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A394446
Triangle T(n,k), where row n lists the coefficients of the n-th palindromic polynomial p(n) defined by a recurrence of order 2.
1
1, 1, 3, 1, 1, 7, 13, 7, 1, 1, 12, 45, 69, 45, 12, 1, 1, 18, 108, 296, 411, 296, 108, 18, 1, 1, 25, 215, 890, 2015, 2633, 2015, 890, 215, 25, 1, 1, 33, 381, 2180, 7110, 14148, 17739, 14148, 7110, 2180, 381, 33, 1, 1, 42, 623, 4662, 20391, 56238, 101941, 124029, 101941, 56238, 20391, 4662, 623, 42, 1
OFFSET
0,3
COMMENTS
This should be the h^* vectors of some reflexive polytopes.
LINKS
FORMULA
T(n,k) = [x^k] p(n) with p(0) = 1; p(1) = 1+3*x+x^2; (n+2)*p(n) = p(1)*(2*n+1)*p(n-1)-(1+x+x^2)^2*(n-1)*p(n-2).
EXAMPLE
The first few polynomials are:
1,
x^2 + 3*x + 1,
x^4 + 7*x^3 + 13*x^2 + 7*x + 1,
x^6 + 12*x^5 + 45*x^4 + 69*x^3 + 45*x^2 + 12*x + 1.
Triangle T(n,k) begins:
1;
1, 3, 1;
1, 7, 13, 7, 1;
1, 12, 45, 69, 45, 12, 1;
1, 18, 108, 296, 411, 296, 108, 18, 1;
1, 25, 215, 890, 2015, 2633, 2015, 890, 215, 25, 1;
...
MAPLE
b:= proc(n) option remember; `if`(n=0, 1, `if`(n=1, 1+3*x+x^2,
expand((b(1)*(2*n+1)*b(n-1)-(1+x+x^2)^2*(n-1)*b(n-2))/(n+2))))
end:
T:= n-> coeffs(b(n)):
seq(T(n), n=0..7); # Alois P. Heinz, Mar 20 2026
MATHEMATICA
b[n_] := b[n] = If[n == 0, 1, If[n == 1, 1+3*x + x^2, Expand[(b[1]*(2*n+1)*b[n-1] - (1+x+x^2)^2*(n-1)*b[n- 2])/(n+2)]]];
T[n_] := CoefficientList[b[n], x];
Table[T[n], {n, 0, 7}] // Flatten (* Jean-François Alcover, Jun 14 2026, after Alois P. Heinz *)
PROG
(SageMath)
@cached_function
def a(n):
R = PolynomialRing(ZZ, 'x')
if n == 0:
return R.one()
p5 = R([1, 3, 1])
if n == 1:
return p5
p9 = R([1, 1, 1])**2
return (p5*(2*n+1)*a(n-1)-p9*(n-1)*a(n-2)) / (n + 2)
CROSSREFS
Row sums give A059231(n+1).
Main diagonal gives A128079.
Columns k=0-1 give: A000012, A055998.
Sequence in context: A353532 A049290 A297191 * A147990 A134567 A131932
KEYWORD
nonn,tabf
AUTHOR
F. Chapoton, Mar 20 2026
STATUS
approved