 29, 13, 21, 1, 14, 72, 188, 347, 545, 680, 861

Offset: 1

Author

Brian Drake, Mar 06 2007

Keywords

• nonn
• tabf

Comments

The n-th row has binomial(n-1,2)+1 entries for n>=1. Each 1 starts a new row.

The last entry in the n-th row is the Fibonacci number F(n) = A000045(n). The second to last entry is F(n-1). The third from the end is the Lucas number L(n-1) = A000032(n-1).

The first entry in the n-th row is 1: A000012. The second entry is 2(n-2) = A005843(n-2). The third entry is 2(n-3)^2 = A001105(n-3).

The n-th row sum is A001002(n-1).

Examples

			1; 1; 1, 2; 1, 4, 2, 3; 1, 6, 8, 8, 7, 3, 5; 1, 8, 18, 21, 28, 17, 23, 14, 11, 5, 8; ...
a(4,1)=1, a(4,2)=4, a(4,3)=2, a(4,4)=3. f_4(q) = q^3 + 4 q^4 + 2 q^5 + 3 q^6. g_4(q) = 1 + 4 q + 2 q^2 + 3 q^3.
		

Crossrefs

Cf. A001002, A000045, A000032, A005843, A000012, A001105.

Programs

Maple

a:= proc(n)local i, j; option remember; if n=1 then 1 else add(q^i*a(i)*a(n-i), i=1..n-1) + add(a(i) *add(q^(2*n-2*i-j)*a(j)*a(n-i-j), j=1..n-i-1), i=1..n-2); fi; expand(%); sort(%); end;

Mathematica

g[1]=1; g[n_]:=g[n]=Expand[Sum[q^(i-1)g[i]g[n-i],{i,1,n-1}]+Sum[q^(2i+j-2)g[i]g[j]g[n-i-j],{i,1,n-2},{j,1,n-i-1}]]; a[n_,i_]:=Coefficient[g[n],q,i-1]

Formula

Let a(n,i) be the i-th entry in the n-th row, for n >= 1 and 1 <= i <= binomial(n-1,2)+1. Let f_n(q) = Sum_{i=1..binomial(n-1,2)+1} a(n,i) q^(n+i-2) and f(x) = Sum_{n>=1} f_n(q) x^n. Then f satisfies x = f(x) - f(x) f(q x) - f(x) f(q x) f(q^2 x).

The functions f_n satisfy the recurrence f_1 = 1, f_n = Sum_{i=1..n-1}(q^i f_i f_{n-i}) + Sum_{i=1..n-2}( f_i Sum_{j=1..n-i-1}(q^(2i+j) f_j f_{n-i-j})).

Equivalently, let g_n(q) = f_n(q)/q^(n-1) = Sum_{i=0..binomial(n-1,2)} a(n,i) q^(i-2) and g(x) = q f(x/q) = Sum_{n>=1} g_n(q) x^n. Then g satisfies q^2 x = q^2 g(x) - q g(x) g(q x) - g(x) g(q x) g(q^2 x). The functions g_n satisfy g_1 = 1, g_n = Sum_{i=1..n-1} (q^(i-1) g_i g_{n-i}) + Sum_{i=1..n-2} Sum_{j=1..n-i-1} (q^(2i+j-2) g_i g_j g_{n-i-j}).

Extensions

Edited by Dean Hickerson, Mar 09 2007