A372215 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 3*x - sqrt(1 - 10*x + 9*x^2))/(2*x) is the g.f. of A082298.

1, 5, 65, 1013, 16897, 292005, 5157569, 92456341, 1675300865, 30604622405, 562675808065, 10398428960309, 192983418877441, 3594314403564773, 67146652988725697, 1257656071818605013, 23609209650223284225, 444081728926392461445, 8367715805572617168449

Offset: 0

Author

Peter Bala, Apr 23 2024

Keywords

• nonn
• easy

Comments

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes p and positive integers n and k.

We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k. Examples of these supercongruences are given below.

More generally, for each integer m, we conjecture that the sequence {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.

Examples

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 4*x + O(x^2)
  n = 2: G(x)^2 = 1 + 8*x + 56*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 12*x + 108*x^2 + 892 *x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 16*x + 176*x^2 + 1680*x^3 + 15024*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 56 = 65, a(3) = 1 + 12 +  108 + 892 = 1013 and a(4) = 1 + 16 + 176 + 1680 + 15024 = 16897.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
                                            row sums
  n = 0 |     1                                 1
  n = 1 |     4      1                          5
  n = 2 |    56      8     1                   65
  n = 3 |   892    108    12    1            1013
  n = 4 | 15024   1680   176   16   1       16897
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 3594314403564773 - 5 = (2^5)*(3^3)*(13^3)*29*7643*8543 == 0 (mod 13^3).
a(2*7) - a(2) = 67146652988725697 - 65 = (2^7)*3*(7^4)*23*3943*803057 = 0 (mod 7^4).
		

Crossrefs

Cf. A082298, A182959, A333090, A333093, A372214.

Programs

Maple

G := x -> (1/2)*(1 - 3*x - sqrt(1 - 10*x + 9*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 21):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);

Mathematica

Table[SeriesCoefficient[(2*(1 + x)^2/(1 - 2*x + Sqrt[1 - 8*x]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)

PARI

G(x) = (1 - 3*x - sqrt(1 - 10*x + 9*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024

Formula

a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.

O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 4*x + 36*x^2 + 420*x^3 + 5572*x^4 + ....

Row sums of the Riordan array ( 1 + x*F'(x)/F(x), x*F(x) ) belonging to the Hitting time subgroup of the Riordan group.

a(n) ~ 3^(3*n/2 + 3/4) * (1 + sqrt(3))^(2*n-1) / (sqrt(Pi*n) * 2^(n+1)). - Vaclav Kotesovec, May 02 2024

a(n) = [x^n] H(x)^n, where H(x) = 2*(1 + x)^2/(1 - 2*x + sqrt(1 - 8*x)), the g.f. of A182959, satisfies [x^(n)] H(x)^(3*n) = binomial(6*n, 2*n). - Peter Bala, Nov 07 2024