A277930 Array of coefficients a(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = ((2*k+1)*x+sqrt(1+4*k*(k+1)*x^2))/(1-x^2), k>=0.
1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 3, 1, 1, 9, 25, 5, -3, 1, 1, 11, 41, 7, -59, 3, 1, 1, 13, 61, 9, -263, 5, 29, 1, 1, 15, 85, 11, -759, 7, 805, 3, 1, 1, 17, 113, 13, -1739, 9, 6649, 5, -131, 1, 1, 19, 145, 15, -3443, 11, 31241, 7, -12155, 3, 1, 1, 21, 181, 17, -6159, 13, 106261, 9, -200711, 5, 765, 1
Offset: 0
Examples
The terms define the array a(k,n) for k >= 0 and n >= 0, i.e., k\n 0 1 2 3 4 5 6 7 8 9 10 11 ... 0: 1 1 1 1 1 1 1 1 1 1 1 1 ... 1: 1 3 5 3 -3 3 29 3 -131 3 765 3 ... 2: 1 5 13 5 -59 5 805 5 -12155 5 205573 5 ... 3: 1 7 25 7 -263 7 6649 7 -200711 7 6766585 7 ... 4: 1 9 41 9 -759 9 31241 9 -1568759 9 88031241 9 ... 5: 1 11 61 11 -1739 11 106261 11 -7993739 11 672406261 11 ... 6: 1 13 85 13 -3443 13 292909 13 -30824051 13 ... 7: 1 15 113 15 -6159 15 696305 15 -97648655 15 ... 8: 1 17 145 17 -10223 17 1482769 17 -267255791 17 ... 9: 1 19 181 19 -16019 19 2899981 19 ... 10: 1 21 221 21 -23979 21 5300021 21 ... etc. The formal power series corresponding to row 2 is A(2,x) = 1+5*x+13*x^2+5*x^3 .. The terms define the triangle T(k,n) = a(k-n,n) for 0 <= n <=k, i.e., k\n 0 1 2 3 4 5 ... 0: 1 1: 1 1 2: 1 3 1 3: 1 5 5 1 4: 1 7 13 3 1 5: 1 9 25 5 -3 1 etc.
Programs
-
Mathematica
A[k_, n_]:=If[n==0, 1, If[EvenQ[n], 1 - 2 Sum[CatalanNumber[i] (-k(k + 1))^(i + 1), {i, 0, (n - 2)/2}], 2k + 1]]; Table[A[n - k, k], {n, 0, 12}, {k, 0, n}]//Flatten (* Indranil Ghosh, Aug 03 2017 *) -
Python
from sympy import catalan def A(k, n): return 1 if n==0 else 1 - 2*sum([catalan(i)*(-k*(k + 1))**(i + 1) for i in range(n//2)]) if n%2==0 else 2*k + 1 for n in range(13): print([A(n - k, k) for k in range(n + 1)]) # Indranil Ghosh, Aug 03 2017
Formula
a(k,0) = 1 and a(k,2*n+2) = 1-2*(Sum_{i=0..n} A000108(i)*(-k*(k+1))^(i+1)) and a(k,2*n+1) = 2*k+1 for k >= 0 and n >= 0.
A(k,x) = (1+(2*k+1)*x+2*k*(k+1)*x^2*C(-k*(k+1)*x^2))/(1-x^2) for k >= 0, where C is the o.g.f. of A000108.
A(k,x)*A(k,-x) = 1/(1-x^2) for k >= 0.
Conjecture: a(k,2*n+2) = 1+2*k+2*(-k)^(n+2)*(Sum_{i=0..n} A234950(n,i)*k^i) for k>=0 and n>=0. - Werner Schulte, Aug 03 2017
Comments