A370102 -id:A370102 - cp's OEIS frontend

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

1, 2, 16, 146, 1408, 14002, 142000, 1459810, 15158272, 158611106, 1669752016, 17664712562, 187641279616, 2000029880786, 21380213588848, 229129634462146, 2460955893981184, 26482855453375042, 285475524009208720, 3082024598888203090, 33319523640218177408

Offset: 0

Ralf Stephan, Feb 20 2005

From Peter Bala, Mar 01 2020: (Start)
The recurrence given below can be rewritten in the form
(2*n+1)*(2*n+2)*P(2,n)*a(n+1) - (2*n-1)*(2*n-2)*P(2,-n)*a(n-1) = Q(2,n^2)*a(n), where the polynomial Q(2,n) = 4*(55*n^2 - 34*n + 3) and the polynomial P(2,n) = 5*n^2 - 5*n + 1 satisfies the symmetry condition P(2,n) = P(2,1-n) and has real zeros.
More generally, for fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form ( Product_{k = 1..2*m} (2*m*n + k) ) * P(2*m,n)*b(n+1) + (-1)^m*( Product_{k = 1..2*m} (2*m*n - k) ) * P(2*m,-n)*b(n-1) = Q(2*m,n^2)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial P(2*m,n) = P(2*m,1-n) and has real zeros in the interval [0, 1]. The 4*m zeros of the polynomial Q(2*m,n^2) seem to belong to the interval [-1, 1] and 4*m - 2 of these zeros appear to be approximated by the rational numbers +- k/(3*m), where 1 <= k <= 3*m - 2, k not a multiple of 3. (End)

G. C. Greubel, Table of n, a(n) for n = 0..950 [a(0) = 1 inserted by _Georg Fischer_, Apr 03 2020]
Peter Bala, Notes on A103885
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. A002003, A006318, A027307, A103882, A103884, A123164, A144097, A156894, A266213, A370102.

A103885:= func< n | n eq 0 select 1 else (&+[ Binomial(n, k)*Binomial(2*n+k-1, n-1): k in [0..n]]) >;
[A103885(n): n in [0..40]]; // G. C. Greubel, Oct 27 2024

a := n -> `if`(n=0, 1, 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Dec 30 2019
# Alternative (after Peter Bala ):
gf := n -> ( (1 + x)/(1 - x) )^n: ser := n -> series(gf(n), x, 40):
seq(coeff(ser(n), x, 2*n), n=0..17); # Peter Luschny, Mar 20 2020

Prepend[Table[Sum[2^i Binomial[n, i] Binomial[2n-1, i-1], {i, 1, 2n}], {n,1,20}], 1] (* Vaclav Kotesovec, Jul 01 2015 *)

a(n) = if (n==0, 1, sum(i=0, n, 2^i * binomial(n, i) * binomial(2*n-1, i-1))); \\ Michel Marcus, Mar 21 2020

def A103885(n): return 1 if n==0 else sum(binomial(n, k)*binomial(2*n+k-1, n-1) for k in range(n+1))
[A103885(n) for n in range(41)] # G. C. Greubel, Oct 27 2024

a(n) = Sum_{i=0..n} 2^i * binomial(n,i) * binomial(2*n-1,i-1). [Original definition, with summation range {i=1..n}.]
a(n) = A103884(n, n).
G.f.: A(x) = x*B(x)'/B(x), where B(x) is g.f. of A027307. - Vladimir Kruchinin, Jun 30 2015
From Vaclav Kotesovec, Jul 01 2015: (Start)
Recurrence: n*(2*n-1)*(5*n^2 - 15*n + 11)*a(n) = 2*(55*n^4 - 220*n^3 + 296*n^2 - 152*n + 24)*a(n-1) + (n-2)*(2*n-3)*(5*n^2 - 5*n + 1)*a(n-2).
a(n) ~ ((11 + 5*sqrt(5))/2)^n / (2 * 5^(1/4) * sqrt(Pi*n)). (End)
a(n) = [x^n] (1/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ...))))))^n, a continued fraction. - Ilya Gutkovskiy, Sep 29 2017
a(n) = 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2) for n >= 1. - Peter Luschny, Dec 30 2019
From Peter Bala, Mar 01 2020: (Start)
a(n) = Sum_{k = 0..n} C(n, k)*C(2*n+k-1, n-1), with a(0) = 1.
a(n) = Sum_{k = 0..n} C(2*n, 2*k)*C(2*n-k-1, n-1), with a(0) = 1.
a(n) = (1/2)*Sum_{k = 0..n} C(2*n, n-k)*C(2*n+k-1, k). Cf. A156894.
a(n) = [x^n] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
a(n) = (1/2) * [x^(n)] ( (1 + x)/(1 - x) )^(2*n). Cf. A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n.
Conjecture: a(n) = - [x^n] G(x)^(-n), where G(x) = 1 + 2*x + 14*x^2 + 134*x^3 + 1482*x^4 + ... is the o.g.f. of A144097.
a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
From Peter Bala, Sep 22 2021: (Start)
a(n) = Sum_{k = 0..n} 4^k*binomial(n+k-1,n)*binomial(n,k)^2 / binomial(2*k,k).
Equivalently, a(n) = [x^n] T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A103882. (End)
For n>0, a(n) = (1/3) * [x^n] (1/S(-x))^(3*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318. Cf. A370102. - Peter Bala, Jul 29 2024

a(0) = 1 added and new name by Peter Bala, Mar 01 2020

A080247 Formal inverse of triangle A080246. Unsigned version of A080245.

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 22, 16, 6, 1, 90, 68, 30, 8, 1, 394, 304, 146, 48, 10, 1, 1806, 1412, 714, 264, 70, 12, 1, 8558, 6752, 3534, 1408, 430, 96, 14, 1, 41586, 33028, 17718, 7432, 2490, 652, 126, 16, 1, 206098

Offset: 0

Paul Barry, Feb 15 2003

Row sums are little Schroeder numbers A001003. Diagonal sums are generalized Fibonacci numbers A006603. Columns include A006318, A006319, A006320, A006321.
T(n,k) is the number of dissections of a convex (n+3)-gon by nonintersecting diagonals with exactly k diagonals emanating from a fixed vertex. Example: T(2,1)=4 because the dissections of the convex pentagon ABCDE having exactly one diagonal emanating from the vertex A are: {AC}, {AD}, {AC,EC} and {AD,BD}. - Emeric Deutsch, May 31 2004
For more triangle sums, see A180662, see the Schroeder triangle A033877 which is the mirror of this triangle. - Johannes W. Meijer, Jul 15 2013

			Triangle starts:
[0]    1
[1]    2,    1
[2]    6,    4,   1
[3]   22,   16,   6,   1
[4]   90,   68,  30,   8,  1
[5]  394,  304, 146,  48, 10,  1
[6] 1806, 1412, 714, 264, 70, 12, 1
...
From _Gary W. Adamson_, Jul 25 2011: (Start)
n-th row = top row of M^n, M = the following infinite square production matrix:
  2, 1, 0, 0, 0, ...
  2, 2, 1, 0, 0, ...
  2, 2, 2, 1, 0, ...
  2, 2, 2, 2, 1, ...
  ... (End)

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
Paul Barry, Laurent Biorthogonal Polynomials and Riordan Arrays, arXiv preprint arXiv:1311.2292 [math.CA], 2013.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 21, 29.
James East and Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.
P. Flajolet and M. Noy, Analytic combinatorics of noncrossing configurations, Discrete Math. 204 (1999), 203-229.
Shishuo Fu and Yaling Wang, Bijective recurrences concerning two Schröder triangles, arXiv:1908.03912 [math.CO], 2019.
W.-j. Woan, The Lagrange inversion formula and divisibility properties, JIS 10 (2007) 07.7.8, example 5.

Cf. A000007, A033877 (mirror), A084938.

A080247:=(n,k)->(k+1)*add(binomial(n+1,k+j+1)*binomial(n+j,j),j=0..n-k)/(n+1):
seq(seq(A080247(n,k),k=0..n),n=0..9);

Clear[w] w[n_, k_] /; k < 0 || k > n := 0 w[0,0]=1 ; w[n_, k_] /; 0 <= k <= n && !n == k == 0 := w[n, k] = w[n-1,k-1] + w[n-1,k] + w[n,k+1] Table[w[n,k],{n,0,10},{k,0,n}] (* David Callan, Jul 03 2006 *)
T[n_, k_] := Binomial[n, k] Hypergeometric2F1[k - n, n + 1, k + 2, -1];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Peter Luschny, Jan 08 2018 *)

T(n,k):=((k+1)*sum(2^m*binomial(n+1,m)*binomial(n-k-1,n-k-m),m,0,n-k))/(n+1); /* Vladimir Kruchinin, Jan 10 2022 */

def A080247_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return prec(n-1,k-1)-2*sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k)*prec(n, k) for k in (1..n)]
for n in (1..10): print(A080247_row(n)) # Peter Luschny, Mar 16 2016

G.f.: 2/(2+y*x-y+y*(x^2-6*x+1)^(1/2))/y/x. - Vladeta Jovovic, Feb 16 2003
Essentially same triangle as triangle T(n,k), n > 0 and k > 0, read by rows; given by [0, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] DELTA A000007 where DELTA is Deléham's operator defined in A084938.
T(n, k) = T(n-1, k-1) + 2*Sum_{j>=0} T(n-1, k+j) with T(0, 0) = 1 and T(n, k)=0 if k < 0. - Philippe Deléham, Jan 19 2004
T(n, k) = (k+1)*Sum_{j=0..n-k} (binomial(n+1, k+j+1)*binomial(n+j, j))/(n+1). - Emeric Deutsch, May 31 2004
Recurrence: T(0,0)=1; T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n,k+1). - David Callan, Jul 03 2006
T(n, k) = binomial(n, k)*hypergeom([k - n, n + 1], [k + 2], -1). - Peter Luschny, Jan 08 2018
T(n,k) = (k+1)/(n+1)*Sum_{m=0..n-k} 2^m*binomial(n+1,m)*binomial(n-k-1,n-k-m). - Vladimir Kruchinin, Jan 10 2022
From Peter Bala, Sep 16 2024: (Start)
Riordan array (S(x), x*S(x)), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the g.f. of the large Schröder numbers A006318.
For integer m and n >= 1, (m + 2)*[x^n] S(x)^(m*n) = m*[x^n] (1/S(-x))^((m+2)*n). For cases of this identity see A103885 (m = 1), A333481 (m = 2) and A370102 (m = 3). (End)

A370101 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(4*n-k-1,n-k).

Original entry on oeis.org

1, 7, 97, 1519, 25089, 427007, 7408897, 130287871, 2313945089, 41409732607, 745530884097, 13488086405119, 245014271688705, 4465915098890239, 81637668328243201, 1496095489290731519, 27477504726883368961, 505627095685486608383, 9320167322334416338945

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A001448, A370100, A370102.
Cf. A119259, A370097.
Cf. A365846.

Table[Sum[Binomial[4n,k]Binomial[4n-k-1,n-k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Dec 09 2024 *)
Table[Sum[2^k*(-1)^(n-k)*Binomial[4*n, k], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 31 2025 *)

a(n) = sum(k=0, n, binomial(4*n, k)*binomial(4*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^4/(1-x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^3/(1+x)^4 ). See A365846.
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k). - Seiichi Manyama, Jul 31 2025
a(n) ~ 2^(9*n + 3/2) / (7 * sqrt(Pi*n) * 3^(3*n - 1/2)). - Vaclav Kotesovec, Jul 31 2025
a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k). - Seiichi Manyama, Aug 01 2025
a(n) = [x^n] 1/((1-x) * (1-2*x)^(3*n)). - Seiichi Manyama, Aug 09 2025

A378238 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(3n+r+k,n)/(3n+r+k) for k > 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 14, 0, 1, 6, 32, 134, 0, 1, 8, 54, 324, 1482, 0, 1, 10, 80, 578, 3696, 17818, 0, 1, 12, 110, 904, 6810, 45316, 226214, 0, 1, 14, 144, 1310, 11008, 85278, 583152, 2984206, 0, 1, 16, 182, 1804, 16490, 140936, 1113854, 7769348, 40503890, 0

Offset: 0

Seiichi Manyama, Nov 20 2024

			Square array begins:
  1,      1,      1,       1,       1,       1,       1, ...
  0,      2,      4,       6,       8,      10,      12, ...
  0,     14,     32,      54,      80,     110,     144, ...
  0,    134,    324,     578,     904,    1310,    1804, ...
  0,   1482,   3696,    6810,   11008,   16490,   23472, ...
  0,  17818,  45316,   85278,  140936,  216002,  314700, ...
  0, 226214, 583152, 1113854, 1870352, 2914790, 4320608, ...

Columns k=0..3 give A000007, A144097, A371675, A365843.
T(n,n) gives 1/4 * A370102(n) for n > 0.
Cf. A033877, A071949, A378236, A378237, A378239, A378240.

T(n, k, t=3, u=1) = if(k==0, 0^n, k*sum(r=0, n, binomial(n, r)*binomial(t*n+u*r+k, n)/(t*n+u*r+k)));
matrix(7, 7, n, k, T(n-1, k-1))

G.f. A_k(x) of column k satisfies A_k(x) = ( 1 + x * A_k(x)^(3/k) * (1 + A_k(x)^(1/k)) )^k for k > 0.
G.f. of column k: B(x)^k where B(x) is the g.f. of A144097.
B(x)^k = B(x)^(k-1) + x * B(x)^(k+2) + x * B(x)^(k+3). So T(n,k) = T(n,k-1) + T(n-1,k+2) + T(n-1,k+3) for n > 0.

A370098 a(n) = Sum_{k=0..n} binomial(3n,k) binomial(4*n-k-1,n-k).

Original entry on oeis.org

1, 6, 72, 978, 14016, 207006, 3116952, 47568618, 733189632, 11387193846, 177923724072, 2793666465090, 44042615547456, 696708049377294, 11053262513080440, 175800225426741978, 2802193910116429824, 44752001810800994022, 715924864099841086728

Offset: 0

Seiichi Manyama, Feb 10 2024

nonn

Cf. A091527, A370097.
Cf. A370099, A370102.
Cf. A365843.

a(n) = sum(k=0, n, binomial(3*n, k)*binomial(4*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^3/(1-x)^3 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^3/(1+x)^3 ). See A365843.
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(3*n).
a(n) = Sum_{k=0..n} 2^k * binomial(3*n,k) * binomial(n-1,n-k).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n+k-1,k) * binomial(n-1,n-k). (End)

A370100 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(2*n-k-1,n-k).

Original entry on oeis.org

1, 5, 47, 500, 5615, 65005, 767396, 9183144, 110995695, 1351922495, 16566597047, 204010570296, 2522556212228, 31298015910140, 389458822888280, 4858487926378000, 60742838865326319, 760901358321592611, 9547848458062427405, 119990407515367475700

Offset: 0

Seiichi Manyama, Feb 10 2024

Cf. A001448, A352373, A370101, A370102.

Cf. A365754.

Table[Sum[Binomial[4*n, k]*Binomial[2*n - k - 1, n - k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 12 2024 *)

a(n) = sum(k=0, n, binomial(4*n, k)*binomial(2*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^4/(1-x) )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)/(1+x)^4 ). See A365754.
From Peter Bala, Jun 08 2024: (Start)
2*n*(n - 1)*(2*n - 1)*(51*n^2 - 144*n + 100)*a(n) = -(n - 1)*(5457*n^4 - 20865*n^3 + 26366*n^2 - 12172*n + 1560)*a(n-1) + 64*(2*n - 3)*(4*n - 5)*(4*n - 7)*(51*n^2 - 42*n + 7)*a(n-2) with a(0) = 1 and a(1) = 5.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. See A352373 for a more general conjecture. (End)
a(n) ~ sqrt(3 + 5/sqrt(17)) * (51*sqrt(17) - 107)^n / (sqrt(Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, Jun 12 2024
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(2*n+1) * (1-2*x)^n).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(3*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(3*n-k,n-k). (End)

A370099 a(n) = Sum_{k=0..n} binomial(2n,k) binomial(3*n-k-1,n-k).

Original entry on oeis.org

1, 4, 32, 292, 2816, 28004, 284000, 2919620, 30316544, 317222212, 3339504032, 35329425124, 375282559232, 4000059761572, 42760427177696, 458259268924292, 4921911787962368, 52965710906750084, 570951048018417440, 6164049197776406180, 66639047280436354816

Offset: 0

Seiichi Manyama, Feb 10 2024

Cf. A370098, A370102.

Cf. A032349, A103885.

a(n) = sum(k=0, n, binomial(2*n, k)*binomial(3*n-k-1, n-k));

a(n) = [x^n] ( (1+x)^2/(1-x)^2 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^2/(1+x)^2 ).
a(n) = 2 * A103885(n) for n >= 1. - Peter Bala, Sep 16 2024
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(2*n).
a(n) = Sum_{k=0..n} 2^k * binomial(2*n,k) * binomial(n-1,n-k).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n+k-1,k) * binomial(n-1,n-k). (End)

cp's OEIS Frontend

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

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A080247 Formal inverse of triangle A080246. Unsigned version of A080245.

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A370101 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(4*n-k-1,n-k).

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A378238 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(3*n+r+k,n)/(3*n+r+k) for k > 0.

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A370098 a(n) = Sum_{k=0..n} binomial(3*n,k) * binomial(4*n-k-1,n-k).

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A370100 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(2*n-k-1,n-k).

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A370099 a(n) = Sum_{k=0..n} binomial(2*n,k) * binomial(3*n-k-1,n-k).

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A370101 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(4*n-k-1,n-k).

A378238 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(3n+r+k,n)/(3n+r+k) for k > 0.

A370098 a(n) = Sum_{k=0..n} binomial(3n,k) binomial(4*n-k-1,n-k).

A370100 a(n) = Sum_{k=0..n} binomial(4n,k) binomial(2*n-k-1,n-k).

A370099 a(n) = Sum_{k=0..n} binomial(2n,k) binomial(3*n-k-1,n-k).