OFFSET
0,4
FORMULA
Let P_{n}(x) = Sum_{k=0..n} p_{n,k}(x) then
2^n*P_{n}(1/2) = A298611(n).
The general case: for fixed k the sequence P_{n}(k) with n >= 0 has the generating function ogf(k, x) = (1-2*(k+1)*x + (k^2+2*k-1)*x^2 + 2*(k-1)*x^3 + x^4)^(-1/2). The example section shows the start of this square array of sequences.
These sequences can be computed by the recurrence P(n,k) = ((2-n)*P(n-4,k)+(3-2*n)*(k-1)*P(n-3,k)+(k^2+2*k-1)*(1-n)*P(n-2,k)+(2*n-1)*(k+1)*P(n-1,k))/n with initial values 1, k+1, (k+1)^2+1 and (k+1)^3+2*k+4.
The partial polynomials p_{n,k}(x) reduce for x = 1 to A108625 (seen as a triangle).
EXAMPLE
The partial polynomials p_{n,k}(x) start:
[0] 1
[1] 1, x
[2] 1, 2*x+ 1, x^2
[3] 1, 3*x+ 4, 3*x^2+ 2*x, x^3
[4] 1, 4*x+ 9, 6*x^2+12*x+1, 4*x^3+ 3*x^2, x^4
[5] 1, 5*x+16, 10*x^2+36*x+9, 10*x^3+24*x^2+3*x, 5*x^4+4*x^3, x^5
.
The polynomials P_{n}(x) start:
[0] 1
[1] 1 + x
[2] 2 + 2*x + x^2
[3] 5 + 5*x + 3*x^2 + x^3
[4] 11 + 16*x + 9*x^2 + 4*x^3 + x^4
[5] 26 + 44*x + 34*x^2 + 14*x^3 + 5*x^4 + x^5
.
The triangle starts:
[0] 1
[1] 1, 1
[2] 2, 2, 1
[3] 5, 5, 3, 1
[4] 11, 16, 9, 4, 1
[5] 26, 44, 34, 14, 5, 1
[6] 63, 122, 111, 60, 20, 6, 1
[7] 153, 341, 351, 225, 95, 27, 7, 1
[8] 376, 940, 1103, 796, 400, 140, 35, 8, 1
[9] 931, 2581, 3384, 2764, 1561, 651, 196, 44, 9, 1
.
The square array P_{n}(k) near k=0:
...... [k=-2] 1, -1, 2, -1, -1, 10, -25, 51, -68, 41, ...
A182883 [k=-1] 1, 0, 1, 2, 1, 6, 7, 12, 31, 40, ...
A051286 [k=0] 1, 1, 2, 5, 11, 26, 63, 153, 376, 931, ...
A108626 [k=1] 1, 2, 5, 14, 41, 124, 383, 1200, 3799, 12122, ...
A299443 [k=2] 1, 3, 10, 35, 127, 474, 1807, 6999, 27436, 108541, ...
...... [k=3] 1, 4, 17, 74, 329, 1490, 6855, 31956, 150607, 716236, ...
MAPLE
CoeffList := p -> op(PolynomialTools:-CoefficientList(p, x)):
PrintPoly := p -> print(sort(expand(p), x, ascending)):
T := (n, k) -> x^k*binomial(n, k)*hypergeom([-k, k-n, k-n], [1, -n], 1/x):
P := [seq(add(simplify(T(n, k)), k=0..n), n=0..10)]:
seq(CoeffList(p), p in P); seq(PrintPoly(p), p in P);
R := proc(n, k) option remember; # Recurrence
if n < 4 then return [1, k+1, (k+1)^2+1, (k+1)^3+2*k+4][n+1] fi; ((2-n)*R(n-4, k)+
(3-2*n)*(k-1)*R(n-3, k)+(k^2+2*k-1)*(1-n)*R(n-2, k)+(2*n-1)*(k+1)*R(n-1, k))/n end:
for k from -2 to 3 do lprint(seq(R(n, k), n=0..9)) od;
MATHEMATICA
nmax = 10;
p[n_, k_, x_] := x^k*Binomial[n, k]*HypergeometricPFQ[{-k, k-n, k-n}, {1, -n}, 1/x];
p[n_, x_] := Sum[p[n, k, x], {k, 0, n}];
Table[CoefficientList[p[n, x], x], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Feb 26 2018 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, Feb 11 2018
STATUS
approved