A333097 -id:A333097 - cp's OEIS frontend

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 - 6x + x^2))/(2x) 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

Peter Bala, Mar 22 2020

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.

			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 ).

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

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);

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 *)

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

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 2, 8, 41, 232, 1377, 8399, 52138, 327656, 2077934, 13270633, 85226594, 549837391, 3560702069, 23132584742, 150695482041, 984021596136, 6438849555963, 42208999230224, 277144740254566, 1822379123910857, 11998811140766701, 79095365076843134

Offset: 0

Peter Bala, Mar 07 2020

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 prime 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 c(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. For cases, see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^n:
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^1 = 1 + x + O(x^2)
  n = 2: c(x)^2 = 1 + 2*x + 5*x^2 + O(x^3)
  n = 3: c(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + O(x^4)
  n = 4: c(x)^4 = 1 + 4*x + 14*x^2 + 48*x^3 + 165*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 1 = 2, a(2) = 1 + 2 + 5 = 8, a(3) = 1 + 3 + 9 + 28 = 41 and a(4) = 1 + 4 + 14 + 48 + 165 = 232.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                        row sums
  n = 0 |   1                               1
  n = 1 |   1   1                           2
  n = 2 |   5   2    1                      8
  n = 3 |  28   9    3   1                 41
  n = 4 | 165  48   14   4   1            232
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 1, 5, 28, 165, ...] = [x^n] c(x)^n = A025174(n).
Examples of supercongruences:
a(13) - a(1) = 3560702069 - 2 = (3^2)*(13^3)*31*37*157 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11998811140766701 - 41 = (2^2)*5*(7^4)*32213*7756841 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 22794614296746579502 - 1377 = (5^6)*7*53*6491*605796421 == 0 ( mod 5^6 ).

Peter Bala, Notes on A333093

Cf. A000108, A001006, A001764, A005554, A025174, A333090 through A333097, A372214, A372215.

seq(add(n/(n+k)*binomial(n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) -> series(c(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - Sqrt[(1 - 3*x)/(1 + x)]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} n/(n+k)*binomial(n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c(x/(1 + x)) )^n = [x^n] ( (1 + x)*(1 + x*M(x)) )^n, where M(x) = ( 1 - x - sqrt(1 - 2*x - 3*x^2) ) / (2*x^2) is the o.g.f. of the Motzkin numbers A001006.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... = (1/x)*Revert( x/c(x) ) is the o.g.f. of A001764.
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 + 3/2) / (7 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(n+2*k)*binomial(n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024
D-finite with recurrence 2*n*(2*n-1)*(3991*n -21664)*a(n) +(-1329757*n^3 +9119565*n^2 -18270518*n +10657440)*a(n-1) +10*(947050*n^3 -6943257*n^2 +15944396*n -11260008)*a(n-2) +12*(-787878*n^3 +5778161*n^2 -13283386*n +9383340)*a(n-3) +9*(3*n-10)*(3*n-8)*(100503*n -141587)*a(n-4)=0, n>=5. - R. J. Mathar, Nov 22 2024

A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 3, 19, 144, 1171, 9878, 85216, 746371, 6609043, 59008563, 530279894, 4790262348, 43458522976, 395683988547, 3613641184739, 33088666355144, 303670285138067, 2792497004892302, 25724693177503987, 237350917999324431, 2193027397174233046, 20288470364637624223

Offset: 0

Peter Bala, Mar 15 2020

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 prime 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) defined as the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases A099837(m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(2*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^2 = 1 + 2*x + O(x^2)
  n = 2: c(x)^4 = 1 + 4*x + 14*x^2 + O(x^3)
  n = 3: c(x)^6 = 1 + 6*x + 27*x^2 + 110*x^3 + O(x^4)
  n = 4: c(x)^8 = 1 + 8*x + 44*x^2 + 208*x^3 + 910*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 14 = 19, a(3) = 1 + 6 + 27 + 110 = 144 and a(4) = 1 + 8 + 44 + 208 + 910 = 1171.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |   1                                 1
  n = 1 |   2    1                            3
  n = 2 |  14    4    1                      19
  n = 3 | 110   27    6   1                 144
  n = 4 | 910  208   44   8   1            1171
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 395683988547 - 3 = (2^6)*(3^2)*(13^3)*312677 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 20288470364637624223 - 144 = (7^3)*17*269*12934629208861 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150194008594715226556753 - 9878 = (5^6)*2593*5471* 677584325533 == 0 ( mod 5^6 ).

Cf. A000108, A069271, A333090 through A333097.

seq(add(2*n/(2*n+k)*binomial(2*n+2*k-1, k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x, n) -> series(c(x)^(2*n), x, 76):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - x - Sqrt[(1 - 3*x)*(1 + x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 2*n/(2*n+k)*binomial(2*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^2(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + ... = (1/x)*Revert( x/c^2(x) ) is the o.g.f. of A069271.
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) ~ 2^(8*n + 7/2) / (13 * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(2*n+2*k)*binomial(2*n+2*k, k) for n >= 1. - Peter Bala, Apr 19 2024

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 4, 34, 337, 3554, 38754, 431521, 4874377, 55639010, 640177033, 7412165034, 86256322816, 1007980394849, 11820510331777, 139032549536551, 1639506780365337, 19376785465043938, 229458302589724067, 2721958273545613513, 32339465512495259708, 384758834631081248554

Offset: 0

Peter Bala, Mar 15 2020

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 prime 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) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(3*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^3 = 1 + 3*x + O(x^2)
  n = 2: c(x)^6 = 1 + 6*x + 27*x^2 + O(x^3)
  n = 3: c(x)^9 = 1 + 9*x + 54*x^2 + 273*x^3 + O(x^4)
  n = 4: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + 2907*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 27 = 34, a(3) = 1 + 9 + 54 + 273 = 337 and a(4) = 1 + 12 + 90 + 544 + 2907 = 3554.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                              row sums
  n = 0 |    1                                    1
  n = 1 |    3       1                            4
  n = 2 |   27       6    1                      34
  n = 3 |  273      54    9   1                 337
  n = 4 | 2907     544   90  12   1            3554
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 11820510331777 - 4 = 3*11*(13^3)*(43^2)*88177 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 4583419703934987639046 - 337 = (3^2)*(7^4)*2441* 86893477573061 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 93266278848727959965820004 - 38754 = 2*(5^7)*19* 31416009717466260199 == 0 ( mod 5^6 ).

Cf. A000108, A118970, A333090 through A333097.

seq(add(3*n/(3*n+k)*binomial(3*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[3*Binomial[5*n-1, n] * HypergeometricPFQ[{1, -4*n, -n}, {1/2 - 5*n/2, 1 - 5*n/2}, 1/4]/4, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 3*n/(3*n+k)*binomial(3*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + ... = (1/x)*Revert( x/c^3(x) ) is the o.g.f. of A118970.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5^(5*n + 3/2) / (7 * 2^(8*n + 3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 3*n/(3*n+2*k)*binomial(3*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 5, 53, 647, 8373, 111880, 1525511, 21093476, 294663349, 4148593604, 58770091928, 836722722951, 11961868391175, 171601856667701, 2469036254872996, 35615467194043147, 514888180699419829, 7458193213805231529, 108219144962546395364, 1572690742149983040857

Offset: 0

Peter Bala, Mar 15 2020

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 prime 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) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(4*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^4 = 1 + 4*x + O(x^2)
  n = 2: c(x)^8 = 1 + 8*x + 44*x^2 + O(x^3)
  n = 3: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + O(x^4)
  n = 4: c(x)^16 = 1 + 16*x + 152*x^2 + 1120*x^3 + 7084*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 44 = 53, a(3) = 1 + 12 + 90 + 544 = 647 and a(4) = 1 + 16 + 152 + 1120 + 7084 = 8373.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(4*n), n >= 0, in descending powers of x begins
                                         row sums
  n = 0 |    1                               1
  n = 1 |    4     1                         5
  n = 2 |   44     8     1                  53
  n = 3 |  544    90    12     1           647
  n = 4 | 7084  1120   152    16   1      8373
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of congruences:
a(13) - a(1) = 171601856667701 - 5 = (2^4)*3*(7^2)*(13^3)*33208909 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 333475516822140871773101 - 647 = 2*(3^2)*(7^3)* 54012879303877692221 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 15187725485911657497382846255 - 111880 = (3^3)*(5^7)*29* 248279548173268475053 == 0 ( mod 5^6 ).

Cf. A000108, A212073, A333090 through A333097.

seq(add(4*n/(4*n+k)*binomial(4*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(4*n), x, 126):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^4(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + ... = (1/x)*Revert( x/c^4(x) ) is the o.g.f. of A212073.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 4*n/(4*n+2*k)*binomial(4*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 5, 57, 761, 10817, 159005, 2386857, 36348401, 559362561, 8676917429, 135445348057, 2125030235113, 33479772021953, 529326516063181, 8393856020704841, 133449301759137761, 2126391547960594433, 33948765589280671589, 542950968447834265209, 8697032976559212532953

Offset: 0

Peter Bala, Mar 22 2020

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 prime 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), defined as the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases see A333090 (m = 1) and A333092 (m = 3). For similarly defined sequences see A333093 through A333097.

			n-th order Taylor polynomial of S(x)^(2*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^2 = 1 + 4*x + O(x^2)
  n = 2: S(x)^4 = 1 + 8*x + 48*x^2 + O(x^3)
  n = 3: S(x)^6 = 1 + 12*x + 96*x^2 + 652*x^3 + O(x^4)
  n = 4: S(x)^8 = 1 + 16*x + 160*x^2 + 1296*x^3 + 9344*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 48 = 57, a(3) = 1 + 12 + 96 + 652 = 761 and a(4) = 1 + 16 + 160 + 1296 + 9344 = 10817.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    4    1                           5
  n = 2 |   48    8    1                     57
  n = 3 |  652   96   12   1                761
  n = 4 | 9344 1296  160  16   1          10817
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 4, 48, 652, 9344, 138004, ...]  = [x^n] S(x)^(2*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 529326516063181 - 5 = (2^3)*(13^3)*30116438101 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 2240508640665255893197949 - 761 = (2^2)*3*(7^3)*11* 49485569411283149863 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150633078429259494145205034005 - 159005 = (2^3)*(3^3)*(5^6)*11*23*61*2663*28097*119633*323083 == 0 ( mod 5^6 ).

A006318, A333090 through A333097.

S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^(2*n), x, 76):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

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

a(n) = [x^n] ( (1 + x)*S^2(x/(1 + x)) )^n
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 32*x^2 + 324*x^3 + 3696*x^4 + ... = (1/x)*Revert( x/S^2(x) ).
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) ~ sqrt(120 + 39*sqrt(10)) * (223 + 70*sqrt(10))^n / (30*sqrt(Pi*n) * 3^(3*n)). - Vaclav Kotesovec, Mar 28 2020

A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 7, 109, 1951, 36993, 724007, 14457421, 292732671, 5987886081, 123440423047, 2560421160109, 53373725431583, 1117198199782785, 23465732683090471, 494330214846965389, 10440064992542621951, 220978578227187097601, 4686426367646858888711, 99559270036968523118317

Offset: 0

Peter Bala, Mar 22 2020

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 prime 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) defined as the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases, see A333090 (m = 1) and A333091 (m = 2). For similarly defined sequences see A333093 through A333097.

			n-th order Taylor polynomial of S(x)^(3*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^3 = 1 + 6*x + O(x^2)
  n = 2: S(x)^6 = 1 + 12*x + 96*x^2 + O(x^3)
  n = 3: S(x)^9 = 1 + 18*x + 198*x^2 + 1734*x^3 + O(x^4)
  n = 4: S(x)^12 = 1 + 24*x + 336*x^2 + 3608*x^3 + 33024*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 6 = 7, a(2) = 1 + 12 + 96 = 109, a(3) = 1 + 18 + 198 + 1734 = 1951 and a(4) = 1 + 24 + 336 + 3608 + 33024 = 36993.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                           row sums
  n = 0 |     1                                1
  n = 1 |     6    1                           7
  n = 2 |    96   12    1                    109
  n = 3 |  1734  198   18   1               1951
  n = 4 | 33024 3608  336  24   1          36993
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 6, 96, 1734, 33024, 648006, ...]  = [x^n] S(x)^(3*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 23465732683090471 - 7 = (2^5)*(3^4)*(13^3)*83*911*54497 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 962815680123979633351467303 - 1951 = (2^3)*(7^3)*29*41* 1832861*161008076794727  == 0 ( mod 7^3 ).
a(5^2) - a(5) = 201479167004032422703424646224007 - 724007 = (2^5)*(5^6)* 402958334008064845406849291 == 0 ( mod 5^6 ).

Cf. A006318, A333090 through A333097.

S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1+x)*(1 - 3*x*(1+x) + (x^2 + x - 1)*Sqrt[1 - 4*x*(1+x)]) / (2*x^3))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = [x^n] ( (1 + x)*S^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 6*x + 66*x^2 + 902*x^3 + 13794*x^4 + ... = (1/x) * series reversion of ( x/S^3(x) ).
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*sqrt(85 + 21*sqrt(17)) * (349 + 85*sqrt(17))^n / (68 * sqrt(Pi*n) * 2^(5*n)). - Vaclav Kotesovec, Mar 28 2020

cp's OEIS Frontend

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.

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A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2*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.

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A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3*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.

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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 - 6x + x^2))/(2x) is the o.g.f. of the Schröder numbers A006318.

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.