A182959 -id:A182959 - cp's OEIS frontend

A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

1, 5, 95, 2496, 76063, 2524161, 88534548, 3228482908, 121171012431, 4649906785719, 181614908182551, 7196014051078368, 288537887780406468, 11686156771344086156, 477379538839242423528, 19645977861506470428324

Offset: 0

Paul D. Hanna, Dec 31 2010

nonn

The logarithmic derivative of this sequence is a bisection of the logarithmic derivative of A001764 (ternary trees).
To see this, compare the g.f. of this sequence with g.f. of A001764:
exp(Sum_{n>=1} C(3n-1,n-1)*x^n/n) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 +...

			G.f.: A(x) = 1 + 5*x + 95*x^2 + 2496*x^3 + 76063*x^4 + 2524161*x^5 +...
log(A(x)) = 5*x + 165*x^2 + 6188*x^3 + 245157*x^4 + 10015005*x^5 +...+ A025174(2n)*x^n/n +...
G.f. satisfies: A(x) = G(x*A(x)^3) where G(x) begins:
G(x) = 1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 + 19584*x^6 +...

Cf. A182959, A025174, A001764, A060941, A300386 - A300389.

f[n_] := Exp[ Sum[ Binomial[6 i - 1, 2 i - 1]*x^i/i, {i, n}]]; CoefficientList[ Series[f@ 15, {x, 0, 15}], x]  (* Robert G. Wilson v, Dec 31 2010 *)

{a(n)=polcoeff(exp(sum(m=1,n,binomial(6*m-1,2*m-1)*x^m/m)+x*O(x^n)),n)}

{a(n)=polcoeff((serreverse(x*(1-2*x+sqrt(1-8*x+x*O(x^n)))^3/(8*(1+x)^6))/x)^(1/3),n)}

G.f. A(x) satisfies: A(x^2) = F(x)*F(-x) where F(x) = 1 + x*F(x)^3 = Sum_{n>=0} C(3n,n)*x^n/(2n+1) is the g.f. of A001764.
Let G(x) = 2*(1+x)^2/(1-2*x+sqrt(1-8*x)), then g.f. A(x) satisfies:
* A(x) = G(x*A(x)^3) and A(x/G(x)^3) = G(x);
* A(x) = [Series_Reversion( x/G(x)^3 ) / x]^(1/3).
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 5, 215, 10463, 537287, 28435880, 1534398353, 83920389642, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 (checked up to p = 101). - Peter Bala, Sep 14 2021
From Vaclav Kotesovec, Sep 16 2021: (Start)
Recurrence: 16*n*(2*n + 1)*(4*n - 1)*(4*n + 1)*(27*n - 31)*a(n) = 18*(69984*n^5 - 220320*n^4 + 267705*n^3 - 156510*n^2 + 42021*n - 3680)*a(n-1) - 59049*(n-1)*(2*n - 3)*(3*n - 5)*(3*n - 4)*(27*n - 4)*a(n-2).
a(n) ~ sqrt((sqrt(2) - 1)^(2/3) + (sqrt(2) + 1)^(2/3) - 2) * 3^(6*n + 3/2) / (sqrt(Pi) * n^(3/2) * 2^(4*n + 7/2)). (End)

A167422 Expansion of (1+x)*c(x), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 2, 3, 7, 19, 56, 174, 561, 1859, 6292, 21658, 75582, 266798, 950912, 3417340, 12369285, 45052515, 165002460, 607283490, 2244901890, 8331383610, 31030387440, 115948830660, 434542177290, 1632963760974, 6151850548776

Offset: 0

Paul Barry, Nov 03 2009

Hankel transform is A167423.

Apparently a(n) = A071716(n) if n>1. - R. J. Mathar, Nov 12 2009

G. C. Greubel, Table of n, a(n) for n = 0..500
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2

Cf. A000108, A001450, A005807, A071716, A167423, A182959.

A167422List := proc(m) local A, P, n; A := [1, 2]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), A[-1]]);
A := [op(A), P[-1]] od; A end: A167422List(26); # Peter Luschny, Mar 24 2022

Table[If[n < 2, n + 1, Binomial[2 n, n]/(n + 1) + Binomial[2 (n - 1), n - 1]/n], {n, 0, 25}] (* Michael De Vlieger, Oct 05 2015 *)
CoefficientList[Series[(1 + t)*(1 - Sqrt[1 - 4*t])/(2*t), {t, 0, 50}], t] (* G. C. Greubel, Jun 12 2016 *)

a(n) = if (n<2, n+1, binomial(2*n, n)/(n+1) + binomial(2*(n-1), n-1)/n);
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 04 2015

a(n) = Sum_{k=0..n} A000108(k)*C(1,n-k).
a(0)= 1, a(n) = A005807(n-1) for n>0. - Philippe Deléham, Nov 25 2009
(n+1)*a(n) +(-3*n+1)*a(n-1) +2*(-2*n+5)*a(n-2)=0, for n>2. - R. J. Mathar, Feb 10 2015
-(n+1)*(5*n-6)*a(n) +2*(5*n-1)*(2*n-3)*a(n-1)=0. - R. J. Mathar, Feb 10 2015
The o.g.f. A(x) satisfies [x^n] A(x)^(5*n) = binomial(5*n,2*n) = A001450(n). Cf. A182959. - Peter Bala, Oct 04 2015

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 - 3x - sqrt(1 - 10x + 9x^2))/(2x) is the g.f. of A082298.

Original entry on oeis.org

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

Offset: 0

Peter Bala, Apr 23 2024

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 + 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).

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

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);

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 *)

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

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

A259613 a(n) = binomial(6n,2n)/3, n>0, a(0)=1.

Original entry on oeis.org

1, 5, 165, 6188, 245157, 10015005, 417225900, 17620076360, 751616304549, 32308782859535, 1397281501935165, 60727722660586800, 2650087220696342700, 116043807643289338428, 5096278545356362962504, 224377658168860057076688

Offset: 0

Vladimir Kruchinin, Jun 30 2015

G. C. Greubel, Table of n, a(n) for n = 0..600
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A001448, A001450, A182959.

[1] cat [Binomial(6*n,2*n)/3: n in [1..20]]; // Vincenzo Librandi, Jul 01 2015

Join[{1}, Table[Binomial[6 n, 2 n]/3, {n, 30}]] (* Vincenzo Librandi, Jul 01 2015 *)

vector(20,n, n--; if (n==0, 1, binomial(6*n,2*n)/3)) \\ Michel Marcus, Jul 01 2015

G.f.: A(x) = 1 + (x*B(x)')/(B(x)) where  B(x) = 2 * (1 + x*B(x)^2)^2 / (1 - 2*x*B(x)^2 + sqrt(1-8*x*B(x)^2)).
a(n) ~ 3^(6*n-1/2) / (sqrt(Pi*n) * 2^(4*n+3/2)). - Vaclav Kotesovec, Jul 01 2015
a(n) = A025174(2*n), n>0. - R. J. Mathar, Jun 07 2016
From Peter Bala, Jun 08 2024: (Start)
a(n) = (9/2)*(6*n-1)*(6*n-5)*(3*n-1)*(3*n-2)/((4*n-1)*(4*n-3)*(2*n-1)*n) * a(n-1) with a(0) = 1 and a(1) = 5.
Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(5*n-k, 2*n-k) = (1/3)*binomial(6*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(5*n-k, 2*n-k) = binomial(4*n, 2*n). (End)
From Karol A. Penson, Jan 26 2025: (Start)
G.f. for 3*a(n),a(0)=1, denoted A, expressible entirely by radicals: A = A1 + A2 with
A1 = ((4*sqrt(4 - 27*sqrt(z)) + 12*i*sqrt(3)*z^(1/4))^(1/3) + (4*sqrt(4 - 27*sqrt(z)) - 12*i*sqrt(3)*z^(1/4))^(1/3))/(4*sqrt(4 - 27*sqrt(z))), and
A2 = (1/(4*sqrt(4 + 27*sqrt(z)) + 12*sqrt(3)*z^(1/4))^(1/3) + 1/(4*sqrt(4 + 27*sqrt(z)) - 12*sqrt(3)*z^(1/4))^(1/3))/sqrt(4 + 27*sqrt(z)),
where i = sqrt(-1), the imaginary unit. (End)

cp's OEIS Frontend

A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

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A167422 Expansion of (1+x)*c(x), c(x) the g.f. of A000108.

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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.

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A259613 a(n) = binomial(6*n,2*n)/3, n>0, a(0)=1.

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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 - 3x - sqrt(1 - 10x + 9x^2))/(2x) is the g.f. of A082298.

A259613 a(n) = binomial(6n,2n)/3, n>0, a(0)=1.