A372215 -id:A372215 - 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

A182959 Expansion of o.g.f. 2(1+x)^2/(1-2x+sqrt(1-8*x)).

Original entry on oeis.org

1, 5, 20, 96, 528, 3136, 19584, 126720, 841984, 5710848, 39376896, 275185664, 1944821760, 13875707904, 99807723520, 722997411840, 5269761884160, 38620004352000, 284405842575360, 2103530005463040, 15619068033761280

Offset: 0

Paul D. Hanna, Dec 31 2010

			G.f.: A(x) = 1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 +...
where A(x*F(x)^3) = F(x) is the g.f. of A182960:
F(x) = 1 + 5*x + 95*x^2 + 2496*x^3 + 76063*x^4 + 2524161*x^5 +...

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A182960, A001450, A167422, A372215.

CoefficientList[ Series[2 (1 + x)^2/(1 - 2 x + Sqrt[1 - 8 x]), {x, 0, 20}], x]  (* Robert G. Wilson v, Dec 31 2010 *)

{a(n)=polcoeff(2*(1+x)^2/(1-2*x+sqrt(1-8*x+x*O(x^n))),n)}

Let F(x) be the g.f. of A182960, then g.f. of this sequence satisfies:
* A(x) = F(x/A(x)^3) and A(x*F(x)^3) = F(x);
* A(x) = [x/Series_Reversion( x*F(x)^3 )]^(1/3).
G.f.: 1/2/x - 1/2 - x - (1+x)/x/G(0), where G(k)= 1 + 1/(1 - 4*x*(2*k+1)/(4*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
a(n) ~ 9*2^(3*n-2)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 29 2013
From Peter Bala, Oct 04 2015: (Start)
O.g.f. A(x) = (1 + x)*(2*C(2*x) - 1), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
[x^n] A(x)^(3*n) = binomial(6*n,2*n). Cf. with the identity [x^n] ( (1 + x)*C(x) )^(5*n) = binomial(5*n,2*n) = A001450(n). (End)
Conjecture: D-finite with recurrence (n+1)*a(n) +(-7*n+3)*a(n-1) +4*(-2*n+5)*a(n-2)=0. - R. J. Mathar, Jan 22 2020
From Peter Bala, May 15 2023: (Start)
a(n) = 3*(2^n)*(3*n - 1)/(n*(n + 1)) * binomial(2*n-2,n-1) for n >= 2.
(n + 1)*(3*n - 4)*a(n) = 4*(2*n - 3)*(3*n - 1)*a(n-1) for n >= 3 with a(2) = 20. Mathar's conjectured second order recurrence above follows from this. (End)
[x^n] A(x)^n = A372215(n). - Peter Bala, Nov 07 2024

A372214 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 - 2x - sqrt(1 - 8x + 4x^2))/(2x).

Original entry on oeis.org

1, 4, 40, 487, 6376, 86629, 1203823, 16984678, 242274280, 3484593028, 50444222665, 734066291974, 10728052396111, 157349171819155, 2314894133906086, 34145661019248487, 504810905195542504, 7478066502444399439, 110972913533524676080, 1649407167353221551706, 24549982881130265421001

Offset: 0

Peter Bala, Apr 23 2024

x*G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/2 is the o.g.f. of A047891.
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.

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 3*x + O(x^2)
  n = 2: G(x)^2 = 1 + 6*x + 33*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 9*x + 63*x^2 + 414*x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 12*x + 102*x^2 + 768*x^3 + 5493*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 33 = 40, a(3) = 1 + 9 + 63 + 414 = 487 and a(4) = 1 + 12 + 102 + 768 + 5493 = 6376.
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 |    3    1                              4
  n = 2 |   33    6     1                       40
  n = 3 |  414   63     9    1                 487
  n = 4 | 5493  768   102   12   1            6376
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 157349171819155 - 4 = (3^3)*(13^3)*269*9860941 == 0 (mod 13^3).
a(2*7) - a(2) = 2314894133906086 - 40 = 2*(3^4)*(7^3)*11*12119*312509 == 0 (mod 7^3).

Cf. A047891, A219535, A333090, A333093, A372215.

G := x -> (1/2)*(1 - 2*x - sqrt(1 - 8*x + 4*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 41):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);

Table[SeriesCoefficient[(2*(1 + x)^2/(1 - x + Sqrt[1 - 6*x - 3*x^2]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)

G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*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

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 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + ... is the o.g.f. of A219535.
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(1 + 17/sqrt(33)) * (59 + 11*sqrt(33))^n / (sqrt(3*Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, May 02 2024

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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A182959 Expansion of o.g.f. 2*(1+x)^2/(1-2*x+sqrt(1-8*x)).

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A372214 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 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x).

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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.

A182959 Expansion of o.g.f. 2(1+x)^2/(1-2x+sqrt(1-8*x)).

A372214 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 - 2x - sqrt(1 - 8x + 4x^2))/(2x).