A333564 -id:A333564 - cp's OEIS frontend

A014300 Number of nodes of odd outdegree in all ordered rooted (planar) trees with n edges.

1, 2, 7, 24, 86, 314, 1163, 4352, 16414, 62292, 237590, 909960, 3497248, 13480826, 52097267, 201780224, 783051638, 3044061116, 11851853042, 46208337584, 180383564228, 704961896036, 2757926215742, 10799653176704, 42326626862636, 166021623024584, 651683311373788

Offset: 1

Emeric Deutsch

Also total number of blocks of odd size in all Catalan(n) possible noncrossing partitions of [n].
Convolution of the sequence of central binomial coefficients 1,2,6,20,70,... (A000984) and of the sequence of Fine numbers 1,0,1,2,6,18,... (A000957).
Row sums of A119307. - Paul Barry, May 13 2006
Hankel transform is A079935. - Paul Barry, Jul 17 2009
Also for n>=1 the number of unimodal functions f:[n]->[n] with f(i)<>f(i+1). a(3) = 7: [1,2,1], [1,2,3], [1,3,1], [1,3,2], [2,3,1], [2,3,2], [3,2,1]. - Alois P. Heinz, May 23 2013
Also, number of sets of n rational numbers on [0,1) such that if x belongs to the set, the fractional part of 2x also belongs to it. - Jianing Song and Andrew Howroyd, May 18 2018
Let A(i, j) denote the infinite array such that the i-th row of this array is the sequence obtained by applying the partial sum operator i times to the function ((-1)^(n + 1) + 1)/2 for n > 0. Then A(n, n) equals a(n) for all n > 0. - John M. Campbell, Jan 20 2019
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p >= 3 and positive integers n and k. - Peter Bala, Jan 07 2022

Alois P. Heinz, Table of n, a(n) for n = 1..500
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Hacène Belbachir and Abdelghani Mehdaoui, Diagonal sums in Pascal pyramid (1, 2, r), Les Annales RECITS (2019) Vol. 6, 45-52.
N. Dershowitz and S. Zaks, Ordered trees and non-crossing partitions, Discrete Math., 62 (1986), 215-218.
Emeric Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.
Index entries for sequences related to rooted trees

Cf. A059481, A000957, A000984, A072547, A119259, A119307, A079935, A333564.

[(&+[(-1)^(n-k)*Binomial(n+k-1, k-1): k in [0..n]]): n in [1..30]]; // G. C. Greubel, Feb 19 2019

a:= proc(n) a(n):= `if`(n<3, n, ((12-40*n+21*n^2) *a(n-1)+
       2*(3*n-1)*(2*n-3) *a(n-2))/ (2*(3*n-4)*n))
    end:
seq(a(n), n=1..30);  # Alois P. Heinz, Oct 30 2012

Rest[CoefficientList[Series[2x/(1-4x+(1+2x)Sqrt[1-4x]),{x,0,40}],x]]  (* Harvey P. Dale, Apr 25 2011 *)
a[n_] := Sum[Binomial[2k, n-1], {k, 0, n-1}]; Array[a, 30] (* Jean-François Alcover, Dec 25 2015, after Paul Barry *)

a(n) = n--; sum(k=0, n, binomial(2*k,n)); \\ Michel Marcus, May 18 2018

from itertools import count, islice
def A014300_gen(): # generator of terms
    yield from (1,2)
    a, c = 1, 1
    for n in count(1):
        yield (a:=(3*n+5)*(c:=c*((n<<2)+2)//(n+2))-a>>1)
A014300_list = list(islice(A014300_gen(),20)) # Chai Wah Wu, Apr 26 2023

[sum((-1)^(n-k)*binomial(n+k-1, k-1) for k in (0..n)) for n in (1..30)] # G. C. Greubel, Feb 19 2019

a(n) = (binomial(2*n, n) + A000957(n))/3; [simplified by Alexander Burstein, Nov 24 2023]
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1, k-1). - Vladeta Jovovic, Aug 28 2002
G.f.: 2*z/(1-4*z+(1+2*z)*sqrt(1-4*z)).
a(n) = Sum_{j=0..floor((n-1)/2)} binomial(2*n-2*j-2, n-1).
2*a(n) + a(n-1) = (3*n-1)*Catalan(n-1). - Vladeta Jovovic, Dec 03 2004
a(n) = (-1)^n*Sum_{i=0..n} Sum_{j=n..2*n} (-1)^(i+j)*binomial(j, i). - Benoit Cloitre, Jun 18 2005
a(n) = Sum_{k=0..n} C(2*k,n) [offset 0]. - Paul Barry, May 13 2006
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n+k-1,k-1). - Paul Barry, Jul 18 2006
From Paul Barry, Jul 17 2009: (Start)
a(n) = Sum_{k=0..n} C(2*n-k,n-k)*(1+(-1)^k)/2.
a(n) = Sum_{k=0..n} C(n+k,k)*(1+(-1)^(n-k))/2. (End)
a(n) is the coefficient of x^(n+1)*y^(n+1) in 1/(1- x^2*y/((1-2*x)*(1-y))). - Ira M. Gessel, Oct 30 2012
a(n) = -binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2], 2). - Peter Luschny, Jul 25 2014
a(n) = [x^n] x/((1 - x^2)*(1 - x)^n). - Ilya Gutkovskiy, Oct 25 2017
a(n) ~ 4^n / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 25 2017
D-finite with recurrence: 2*n*a(n) +(-3*n-4)*a(n-1) +2*(-9*n+19)*a(n-2) +4*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Feb 20 2020
a(n) = A333564(n)/2^n. - Peter Bala, Apr 09 2020
a(n) = (1/2)*(binomial(2*n,n) - A072547(n)). - Peter Bala, Mar 28 2023

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

Original entry on oeis.org

1, 1, 7, 19, 103, 376, 1825, 7547, 35175, 153838, 708132, 3181091, 14616481, 66582283, 306501377, 1407473269, 6497464679, 29991098982, 138844558150, 643215119214, 2985368996228, 13868212710623, 64508509024241, 300324344452479, 1399598738196897, 6527698842078501

Offset: 0

Peter Bala, Feb 27 2022

Given an integer sequence (g(n))n>=1, there exists a formal power series G(x), with rational coefficients, such that g(n) = [x^n] G(x)^n. The power series G(x) has integer coefficients iff the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r.
The central binomial coefficient binomial(2*n,n) = A000984(n) may be defined using the coefficient extraction operator as binomial(2*n,n) = [x^n] ((1 + x)^2)^n and hence the Gauss congruences hold for A000984. Moreover, it is known that the stronger supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. See Meštrović, equation 39.
We define an infinite family of sequences as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. In particular, G_1(x) = (1 + x)^2 and the sequence u_1 is the sequence of central binomial coefficients. The present sequence is the case k = 2. See A351859 for the case k = 3.
Conjecture: for k >= 2, each sequence u_k satisfies the same supercongruences as the central binomial coefficients.
More generally, if r is a positive integer and s an integer then the sequence defined by  u_k(r,s;n) = [x^(r*n)] G_k(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 376 - 1 = 3*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 306501377 - 7 = 2*5*(7^3)*193*463 == 0 (mod 7^3)
A(5^2) - a(5) = 6527698842078501 - 376 = (5^6)*17*107*229671647 == 0 (mod 5^6)

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351859, A372211, A372212, A372213.

seq(add(add((-1)^(n-k-j)*binomial(n,k)*binomial(3*n,j)* binomial(4*n-2*j-k-1,n-k-j), j = 0..n-k), k = 0..n), n = 0..25);

A351858[n_]:=Sum[(-1)^(n-k-j)Binomial[n,k]Binomial[3n,j]Binomial[4n-2j-k-1,n-k-j],{k,0,n},{j,0,n-k}];Array[A351858,25,0] (* Paolo Xausa, Oct 04 2023 *)
a[n_]:=SeriesCoefficient[(1 + x + x^2)^(3*n)/(1 + x)^(2*n),{x,0,n}]; Array[a,26,0] (* Stefano Spezia, Apr 30 2024 *)

a(n) = Sum_{k = 0..n} Sum_{j = 0..n-k} (-1)^(n-k-j)*C(n,k)*C(3*n,j)*C(4*n-2*j-k-1,n-k-j).
Conjecture: a(n) = Sum_{k = 0..floor(n/2)} C(3*n,k)*C(n-k,k).
The o.g.f. A(x) = 1 + x + 7*x^2 + 19*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2)^3/(1 + x)^2) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x)^2/(1 + x + x^2)^3 ) = 1 + x + 4*x^2 + 10*x^3 + 40*x^4 + 133*x^5 + 536*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
a(n) ~ sqrt(2/9 + 2*sqrt(53/47)*cos(arccos(1259*sqrt(47/53)/1696)/3)/9) * (2*sqrt(164581)*cos(arccos(-90631279/(1316648*sqrt(164581)))/3)/81 - 293/81)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jun 05 2022

A333565 O.g.f.: (1 + 4x)/((1 + x)sqrt(1 - 8*x)).

Original entry on oeis.org

1, 7, 33, 223, 1537, 11007, 80385, 595455, 4456449, 33615871, 255148033, 1946337279, 14908784641, 114597822463, 883479412737, 6828492980223, 52895475040257, 410544577183743, 3191929428770817, 24855137310736383, 193811815161921537, 1513167009951514623, 11827298001565515777

Offset: 0

Peter Bala, Apr 11 2020

This 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, since the power series E(x) := exp( Sum_{n >= 1} a(n)*x^n/n ) has integer coefficients. See Stanley, Ex. 5.2 (a), p. 72, and its solution on p. 104.

We conjecture that this sequence satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 3 and positive integers n and k. The particular case when n = k = 1 follows from the corresponding result for A333564. Some examples of these congruences are given below.

			Examples of congruences:
a(11) - a(1) = 1946337279 - 7 = (2^3)*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 11827298001565515777 - 33 = (2^5)*(3^2)*(11^3)*107* 288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 5680983691406772011007 - 11007 = (2^8)*(3^3)*(5^6)*7* 19*1123*352183001 == 0 ( mod 5^6 ).

R. P. Stanley. Enumerative combinatorics. Vol. 2, (volume 62 of Cambridge Studies in Advanced Mathematics). Cambridge University Press, Cambridge, 1999.

Cf. A000984, A113647, A115137, A119259, A333564.

a := proc (n) option remember; `if`(n = 0, 1, `if`(n = 1, 7, `if`(n = 2, 33, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
end proc:
seq(a(n), n = 0..25);

a[n_] := (-1)^n - 2^(n+2) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
Table[Simplify[a[n]], {n, 0, 22}] (* Peter Luschny, Apr 13 2020 *)
CoefficientList[Series[(1+4x)/((1+x)Sqrt[1-8x]),{x,0,30}],x] (* Harvey P. Dale, Jan 24 2021 *)

a(n) = (2^n)*binomial(2*n,n) + 3*sum_{k = 0..n-1} (-1)^(n+k+1)*2^k* binomial(2*k,k).
a(n) = 4*A333564(n) + (-1)^n for n >= 1.
a(n) = 2*A119259(n) - (-1)^n.
a(n) = (-1)^n + 4*Sum_{k = 1..n} (3*k-1)*2^(k-1)*A000108(k-1).
a(n) ~ 8^n * 4/(3*sqrt(Pi*n)).
Congruences: a(p) == 7 ( mod p^3 ) for all prime p >= 3.
O.g.f. A(x) = 1 + 7*x + 33*x^2 + ... satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) = 0. Cf. A333564.
P-recursive: n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 7.
Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).
Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(0) = 1, a(1) = 7 and a(2) = 33.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 41*x^2 + 247*x^3 + ... is the o.g.f. of the second diagonal of triangle A113647. See also A115137.

A333563 a(n) = [x^n] G(x)^n, where G(x) is the o.g.f. of A079489.

Original entry on oeis.org

1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, 5344683429, 121590541641, 2782821611053, 64001191118956, 1477895865330092, 34243264651422596, 795729752353810824, 18537154747116799056, 432781371485493257637, 10123439350286679005973

Offset: 0

Peter Bala, Apr 12 2020

It can be shown that a(n) satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for prime p and positive integers n and k.
The o.g.f. G(x) of A079489 is given by G(x) = c(sqrt(x))*c(-sqrt(x)), where c(x) = ( (1 - sqrt(1 - 4*x))/(2*x) ) is the o.g.f. of the Catalan numbers A000108. It is known that the sequence b(n) := [x^n] c(x)^n  = 1/3*binomial(3*n,n) satisfies the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k - see Meštrović, equation 39. We conjecture that the present sequence  satisfies the same congruences. Some examples are given below.
More generally, if r > 0 and s are integers then the sequence a(r,s;n) := [x^(r*n)] G(x)^(s*n) may also satisfy the above congruences.

			Examples of congruences:
a(17) - a(1) = 10123439350286679005973 - 3 = 2(3^3)*5*(17^3)* 7631634401766047  == 0 ( mod 17^3 ).
a(3*5) - a(3) = 18537154747116799056 - 1056 = (2^4)*3*(5^3)*13* 237655830091241 == 0 ( mod 5^3 ).
a(5^2) - a(5) = 952866706104433648666617525245628 - 480003 = 3*(5^7)*17* 3642302759*65659247842693913 == 0 ( mod 5^6 ).

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000108, A079489, A333564, A333715.

c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G:= x -> c(sqrt(x))*c(-sqrt(x)):
H:= series(G(x)^n, x, 26):
seq(coeff(H, x, n), n = 0..25);

Join[{1}, Table[n^2 * Sum[(-1)^k/((n + 2*k)*(5*n - 2*k))*Binomial[n + 2*k, k] * Binomial[5*n - 2*k, 2*n - k], {k, 0, 2*n}], {n, 1, 20}]] (* Vaclav Kotesovec, Apr 20 2020 *)
Join[{1}, Table[Binomial[5*n, 2*n] * HypergeometricPFQ[{1/2 + n/2, -3*n, -2*n, n/2}, {1/2 - 5*n/2, 1 - 5*n/2, 1 + n}, -1]/5, {n, 1, 20}]] (* Vaclav Kotesovec, May 16 2020 *)

a(n) = if (n==0, 1, n^2 * sum(k = 0, 2*n, (-1)^k/((n+2*k)*(5*n-2*k))*binomial(n+2*k,k)*binomial(5*n-2*k, 2*n-k))); \\ Michel Marcus, May 16 2020

a(n) = n^2 * Sum_{k = 0..2*n} (-1)^k/((n+2*k)*(5*n-2*k))*C(n+2*k,k)* C(5*n-2*k, 2*n-k) for n >= 1.
a(p) == 3 ( mod p^3) for prime p >= 3, follows from the above formula.
P-recursive: 6*n*(n - 1)*(2*n - 1)*(3*n - 1)*(3*n - 2)*(2117*n^4 - 12615*n^3 + 27976*n^2 - 27348*n + 9936)*a(n) =  -(n - 1)*(27269077*n^8 - 217031969*n^7 + 722440183*n^6 - 1304402267*n^5 + 1384804360*n^4 - 874884704*n^3 + 315932544*n^2 - 57998736*n + 3913920)*a(n-1)  + 48*(6*n - 7)*(6*n - 8)*(6*n - 9)*(6*n - 10)*(6*n - 11)*(2117*n^4 - 4147*n^3 + 2833*n^2 - 773*n + 66)*a(n-2) with a(1) = 3, a(2) = 53.
[We note that the sequence u(n) := n^2 * Sum_{k = 0..2*n} 1/((n+2*k)*(5*n-2*k))*C(n+2*k,k)*C(5*n-2*k, 2*n-k) = (1/3)*C(6*n,2*n) is known to satisfy the congruences u(n*p^k) == u(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k - see Meštrović, equation 39. If, in the binomial sum formulas for a(n) and u(n) given above, we restrict the summation range to k = 0..n then we conjecture that the resulting pair of sequences satisfy the same congruences.]
a(n) ~ sqrt(1/24 + 1/(8*sqrt(73))) * ((2117*sqrt(73) - 12881)/216)^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 20 2020
a(n) = (1/5)*binomial(5*n, 2*n)*hypergeom([1/2 + n/2, -3*n, -2*n, n/2], [1/2 - 5*n/2, 1 - 5*n/2, 1 + n], -1) for n >= 1. - Vaclav Kotesovec, May 16 2020

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).

Original entry on oeis.org

1, 1, 3, 19, 67, 251, 1137, 4803, 20035, 87013, 377753, 1634469, 7134385, 31261114, 137121113, 603206144, 2660097603, 11749336328, 51981371895, 230336544210, 1021976441817, 4539784391763, 20188837618799, 89871081815631, 400427435522737, 1785639575031501

Offset: 0

Peter Bala, Mar 01 2022

This sequence is the third in an infinite family of sequences defined as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k, so that G_1(x) = (1 + x)^2, and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. See A000984, the sequence of central binomial coefficients, for the case k = 1 and A351858 for the case k = 2. The present sequence is the case k = 3.
Given a power series G(x) with integer coefficients it is known that the sequence (g(n))n>=1 defined by g(n) := [x^n] G(x)^n satisfies the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Thus a(n) satisfies the Gauss congruences. Calculation suggests that, in fact, the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. These supercongruences are known to hold for the central binomial coefficients A000984(n) = [x^n] ((1 + x)^2)^n (Meštrović, equation 39).
More generally, if r is a positive integer and s an integer then the sequence defined by a(r,s;n) = [x^(r*n)] G_3(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 251 - 1 = 2*(5^3) == 0 (mod 5^3)
a(2*7) - a(2) = 137121113 - 3 = 2*5*(7^4)*5711 == 0 (mod 7^4)
a(5^2) - a(5) = 1785639575031501 - 251 = 2*(5^6)*1373*3989*10433 == 0 (mod 5^6)

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Paolo Xausa, Table of n, a(n) for n = 0..425
R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351858.

seq(add(add(add((-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)* binomial(3*n+j-1,j)*binomial(j,k), k = 0..j), j = 0..n), i = 0..n), n = 0..25);

A351859[n_] := Sum[(-1)^j*Binomial[4*n, n-2*i-j-k]*Binomial[4*n, i]*Binomial[3*n+j-1, j]*Binomial[j, k], {i, 0, n}, {j, 0, n}, {k, 0, j}];
Array[A351859, 25, 0] (* Paolo Xausa, May 30 2025 *)

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,(-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)*binomial(3*n+j-1,j)*binomial(j,k))));
vector(25,n,a(n-1)) \\ Paolo Xausa, May 04 2022

a(n) = Sum_{i = 0..n} Sum_{j = 0..n} Sum_{k = 0..j} (-1)^j* C(4n,n-2*i-j-k) *C(4n,i)*C(3n+j-1,j)*C(j,k).
The o.g.f. A(x) = 1 + x + 3*x^2 + 19*x^3 + 67*x^4 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2 + x^3)^4/(1 + x + x^2)^3) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x + x^2)^3/(1 + x + x^2 + x^3)^4 ) = 1 + x + 2*x^2 + 8*x^3 + 25*x^4 + 81*x^5 + 305*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).

A349648 Expansion of g.f.: Catalan(x)/Catalan(-x).

Original entry on oeis.org

1, 2, 2, 8, 14, 64, 132, 640, 1430, 7168, 16796, 86016, 208012, 1081344, 2674440, 14057472, 35357670, 187432960, 477638700, 2549088256, 6564120420, 35223764992, 91482563640, 493132709888, 1289904147324, 6979724509184, 18367353072152, 99710350131200

Offset: 0

Alexander Burstein, Nov 23 2021

Cf. A000108, A001622, A048990 (bijection), A052707 (bijection), A006318, A079489, A246062, A333564.

gf:= (c-> c(x)/c(-x))(x-> hypergeom([1/2, 1], [2], 4*x)):
a:= n-> coeff(series(gf, x, n+1), x, n):
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 23 2021

CoefficientList[Series[(1-Sqrt[1-4x])/(Sqrt[1+4x]-1),{x,0,24}],x]

a(2*n) = A048990(n) = A000108(2*n), n>=0.
a(2*n+1) = A052707(n+1) = 2^(2*n+1)*A000108(n), n>=0.
G.f.: A(x) = C(x)/C(-x) = (1 - sqrt(1 - 4*x))/(sqrt(1 + 4*x) - 1), where C(x) is the g.f. of A000108.
G.f.: A(x) = F(x^2) + 2*x*F(x^2)^2 = (C(x) + C(-x))/2 + 2*x*C(4*x^2), where F(x) is the g.f. of A048990.
G.f.: A(-x) = 1/A(x).
G.f.: A(x) = R(x*C(-x)^2) = 1/R(-x*C(x)^2), where R(x) is the g.f. of A006318.
G.f.: A(x) = (1 + x*C(x)*C(-x))/(1 - x*C(x)*C(-x)), see A079489 for the expansion of C(x)*C(-x).
D-finite with recurrence n*(n-1)*(n+1)*a(n) -4*(n-1)*(8*n^2-32*n+35)*a(n-2) +64*(2*n-5)*(2*n-7)*(n-4)*a(n-4)=0. - R. J. Mathar, Mar 06 2022
Sum_{n>=0} 1/a(n) = 28/15 + 2*Pi/(9*sqrt(3)) + 64*arcsin(1/4)/(75*sqrt(15)) - 12*log(phi)/(25*sqrt(5)), where phi is the golden ratio (A001622). - Amiram Eldar, Apr 20 2023
G.f.: A(x) = exp( Sum_{n >= 1} binomial(4*n-2,2*n-1)*x^(2*n-1)/(2*n-1) ). - Peter Bala, Apr 28 2023

cp's OEIS Frontend

A014300 Number of nodes of odd outdegree in all ordered rooted (planar) trees with n edges.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

A351858 a(n) = [x^n] (1 + x + x^2)^(3*n)/(1 + x)^(2*n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A333565 O.g.f.: (1 + 4*x)/((1 + x)*sqrt(1 - 8*x)).

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Programs

Maple

Mathematica

Formula

A333563 a(n) = [x^n] G(x)^n, where G(x) is the o.g.f. of A079489.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4*n)/(1 + x + x^2)^(3*n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A349648 Expansion of g.f.: Catalan(x)/Catalan(-x).

Views

Author

Keywords

Crossrefs

Programs

Maple

Mathematica

Formula

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

A333565 O.g.f.: (1 + 4x)/((1 + x)sqrt(1 - 8*x)).

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).