A002003 -id:A002003 - cp's OEIS frontend

A333715 a(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n.

1, 2, 24, 326, 4672, 69002, 1038984, 15856206, 244396544, 3795731282, 59307908024, 931222155030, 14680871849152, 232236016459098, 3684420837693480, 58600075142247326, 934064636705476608, 14917333936933664674, 238641621366613695576, 3823510794994321546214, 61344017874989324388672

Offset: 0

Peter Bala, Apr 03 2020

a(n) is also equal to [x^n] G(x)^n, where G(x) is the o.g.f. of A027307.
Compare with A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n and A103885(n) = [x^(2*n)] ( (1 + x)/(1 - x) )^n = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318.
If we define an operator I acting on power series f(x) = 1 + f_1*x + f_2*x^2 + ... by I(f(x)) = 1/x * Revert( x/f(x) ) then S(x) = I( (1 + x)/(1 - x) ) and G(x) = (I o I)( (1 + x)/(1 - x )).
It can be shown that a(n) 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 a(n) satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples of these congruences are given below.
The same congruences may hold more generally for the sequences a(r,s,n) := [x^(r*n)]( (1 + x)/(1 - x) )^(s*n), r a positive integer and s an integer. This is the case a(3,1,n).
For fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form P(6*m,n)*b(n+1) + P(6*m,-n)*b(n-1) = Q(3*m,n^2)*b(n), where the polynomials P(6*m,n) and Q(3*m,n^2) have degree 6*m. Conjecturally, the 6*m zeros of the polynomial Q(3*m,n^2) belong to the interval [-1, 1] and 6*m - 4 of these zeros appear to be approximated by the rational numbers +- k/(4*m), where 1 <= k <= 4*m - 3, k not a multiple of 4.

			Examples of congruences:
a(11) - a(1) = 931222155030 - 2 = (2^2)*(11^3)*163*1073069 == ( mod 11^3 )
a(3*7) - a(3) = 985413034951400888962602 - 326 = (2^2)*(7^4)*263* 390130947874776863 == 0 ( mod 7^3 )
a(5^2) - a(5) = 66292579025690123511768694002 - 69002 = (2^3)*(5^6)*39461* 13439614612035199009 == 0 ( mod 5^6 )

Seiichi Manyama, Table of n, a(n) for n = 0..824

Cf. A002003, A027307, A103885, A144097.

seq(add(binomial(n,k)*binomial(3*n+k-1,n-1), k = 0..n), n = 0..20);

Table[Binomial[3*n-1, n-1] * Hypergeometric2F1[-n, 3*n, 2*n+1, -1], {n, 0, 20}] (* Vaclav Kotesovec, Apr 04 2020 *)

a(n) = Sum_{k = 0..n} C(n,k)*C(3*n+k-1,n-1).
a(n) = (1/3) * Sum_{k = 0..n} C(3*n,n-k)*C(3*n+k-1,k) for n >= 1.
a(n) = (1/3) * [x^n] ( (1 + x)/(1 - x) )^(3*n) for n >= 1.
a(n) = Sum_{k = 1..n} (2^k)*C(n,k)*C(3*n-1,k-1) for n >= 1.
P-recursive:
P(6,n)*a(n+1) + P(6,-n)*a(n-1) = Q(3,n^2)*a(n), where P(6,n) = (2*n-1)*(3*n+1)*(3*n+2)*(3*n+3)*(35*n^2 - 35*n + 6) and the polynomial Q(3,n) = 4*(7805*n^3 - 7132*n^2 + 1559*n - 72).
Congruences: a(p) == 2 ( mod p^3 ) for prime p >= 3.
a(n) ~ (223 + 70*sqrt(10))^n / (2^(3/4) * 5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Apr 04 2020
exp( Sum_{n>=1} a(n)*x^n/n ) = B(x) where B(x) = 1 + x*(B(x)^3 + B(x)^4) is the g.f. of A144097. - Paul D. Hanna, May 31 2023

A156894 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2n+k-1,k).

Original entry on oeis.org

1, 3, 19, 138, 1059, 8378, 67582, 552576, 4563235, 37972290, 317894394, 2674398268, 22590697614, 191475925332, 1627653567916, 13870754053388, 118464647799075, 1013709715774130, 8689197042438274, 74594573994750972, 641252293546113434, 5519339268476249676, 47558930664216470628

Offset: 0

Paul Barry, Feb 17 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on A156894

Cf. A002003, A003169, A103885, A114496, A123164.

A156894:= func< n | (&+[ Binomial(n,k)*Binomial(2*n+k-1,k): k in [0..n]]) >;
[A156894(n): n in [0..30]]; // G. C. Greubel, Jan 06 2022

a := n -> hypergeom([-n, 2*n], [1], -1);
seq(round(evalf(a(n),32)), n=0..19); # Peter Luschny, Aug 02 2014

Table[Sum[Binomial[n,k]Binomial[2n+k-1,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Nov 12 2014 *)

a(n) = if (n < 1, 1, sum(k=0, n, binomial(n,k)*binomial(2*n+k-1,k)));
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015

[round( hypergeometric([-n, 2*n], [1], -1) ) for n in (0..30)] # G. C. Greubel, Jan 06 2022

a(n) = [x^n] ((1+x)/(1-x)^2)^n.
a(n) = (4*(n+1)*(2*n+1)*A003169(n+1) - (5*n+1)*(2*n-1)*A003169(n))/(17*n + 5) for n>0. - Mark van Hoeij, Jul 14 2010
a(n) = Hypergeometric2F1([-n, 2*n], [1], -1). - Peter Luschny, Aug 02 2014
Conjecture: 64*n*(2*n-1)*a(n) -16*(89*n^2 -134*n +63)*a(n-1) +4*(661*n^2 -2619*n +2576)*a(n-2) -3*(119*n^2 -713*n +1092)*a(n-3) +6*(2*n-7)*(n-4)*a(n-4) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 16*n*(782*n +5365)*(2*n-1)*a(n) +8*(3128*n^3 -362053*n^2 +593930*n -290328)*a(n-1) -3*(726869*n^3 -5105981*n^2 +11667946*n -8715544)*a(n-2) +158*(2*n-5)*(n-3)*(391*n -764)*a(n-3) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 4*n*(2*n-1)*(17*n^2 -52*n +39)*a(n) -(1207*n^4 -4899*n^3 +6692*n^2 -3504*n +576)*a(n-1) +2*(n-2)*(2*n-3)*(17*n^2 -18*n +4)*a(n-2) = 0. - R. J. Mathar, Feb 05 2015 [the Maple command sumrecursion (binomial(n,k) * binomial(2*n+k-1,k), k, a(n)) verifies this recurrence. - Peter Bala, Oct 05 2015 ]
a(n) ~ sqrt(578 + 306*sqrt(17)) * (71 + 17*sqrt(17))^n / (17 * sqrt(Pi*n) * 2^(4*n+2)). - Vaclav Kotesovec, Feb 05 2015
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 14*x^2 + 79*x^3 + ... is the o.g.f. of A003169 (taken with offset 0). - Peter Bala, Oct 05 2015
From Peter Bala, Mar 20 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A119259.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Calculation suggests that, in fact, stronger congruences may hold.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

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 = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

A110171 Triangle read by rows: T(n,k) (0 <= k <= n) is the number of Delannoy paths of length n that start with exactly k (0,1) steps (or, equivalently, with exactly k (1,0) steps).

Original entry on oeis.org

1, 2, 1, 8, 4, 1, 38, 18, 6, 1, 192, 88, 32, 8, 1, 1002, 450, 170, 50, 10, 1, 5336, 2364, 912, 292, 72, 12, 1, 28814, 12642, 4942, 1666, 462, 98, 14, 1, 157184, 68464, 27008, 9424, 2816, 688, 128, 16, 1, 864146, 374274, 148626, 53154, 16722, 4482, 978, 162, 18, 1

Offset: 0

Emeric Deutsch, Jul 14 2005

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).
Column k for k >= 1 has g.f. z^k*R^(k-1)*g*(1+z*R), where R = 1 + zR + zR^2 = (1 - z - sqrt(1-6z+z^2))/(2z) is the g.f. of the large Schroeder numbers (A006318) and g = 1/sqrt(1-6z+z^2) is the g.f. of the central Delannoy numbers (A001850).
Sum_{k=0..n} k*T(n,k) = A050151(n) (the partial sums of the central Delannoy numbers) = (1/2)*n*R(n), where R(n) = A006318(n) is the n-th large Schroeder number.
From Paul Barry, May 07 2009: (Start)
Riordan array ((1+x+sqrt(1-6x+x^2))/(2*sqrt(1-6x+x^2)), (1-x-sqrt(1-6x+x^2))/2).
Inverse of Riordan array ((1-2x-x^2)/(1-x^2), x(1-x)/(1+x)). (End)

			T(2,1)=4 because we have NED, NENE, NEEN and NDE.
Triangle starts:
    1;
    2,  1;
    8,  4,  1;
   38, 18,  6,  1;
  192, 88, 32,  8,  1;
From _Paul Barry_, May 07 2009: (Start)
Production matrix is
   2, 1,
   4, 2, 1,
   6, 2, 2, 1,
   8, 2, 2, 2, 1,
  10, 2, 2, 2, 2, 1,
  12, 2, 2, 2, 2, 2, 1,
  14, 2, 2, 2, 2, 2, 2, 1,
  16, 2, 2, 2, 2, 2, 2, 2, 1,
  18, 2, 2, 2, 2, 2, 2, 2, 2, 1 (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.
W.-j. Woan, The Lagrange Inversion Formula and Divisibility Properties, JIS 10 (2007) 07.7.8, example 5.

Cf. A001850, A002003, A006318, A050146, A050151, A113139, A118384.

Q:=sqrt(1-6*z+z^2): G:=(1+z+Q)/Q/(2-t+t*z+t*Q): Gser:=simplify(series(G,z=0,13)): P[0]:=1: for n from 1 to 10 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 10 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular form

T[n_, n_] = 1;
T[n_, k_] := Sum[Binomial[n, i] Binomial[2n-k-i-1, n-k-i], {i, 0, n}];
Table[T[n, k], {n, 0, 9}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jun 13 2019 *)

A110171 = lambda n,k : binomial(n, k)*hypergeometric([k-n, n], [k+1], -1)
for n in (0..9): [round(A110171(n,k).n(100)) for k in (0..n)] # Peter Luschny, Sep 17 2014

T(n,0) = A002003(n) for n >= 1.
T(n,1) = A050146(n) for n >= 1.
Row sums are the central Delannoy numbers (A001850).
G.f.: (1+z+Q)/(Q(2-t+tz+tQ)), where Q=sqrt(1-6z+z^2).
T(n,k) = x^(n-k)*((1+x)/(1-x))^n. - Paul Barry, May 07 2009
T(n,k) = C(n, k)*hypergeometric([k-n, n], [k+1], -1). - Peter Luschny, Sep 17 2014
From Peter Bala, Jun 29 2015: (Start)
T(n,k) = Sum_{i = 0..n} binomial(n,i)*binomial(2*n-k-i-1,n-k-i).
Matrix product A118384 * A007318^(-1)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - x - sqrt(1 - 6*x + x^2) )/2 and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan). (End)
T(n,k) = P(n-k, k, -1, 3), where P(n, alpha, beta, x) is the n-th Jacobi polynomial with parameters alpha and beta. Cf. A113139. - Peter Bala, Feb 16 2020

A114655 Triangle read by rows: T(n,k) is the number of Schroeder paths of length 2n and having k weak ascents (1<=k<=n). A Schroeder path of length 2n is a lattice path from (0,0) to (2n,0) consisting of U=(1,1), D=(1,-1) and H=(2,0) steps and never going below the x-axis. A weak ascent in a Schroeder path is a maximal sequence of consecutive U and H steps.

Original entry on oeis.org

2, 4, 2, 8, 12, 2, 16, 48, 24, 2, 32, 160, 160, 40, 2, 64, 480, 800, 400, 60, 2, 128, 1344, 3360, 2800, 840, 84, 2, 256, 3584, 12544, 15680, 7840, 1568, 112, 2, 512, 9216, 43008, 75264, 56448, 18816, 2688, 144, 2, 1024, 23040, 138240, 322560, 338688, 169344

Offset: 1

Emeric Deutsch, Dec 23 2005

Row sums are the large Schroeder numbers (A006318). Sum(k*T(n,k),k=1..n)=A002003(n). T(n,k)=2*A114656(n,k).

			T(3,3)=2 because we have (U)D(U)D(H) and (U)D(U)D(U)D, where U=(1,1), D=(1,-1) and H=(2,0) (the weak ascents are shown between parentheses).
Triangle starts:
2;
4,2;
8,12,2;
16,48,24,2;
32,160,160,40,2.

Cf. A006318, A002003, A114656.

T:=(n,k)->2^(n-k+1)*binomial(n,k)*binomial(n,k-1)/n: for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form

Flatten[Table[2^(n-k+1) Binomial[n,k] Binomial[n,k-1]/n,{n,10}, {k,n}]] (* Harvey P. Dale, Oct 01 2011 *)

T(n, k)=2^(n-k+1)*binomial(n, k)*binomial(n, k-1)/n (1<=k<=n). G.f. G=G(t, z) satisfies G=z(2+G)(t+G).

Keyword tabf changed to tabl by Michel Marcus, Apr 09 2013

A333579 a(n) = [x^n] ( (1 + x + x^2)/(1 - x + x^2) )^n.

Original entry on oeis.org

1, 2, 8, 32, 128, 502, 1904, 6862, 22784, 64832, 120008, -223606, -4311424, -33271366, -205802344, -1142307968, -5919738880, -29159028386, -137718099760, -626077804826, -2740865583872, -11523690799904, -46214332516520, -174358991625134, -601230820510720

Offset: 0

Peter Bala, Mar 29 2020

If F(x) = 1 + f(1)*x + f(2)*x^2 + ... is a formal power series with integer coefficients then the sequence u(n) := [x^n] F(x)^n is an integer sequence satisfying the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. In particular u(p^k) == u(p^(k-1)) ( mod p^k ) for all prime p and positive integer k. For certain power series F(x) we may get stronger congruences.
According to Zhi-Wei Sun (see his comment in A002426 posted Nov 30, 2016), the central trinomial coefficients A002426(n) = [x^n] (1 + x + x^2)^n, satisfy the congruences A002426(p) == 1 ( mod p^2 ) for prime p >= 5. More generally, calculation suggests that the congruences A002426(p^k) == A002426(p^(k-1)) ( mod p^(2*k) ) hold for prime p >= 5 and any positive integer k.
We conjecture that the present sequence satisfies the congruences a(p) == 2 ( mod p^3 ) for prime p >= 5 (checked up to p = 499). Calculation suggests that a(p^k) == a(p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and k > 1.
More generally, if c(m,x) denotes the m-th cyclotomic polynomial then the sequences a(m,n) := [x^n] c(m,x)^n and b(m,n) := [x^n] ( c(m,x)/c(m,-x) )^n may satisfy the congruences a(m,p) == a(m,1) ( mod p^2 ) and b(m,p) == b(m,1) ( mod p^3 ), both for prime p >= 5, p not a divisor of m. The present sequence is b(3,n). Note that b(2,n) = A002003(n).

			Examples of congruences a(p) - a(1) == 0 ( mod p^3 ):
a(11) - a(1) = -223606 - 2 = -(2^3)*3*7*11^3 == 0 ( mod 11^3 )
a(19) - a(1) = -626077804826 - 2 = -(2^2)*7*(19^3)*151*21589 == 0 ( mod 19^3 )

Wikipedia, Cyclotomic polynomial

Cf. A002003, A002426, A187925, A318114.

seq(add(add(add((-1)^(n-k-i-j)*binomial(n, k)*binomial(k, i)*binomial(n+j-1, j)*binomial(j, n-k-i-j), j = 0..n-k-i), i = 0..n-k), k = 0..n), n = 0..25);
#alternative program
G := x -> (1 + x + x^2)/(1 - x + x^2):
H := (x,n) -> series(G(x)^n, x, n+1):
a:= n -> coeff(H(x, n), x, n):
seq(a(n), n = 0..25);

a[n_]:=SeriesCoefficient[((1 + x + x^2)/(1 - x + x^2))^n,{x,0,n}]; Array[a,25,0] (* Stefano Spezia, Apr 30 2024 *)

a(n) = polcoeff(((1 + x + x^2)/(1 - x + x^2))^n+ O(x^(n+1)), n, x); \\ Michel Marcus, Mar 31 2020

a(n) = Sum_{0 <= i,j,k <= n} (-1)^(n-k-i-j)*C(n,k)*C(k,i)*C(n+j-1,j)*C(j,n-k-i-j).

A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

Original entry on oeis.org

1, 2, 8, 44, 256, 1502, 8912, 53510, 324352, 1980332, 12160008, 75015162, 464566144, 2886488906, 17985045464, 112333392044, 703119387648, 4409231140086, 27696141476336, 174229516043630, 1097501783152256, 6921721148337452, 43701895245221848

Offset: 0

Peter Bala, Apr 18 2023

Compare with A228960(n) = [x^n] F(x)^n.
Let k and m be positive integers and let f(x) be a finite product of cyclotomic polynomials. Define b(n) = [x^(k*n)] (f(x)/f(-x))^(m*n). Then we conjecture that the supercongruences a(p) == a(1) (mod p^3) and, for n >= 2, a(n*p) == a(n) (mod p^2) hold for all primes p, with a finite number of exceptions depending on f(x).
The present sequence is the case k = m = 1 and f(x) = (1 + x)*(1 + x^3) = C(2,x)^2 * C(6,x), where C(n,x) denotes the n-th cyclotomic polynomial. See A002003 for the case k = m = 1 and f(x) = (1 + x).

Wikipedia, Cyclotomic polynomial

Cf. A002003, A228960, A333579.

F(x) := (1 + x)*(1 + x^3): G(x) := taylor(F(x)/F(-x),x = 0, 50); seq(coeftayl(G(x)^n, x = 0, n), n = 0..50);

Conjectures: 1) the supercongruence a(p) == 2 (mod p^3) holds for all primes p >= 5 (checked up to p = 47).

2) for n >= 2, the supercongruence a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.

A050147 a(n) = T(n,n-1), array T as in A050143. Also T(2n+1,n), array T as in A055807.

Original entry on oeis.org

1, 3, 12, 56, 280, 1452, 7700, 41456, 225648, 1238420, 6840988, 37986984, 211842696, 1185635388, 6655993380, 37463920608, 211350457824, 1194706644516, 6765300359468, 38370431711000, 217931108199672

Offset: 1

Clark Kimberling

nonn

Cf. A002003, A006318, A050143, A055807.

a[n_]:=Binomial[2*n-3,n-1]*Hypergeometric2F1[-n+1,-n,-2*n+3,-1];
Table[a[n],{n,1,21}] (* Detlef Meya, Dec 04 2023 *)

a(n):=if n=1 then 1 else sum((binomial(n,i+1))*binomial(n+i-2,n-2),i,0,n-2)+binomial(2*n-3,n-2); /* Vladimir Kruchinin, Nov 25 2014 */

From Vladimir Kruchinin, Nov 25 2014: (Start)
G.f.: x*((-x^2 + 4*x + 1)/(2*sqrt(x^2 - 6*x + 1)) -x/2 + 1/2).
For n >= 2, a(n) = C(2*n-3,n-2) + Sum_{i=0..n-2} C(n,i+1)*C(n+i-2,n-2). (End)
a(n) ~ (1 + sqrt(2))^(2*n-2) / (2^(1/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Feb 14 2021
a(n) = binomial(2*n-3, n-1)*hypergeom([-n+1, -n], [-2*n+3], -1). - Detlef Meya, Dec 04 2023
D-finite with recurrence (-n+1)*a(n) +(2*n+3)*a(n-1) +3*(8*n-29)*a(n-2) +(-10*n+49)*a(n-3) +(n-6)*a(n-4)=0. - R. J. Mathar, Jul 31 2025

A361743 Central circular Delannoy numbers: a(n) is the number of Delannoy loops on an n X n toroidal grid.

Original entry on oeis.org

1, 2, 16, 114, 768, 5010, 32016, 201698, 1257472, 7777314, 47800080, 292292946, 1779856128, 10799942322, 65336473104, 394246725570, 2373580947456, 14262064668738, 85546366040592, 512323096241714, 3063932437123840, 18300660294266322, 109183694129335056

Offset: 0

Noah Snyder, Mar 22 2023

nonn

An (n,m) Delannoy loop is an oriented unbased loop on a toroidal grid with points labeled by Z/n x Z/m composed of steps of the form (1,0), (0,1), and (1,1), and which loops around the torus exactly once in each of the x-direction and the y-direction. The central circular Delannoy numbers count the number of (n,n) Delannoy loops. This is a modification of the ordinary central Delannoy numbers A001850.

Dimensions of endomorphism algebras End(S^{{n}}) in the circular Delannoy category attached to the oligomorphic group of order-preserving self-bijections of the circle.

			When n=2 see Figure 3 of "The circular Delannoy Category".

Winston de Greef, Table of n, a(n) for n = 0..1296
Nate Harman, Andrew Snowden, and Noah Snyder, The circular Delannoy Category, arxiv: 2303.10814 [math.RT], 2023.

Circular analog of A001850.
Main diagonal of A361745.
Cf. A008288, A108666, A002003, A047781.

a[n_Integer?Positive] := Sum[k Binomial[n, k] Binomial[n, k] 2^k, {k, 1, n}]

a(n) = if(n == 0, 1, sum(k=0, n, binomial(n, k)^2*k*2^k)) \\ Winston de Greef, Mar 22 2023

from math import comb
def A361743(n): return sum(comb(n,k)**2*k<Chai Wah Wu, Mar 22 2023

a(n) = Sum_{k=0..n} binomial(n,k)^2*k*2^k for n >= 1.
a(n) = n*(D(n,n-1) + D(n-1,n-1)) = n*(D(n,n) - D(n-1,n)) for n >= 1, where D(i,j) = A008288(i,j) are the Delannoy numbers.
a(n) = 2*A108666(n) for n >= 1.
From Alois P. Heinz, Mar 22 2023: (Start)
G.f.: 1 + 2*(1-x)*x/sqrt(x^2-6*x+1)^3.
a(n) = n*A002003(n) for n >= 1.
a(n) = 2*n*A047781(n) for n >= 1. (End)
a(n) = 2*n^2*hypergeom([1 - n, 1 - n], [2], 1) for n >= 1. - Peter Luschny, Mar 22 2023

A363418 Square array read by ascending antidiagonals: T(n,k) = [x^(n*k)] ((1 + x)/(1 - x))^k for n, k >= 1.

Original entry on oeis.org

2, 2, 8, 2, 16, 38, 2, 24, 146, 192, 2, 32, 326, 1408, 1002, 2, 40, 578, 4672, 14002, 5336, 2, 48, 902, 11008, 69002, 142000, 28814, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146

Offset: 1

Peter Bala, Jun 12 2023

The n-th row sequence {T(n, k) : k >= 1} satisfies the Gauss congruences, that is, T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^r ) for all primes p and positive integers m and r.

We conjecture that each row sequence satisfies the stronger supercongruences T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5 and positive integers m and r.

			Square array begins
 n\k |  1   2     3      4        5          6           7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  2   8    38    192     1002       5336       28814   ...   (A002003)
  2  |  2  16   146   1408    14002     142000     1459810   ...   (A103885)
  3  |  2  24   326   4672    69002    1038984    15856206   ...   (A333715)
  4  |  2  32   578  11008   216002    4320608    87588482   ...
  5  |  2  40   902  21440   525002   13104184   331482062   ...
  6  |  2  48  1298  36992  1086002   32497680   985524066   ...
  7  |  2  56  1766  58688  2009002   70097384  2478629134   ...
  8  |  2  64  2306  87552  3424002  136485568  5513464322   ...

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

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 antidiagonals)

A002003 (row 1), A103885 (row 2), A333715 (row 3). Cf. A035607, A362724 - A362733, A363419.

# display as a square array
T := (n,k) -> add( binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j) , j = 0..k): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
#alternative program
seq(print(seq(simplify(2*k*hypergeom([1 - k, 1 - n*k], [2], 2)), k = 1..10)), n = 1..10);
# display as a sequence
seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);

T(n,k) = sum(j=0, k, binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j)) \\ Andrew Howroyd, Jan 05 2024

T(n,k) = Sum_{j = 0..k} binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j).
T(n,k) = 1/n * [x^k] ((1 + x)/(1 - x))*(n*k).
T(n,k) = (1/n)*Sum_{j = 0..k} binomial(n*k, j)*binomial((n + 1)*k - j - 1, k - j).
T(2*n,k) = [x^(n*k)] Chebyshev_T(k,(1 + x)/(1 - x)), where Chebyshev_T(n,x) denotes the n-th Chebyshev polynomial of the first kind. See A053120.
T(n,k) = Sum_{j = 1..k} (2^j)*binomial(k, j)*binomial(n*k - 1, n*k - j).
T(n,k) = (2*k) * hypergeom([1 - k, 1 - n*k], [2], 2).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)/(1 + x))^n.
E(m,x) appears to be the g.f. of the (m + 1)-Schroeder numbers. See A027307 (m = 2) and the cross references there.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = (1 + x)/(1 - x), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

cp's OEIS Frontend

A333715 a(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n.

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A156894 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k-1,k).

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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A110171 Triangle read by rows: T(n,k) (0 <= k <= n) is the number of Delannoy paths of length n that start with exactly k (0,1) steps (or, equivalently, with exactly k (1,0) steps).

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A333579 a(n) = [x^n] ( (1 + x + x^2)/(1 - x + x^2) )^n.

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A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

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A156894 a(n) = Sum_{k=0..n} binomial(n,k)binomial(2n+k-1,k).