A333090 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of S(x)^n evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

1, 3, 21, 183, 1729, 17003, 171237, 1752047, 18130433, 189218451, 1987916021, 20996253479, 222730436161, 2371369720827, 25325636818629, 271189884041183, 2910628489408513, 31302328583021091, 337241582882175189, 3639109029230457751, 39324814984207649729

Offset: 0

Author

Peter Bala, Mar 22 2020

Keywords

• nonn
• easy

Comments

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime 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 congruences 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 S(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. See A333091 for m = 2 and A333092 for m = 3. For similarly defined sequences see A333093 through A333097.

Examples

			n-th order Taylor polynomial of S(x)^n:
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^1 = 1 + 2*x + O(x^2)
  n = 2: S(x)^2 = 1 + 4*x + 16*x^2 + O(x^3)
  n = 3: S(x)^3 = 1 + 6*x + 30*x^2 + 146*x^3 + O(x^4)
  n = 4: S(x)^4 = 1 + 8*x + 48*x^2 + 264*x^3 + 1408*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 16 = 21, a(3) = 1 + 6 + 30 + 146 = 183 and a(4) = 1 + 8 + 48 + 264 + 1408 = 1729.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^n, n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    2    1                           3
  n = 2 |   16    4    1                     21
  n = 3 |  146   30    6   1                183
  n = 4 | 1408  264   48   8   1           1729
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence is [x^n]S(x)^n = A103885(n).
Examples of supercongruences:
a(13) - a(1) = 2371369720827 - 3 = (2^3)*(3^2)*(13^3)*83*180617 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 425495386400395896971 - 183 = (2^2)*(7^3*)*19*47* 347287606554703 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 5894174066435445232142003 - 17003 = (2^3)*(3^4)*(5^6)*17* 41*101*5081*1627513421 == 0 ( mod 5^6 ).
		

Crossrefs

Cf. A006318, A027307, A103885, A333091 through A333097, A372214, A372215.

Programs

Maple

S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);

Mathematica

Table[SeriesCoefficient[((1 + x)*(1 - Sqrt[1 - 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

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

O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 10*x^2 + 66*x^3 + 498*x^4 + ... = (1/x)*Revert( x/S(x) ) is the o.g.f. of A027307.

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) ~ phi^(5*n+2) / (2*5^(3/4)*sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 28 2020