A208675 -id:A208675 - cp's OEIS frontend

A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1

Offset: 0

Paul D. Hanna, Jun 12 2005

Compare to the corresponding array A108553 of crystal ball sequences for D_n lattice.
From Peter Bala, Jul 18 2008: (Start)
Row reverse of A099608.
This array has a remarkable relationship with the constant zeta(2). The row, column and diagonal entries of the array occur in series acceleration formulas for zeta(2).
For the entries in row n we have zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k)). For example, n = 4 gives zeta(2) = 2*(1 - 1/4 + 1/9 - 1/16) + 1/(1*21) + 1/(4*21*131) + 1/(9*131*471) + ... . See A142995 for further details.
For the entries in column k we have zeta(2) = (1 + 1/4 + 1/9 + ... + 1/k^2) + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k)). For example, k = 4 gives zeta(2) = (1 + 1/4 + 1/9 + 1/16) + 2*(1/(1*9) - 1/(4*9*61) + 1/(9*61*309) - ... ). See A142999 for further details.
Also, as consequence of Apery's proof of the irrationality of zeta(2), we have a series acceleration formula along the main diagonal of the table: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n,n)*T(n-1,n-1)) = 5*(1/3 - 1/(2^2*3*19) + 1/(3^2*19*147) - ...).
There also appear to be series acceleration results along other diagonals. For example, for the main subdiagonal, calculation supports the result zeta(2) = 2 - Sum_{n >= 1} (-1)^(n+1)*(n^2+(2*n+1)^2)/(n^2*(n+1)^2*T(n,n-1)*T(n+1,n)) = 2 - 10/(2^2*7) + 29/(6^2*7*55) - 58/(12^2*55*471) + ..., while for the main superdiagonal we appear to have zeta(2) = 1 + Sum_{n >= 1} (-1)^(n+1)*((n+1)^2 + (2*n+1)^2)/(n^2*(n+1)^2*T(n-1,n)*T(n,n+1)) = 1 + 13/(2^2*5) - 34/(6^2*5*37) + 65/(12^2*37*309) - ... .
Similar series acceleration results hold for Apery's constant zeta(3) involving the crystal ball sequences for the product lattices A_n x A_n; see A143007 for further details. Similar results also hold between the constant log(2) and the crystal ball sequences of the hypercubic lattices A_1 x...x A_1 and between log(2) and the crystal ball sequences for lattices of type C_n ; see A008288 and A142992 respectively for further details. (End)
This array is the Hilbert transform of triangle A008459 (see A145905 for the definition of the Hilbert transform). - Peter Bala, Oct 28 2008

			Square array begins:
  1,   1,    1,     1,      1,       1,       1, ... A000012;
  1,   3,    5,     7,      9,      11,      13, ... A005408;
  1,   7,   19,    37,     61,      91,     127, ... A003215;
  1,  13,   55,   147,    309,     561,     923, ... A005902;
  1,  21,  131,   471,   1251,    2751,    5321, ... A008384;
  1,  31,  271,  1281,   4251,   11253,   25493, ... A008386;
  1,  43,  505,  3067,  12559,   39733,  104959, ... A008388;
  1,  57,  869,  6637,  33111,  124223,  380731, ... A008390;
  1,  73, 1405, 13237,  79459,  350683, 1240399, ... A008392;
  1,  91, 2161, 24691, 176251,  907753, 3685123, ... A008394;
  1, 111, 3191, 43561, 365751, 2181257, ...      ... A008396;
  ...
As a triangle:
  [0]  1
  [1]  1,  1
  [2]  1,  3,   1
  [3]  1,  7,   5,    1
  [4]  1, 13,  19,    7,    1
  [5]  1, 21,  55,   37,    9,    1
  [6]  1, 31, 131,  147,   61,   11,   1
  [7]  1, 43, 271,  471,  309,   91,  13,   1
  [8]  1, 57, 505, 1281, 1251,  561, 127,  15,  1
  [9]  1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
       ...
Inverse binomial transform of rows yield rows of triangle A063007:
  1;
  1,  2;
  1,  6,   6;
  1, 12,  30,  20;
  1, 20,  90, 140,  70;
  1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
  1;
  1,  1;
  1,  4,   1;
  1,  9,   9,   1;
  1, 16,  36,  16,  1;
  1, 25, 100, 100, 25, 1;
  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
R. Bacher, P. de la Harpe, and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Joseph T. Iosue, T. C. Mooney, Adam Ehrenberg, and Alexey V. Gorshkov, Projective toric designs, difference sets, and quantum state designs, arXiv:2311.13479 [quant-ph], 2023. See page 6.
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.
A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report. Math. Intelligencer 1 (1978/79), no 4, 195-203.
Eric Weisstein's World of Mathematics, Apéry number.

Rows include: A003215 (row 2), A005902 (row 3), A008384 (row 4), A008386 (row 5), A008388 (row 6), A008390 (row 7), A008392 (row 8), A008394 (row 9), A008396 (row 10).
Cf. A063007, A099601 (n-th term of A_{2n} lattice), A108553.
Cf. A008459 (h-vectors type B associahedra), A145904, A145905.
Cf. A005258 (main diagonal), A108626 (antidiagonal sums).
Cf. A108628, A208675, A363867, A363868, A363869, A363870, A363871.

T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array
A108625:= func< n,k | T(n-k,k) >; // antidiagonals
[A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023

T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1):
seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018

T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *)

T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))

def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array
def A108625(n,k): return T(n-k, k) # antidiagonals
flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
The square array = A063007 * transpose(A007318).
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A108628 n-th term of the crystal ball sequence for A_{n+1} lattice for n >= 0.

Original entry on oeis.org

1, 7, 55, 471, 4251, 39733, 380731, 3716695, 36808723, 368750757, 3728940249, 38003358693, 389866749975, 4022124746409, 41697566691555, 434124925278807, 4536783726146499, 47569453938399445, 500266519237489357, 5275183203229043221, 55760274296452936741

Offset: 0

Paul D. Hanna, Jun 14 2005

Equals the secondary diagonal of square array A108625, in which row n equals the crystal ball sequence for A_n lattice. Main diagonal of square array A108625 equals the Apery numbers (A005258).

G. C. Greubel, Table of n, a(n) for n = 0..950
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Cf. A005258, A103882, A108625, A143007, A208675.

A108628:= func< n | (&+[Binomial(n+1,k)^2*Binomial(n+k,k-1): k in [0..n+1]]) >;
[A108628(n): n in [0..30]]; // G. C. Greubel, Oct 06 2023

seq(add(binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k), k = 0..n), n = 0..20); # Peter Bala, Apr 14 2022

Table[Sum[Binomial[n+1,k]^2 Binomial[n+k,k-1],{k,0,n+1}],{n,0,20}] (* Harvey P. Dale, Apr 01 2013 *)

a(n)=sum(k=0,n+1,binomial(n+1,k)^2*binomial(n+k,k-1))

def A108628(n):
    m, g = 1, 0
    for k in range(n+1):
        g += m
        m *= (n+k+2)*(n-k)*(n-k+1)
        m //= (k+1)**3
    return g # Chai Wah Wu, Oct 03 2022

def A108628(n): return sum(binomial(n+1,k)^2*binomial(n+k,k-1) for k in range(n+2))
[A108628(n) for n in range(31)] # G. C. Greubel, Oct 06 2023

a(n) = Sum_{k = 0..n+1} C(n+1, k)^2 * C(n+k, k-1).
a(n) = A108625(n+1, n).
exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + 144*x^4 + 1048*x^5 + 8189*x^6 + 67325*x^7 + 574999*x^8 + ... appears to have integer coefficients. Cf. A208675. - Peter Bala, Jan 12 2016
Recurrence: (n+1)^2*(5*n^2 - 6*n + 2)*a(n) = (55*n^4 - 11*n^3 - 26*n^2 + 5*n + 5)*a(n-1) + (n-1)^2*(5*n^2 + 4*n + 1)*a(n-2). - Vaclav Kotesovec, Jan 13 2016
a(n) ~ sqrt(89/8 + 199/(8*sqrt(5))) * ((1+sqrt(5))/2)^(5*n) / (Pi*n). - Vaclav Kotesovec, Jan 13 2016
Equivalently, a(n) ~ phi^(5*n + 11/2) / (2*5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021
From Peter Bala, Mar 24 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+1,k)*binomial(n+k+1,k).
Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k) for nonnegative k, we have a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+1,k)*binomial(-n+k+1,k) = Sum_{k} (-1)^k* binomial(n+k-1,k)*binomial(n+k-2,k)*binomial(n-2,k) = (-1)^n*A208675(n-1) for n >= 2.
a(n-1) = (1/2)*Sum_{k = 0..floor(n/2)} binomial(n,k)^2 * binomial(3*n-2*k-1,n-2*k) for n >= 1. Cf. A103882.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,k)^2.
Equivalently, a(n) = [(x*z)^(n+1)*y^n] ( (x + y + z)^(n+1) * (x + y)^n * (y + z)^(n+1) ).
a(n) = (1/5)*( 2*A005258(n+1) - A005258(n) ).
a(n) = hypergeometric3F2([n + 2, -n, -n - 1], [1, 1], 1).
a(n) = (-1)^n*(n+1)*hypergeom([n + 2, n + 2, -n ], [1, 2], 1).
a(n) = [x^n] 1/(1 - x)*P(n+1,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A208675(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1 and A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).
a(n) = B(n+1,n,n+1) in the notation of Straub, equation 24. Hence
a(n) = [(x*z)^(n+1)*y^n] 1/(1 - x - y - z + x*z + y*z - x*y*z).
The following takes the sequence offset to be 1: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.
Conjectures: a((p-1)/2) == 0 (mod p) and a((p^3-1)/2) == 0 (mod p^3) for primes p of the form 4*m + 1; a((p^3-1)/2) == 0 (mod p^2) for primes p of the form 4*m + 3; a((p^2-1)/2) == 0 (mod p^2) for primes p >= 5. (End)
From Peter Bala, Sep 11 2024: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(n+k+1)*binomial(n+1, k)*binomial(n+k, k)* binomial(n+k+1, k).
a(n) = (-1)^(n+1) * hypergeom([n+1, n+2, -n-1], [1, 1], 1). (End)
a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n-1, k). Cf. A005258(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Sep 25 2024

A352655 a(n) = (1/2)*(A005258(n) + A005258(n-1)).

Original entry on oeis.org

2, 11, 83, 699, 6252, 58106, 554633, 5399099, 53356322, 533627511, 5388927513, 54859837434, 562267554552, 5796123147756, 60047675871333, 624801952898619, 6526036790730942, 68395815476047901, 718992874207884953, 7578808590187108199

Offset: 1

Peter Bala, Apr 17 2022

The Apéry numbers A005258 satisfy the supercongruences A005258(p) == 3 (mod p^3) and A005258(p-1) == 1 (mod p^3) for primes p >= 5. It easily follows that a(p) == 2 (mod p^3) for primes p >= 3. We conjecture that the stronger supercongruences a(p) == 2 (mod p^5) hold for primes p >= 5. See A212334 for the corresponding conjecture for the Apéry numbers A005259.

Conjecture: for r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ). - Peter Bala, Oct 13 2022

			Examples of superconguences:
a(5) - 2 = 6252 - 2 = 2*(5^5) == 0 (mod 5^5).
a(7) - 2 = 554633 - 2 = 3*(7^5)*11 == 0 (mod 7^5).
a(11) - 2 = 5388927513 - 2 = (11^5)*33461 == 0 (mod 11^5).

Cf. A005258, A103882, A108628, A208675, A212334, A352654.

seq((1/2)*add((2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n, k)^2 * binomial(n + k, k), k = 0..n), n = 1..20);

f(n) = sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)); \\ A005258
a(n) = (f(n) + f(n-1))/2; \\ Michel Marcus, Apr 20 2022

a(n) = (1/2)*Sum_{k = 0..n} (2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n,k)^2 * binomial(n + k,k).
a(n) = (1/2)*Sum_{k = 0..n-1} (4*n + k)*(n - k)/(n*(n + k)) * binomial(n,k)^2* binomial(n + k,k) for n >= 1.
a(n) = (1/2)*(A108628(n-1) + 3*A208675(n)) for n >= 1.
a(n) = (1/2)*(2*A103882(n) - A352654(n)).
a(n) ~ 5^(3/4)*(13 + 5*sqrt(5))/(20*sqrt(22 + 10*sqrt(5))*Pi*n) * ((11 + 5*sqrt(5))/2)^n.
(11*n^2 - 31*n + 22)*n^2*a(n) = (121*n^4 - 462*n^3 + 607*n^2 - 322*n + 64)*a(n-1) + (11*n^2 - 9*n + 2)*(n - 2)^2*a(n-2) with a(1) = 2 and a(2) = 11.
The g.f. A(x) = 2*x + 11*x^2 + 83*x^3 + ... satisfies the differential equation
(x^5 + 13*x^4 + 22*x^3 + 9*x^2 - x)*A''(x) + (x^4 + 4*x^3 + 26*x^2 + 22*x - 1)*A'(x) + (2*x^2 - 16*x + 4)*A(x) + x^2 - 8*x + 2 = 0, with A(0) = 2 and A'(0) = 11.

A208673 Number of words A(n,k), either empty or beginning with the first letter of the k-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 5, 10, 1, 1, 1, 1, 9, 37, 35, 1, 1, 1, 1, 15, 163, 309, 126, 1, 1, 1, 1, 25, 640, 3593, 2751, 462, 1, 1, 1, 1, 41, 2503, 36095, 87501, 25493, 1716, 1, 1, 1, 1, 67, 9559, 362617, 2336376, 2266155, 242845, 6435, 1, 1

Offset: 0

Alois P. Heinz, Feb 29 2012

Also the number of (n*k-1)-step walks on k-dimensional cubic lattice from (1,0,...,0) to (n,n,...,n) with positive unit steps in all dimensions such that the absolute difference of the dimension indices used in consecutive steps is <= 1.

All rows are linear recurrences with constant coefficients and for n > 0 the order of the recurrence is bounded by 2*n-1. For n up to at least 20 this upper bound is exact. - Andrew Howroyd, Feb 22 2022

			A(0,0) = A(n,0) = A(0,k) = 1: the empty word.
A(2,3) = 5:
  +------+   +------+   +------+   +------+   +------+
  |aabbcc|   |aabcbc|   |aabccb|   |ababcc|   |abccba|
  +------+   +------+   +------+   +------+   +------+
  |122222|   |122222|   |122222|   |112222|   |111112|
  |001222|   |001122|   |001112|   |011222|   |011122|
  |000012|   |000112|   |000122|   |000012|   |001222|
  +------+   +------+   +------+   +------+   +------+
  |xx    |   |xx    |   |xx    |   |x x   |   |x    x|
  |  xx  |   |  x x |   |  x  x|   | x x  |   | x  x |
  |    xx|   |   x x|   |   xx |   |    xx|   |  xx  |
  +------+   +------+   +------+   +------+   +------+
Square array A(n,k) begins:
  1,  1,    1,     1,       1,         1,           1, ..
  1,  1,    1,     1,       1,         1,           1, ..
  1,  1,    3,     5,       9,        15,          25, ..
  1,  1,   10,    37,     163,       640,        2503, ..
  1,  1,   35,   309,    3593,     36095,      362617, ..
  1,  1,  126,  2751,   87501,   2336376,    62748001, ..
  1,  1,  462, 25493, 2266155, 164478048, 12085125703, ..

Andrew Howroyd, Table of n, a(n) for n = 0..1325

Columns k=0+1, 2-4 give: A000012, A088218, A208675, A212334.
Rows n=0+1, 2-3 give: A000012, A001595, A208674.
Main diagonal gives A351759.
Cf. A208879 (cyclic alphabet), A331562.

b:= proc(t, l) option remember; local n; n:= nops(l);
     `if`(n<2 or {0}={l[]}, 1,
     `if`(l[t]>0, b(t, [seq(l[i]-`if`(i=t, 1, 0), i=1..n)]), 0)+
     `if`(t0,
                  b(t+1, [seq(l[i]-`if`(i=t+1, 1, 0), i=1..n)]), 0)+
     `if`(t>1 and l[t-1]>0,
                  b(t-1, [seq(l[i]-`if`(i=t-1, 1, 0), i=1..n)]), 0))
    end:
A:= (n, k)-> `if`(n=0 or k=0, 1, b(1, [n-1, n$(k-1)])):
seq(seq(A(n, d-n), n=0..d), d=0..10);

b[t_, l_List] := b[t, l] = Module[{n = Length[l]}, If[n < 2 || {0} == Union[l], 1, If[l[[t]] > 0, b[t, Table[l[[i]] - If[i == t, 1, 0], {i, 1, n}]], 0] + If[t < n && l[[t + 1]] > 0, b[t + 1, Table[l[[i]] - If[i == t + 1, 1, 0], {i, 1, n}]], 0] + If[t > 1 && l[[t - 1]] > 0, b[t - 1, Table[l[[i]] - If[i == t - 1, 1, 0], {i, 1, n}]], 0]]]; A[n_, k_] := If[n == 0 || k == 0, 1, b[1, Join[{n - 1}, Array[n&, k - 1]]]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

F(u)={my(n=#u); sum(k=1, n,u[k]*binomial(n-1,k-1))}
step(u, c)={my(n=#u); vector(n, k, sum(i=max(0, 2*k-c-n), k-1, sum(j=0, n-2*k+i+c, u[k-i+j]*binomial(n-1, 2*k-1-c-i+j)*binomial(k-1, k-i-1)*binomial(k-i+j-c, j) ))) }
R(n,k)={my(r=vector(n+1), u=vector(k), v=vector(k)); u[1]=v[1]=r[1]=r[2]=1; for(n=3, #r, u=step(u,1); v=step(v,0)+u; r[n]=F(v)); r}
T(n,k)={if(n==0||k==0, 1, R(k,n)[1+k])} \\ Andrew Howroyd, Feb 22 2022

A352654 a(0) = 0, a(n) = [x^n] P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.

Original entry on oeis.org

0, 0, 2, 18, 162, 1500, 14240, 137886, 1356194, 13507416, 135916002, 1379301990, 14097919968, 144977296932, 1498750896708, 15564971674518, 162298598439330, 1698353774374704, 17828728267167326, 187693442844729174, 1981038544180652162, 20957881615473229620

Offset: 0

Peter Bala, Apr 14 2022

Compare with A103882(n) = [x^n] P(n,(1 + x)/(1 - x)).

Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k), valid for nonnegative k, we can extend the binomial sum a(n) = Sum_{k} binomial(n-1,k)*binomial(n-1,k-1)*binomial(n+k-1,k) to negative values of n to find a(-n) = Sum_{k} binomial(-n-1,k)*binomial(-n-1,k-1)* binomial(-n+k-1,k) = Sum_{k} (-1)^(k-1)*binomial(n+k,k)*binomial(n+k-1,k-1)*binomial(n,k) = (-1)^(n+1)*A103882(n) (n != 0).

Cf. A005258, A103882, A108628, A208675.

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

a(n) = polcoef(pollegendre(n-1, (1 + x)/(1 - x)) + O(x^(n+1)), n); \\ Michel Marcus, Apr 17 2022

def A352654(n):
    if n == 0: return 0
    m, g = 1, 0
    for k in range(n+1):
        g += -m*n//(n+k) if (n+k)&1 else m*n//(n+k)
        m *= (n+k+1)*(n-k)*(n+k)
        m //= (k+1)**3
    return g # Chai Wah Wu, Oct 03 2022

def a(n): return n * (n-1) * hypergeometric([2-n, 1-n, 1+n], [2, 2], 1)
print([simplify(a(n)) for n in range(22)]) # Peter Luschny, Jan 03 2024

a(n) = Sum_{k = 1..n-1} binomial(n-1,k)*binomial(n-1,k-1)*binomial(n+k-1,k).
a(n) = Sum_{k = 1..n-1} (-1)^(n+k-1)*binomial(n-1,k)*binomial(n+k-1,k)* binomial(n+k-1,k-1).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(n+k-1,k)^2 for n >= 1.
a(n) = (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1. It follows that the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for primes p >= 5 and positive integers n and k.
a(n) = A108628(n-1) - A208675(n) for n >= 1.
a(n) ~ (1/10)*5^(3/4)*(sqrt(5) + 1)/(Pi*sqrt(22 + 10*sqrt(5))*n)*((1/2)*(11 + 5*sqrt(5)))^n.
n^2*(5*n - 7)*(n - 2)*a(n) = (n - 1)*(55*n^3 - 187*n^2 + 190*n - 48)*a(n-1) + (n - 2)^2*(5*n -2 )*(n - 1)*a(n-2) with a(1) = 0 and a(2) = 2.
Conjecture: a(n) = [(x*y)^(n-1)*z^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) = [(x*z)^(n-1)*y^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) for n >= 1.
a(n) = n*(n-1)*hypergeom([2-n, 1-n, 1+n], [2, 2], 1). - Peter Luschny, Jan 03 2024

A271777 a(n) = Sum_{k=1..n} ((-1)^(n-k) * k / ((n+1)^2 + (k-1)(n+1))) binomial(n+1, k+1) * binomial(n+k, k)^2.

Original entry on oeis.org

0, 1, 3, 16, 105, 771, 6083, 50464, 434493, 3849715, 34895685, 322239204, 3021922137, 28710585099, 275827551795, 2675584398912, 26173225402437, 257940602058051, 2558852771578817, 25534696636443160, 256164209036422239, 2582189471692118101, 26142642319644094293

Offset: 0

Vladimir Kruchinin, Apr 14 2016

a(n) is the number of Motzkin paths (A001006) with n upsteps U = (1,1), n flatsteps F = (1,0), and n downsteps D = (1,-1) that begin with an F, contain no DUs and no FDs, and end with D. For example, a(2) = 3 counts FFUUDD, FUDFUD, FUFUDD. Proof. Such a path is obtained from a Dyck path (A000108) of semilength n by inserting zero or more F's before each U as follows. There are n "spaces" available for the F's, one immediately preceding each U. An F must be inserted before the initial U, and before each valley U (else a DU would be present). If there are k peaks, hence k-1 valleys, this leaves n-k F's to be distributed arbitrarily among the n "spaces" -- binomial(2*n-1-k,n-k) choices. There are Narayana(n,k) (A001263) Dyck n-paths with k peaks, hence the total number of Motzkin paths meeting the specifications is Sum_{k=1..n} binomial(2*n-1-k, n-k)*Narayana(n,k). Peter Bala has shown that Maple's sumtools:-sumrecursion command produces the same second-order recurrence for this sum and for the titular binomial sum. QED - David Callan, Feb 15 2022

Cf. A000108, A000984, A001622, A208675.

a := proc(n) option remember; if n < 3 then return [0, 1, 3][n+1] fi;
(3*(-n^3 + 8*n^2 - 19*n + 12)*a(n-3) + (-32*n^3 + 194*n^2 - 386*n + 252)* a(n-2) + (2*n - 2)*(7*n^2 - 10*n + 1)*a(n-1))/(n^3 - n) end:
seq(a(n), n = 0..20); # Peter Luschny, Feb 15 2022

a[n_] := Sum[(-1)^(n-k)*k / ((n+1)^2 + (k-1)*(n+1))*Binomial[n+1, k+1] * Binomial[n+k, k]^2, {k, 1, n}];
Table[a[n], {n, 0, 20}] (* Vaclav Kotesovec, Apr 14 2016 *)
h[n_, k_] := HypergeometricPFQ[{n, -n, n + 1}, {1, k}, 1];
A271777[n_] := If[n == 0, 0, (-1)^n (h[n, 1] - h[n, 2]) / n];
Table[A271777[n], {n, 0, 22}] (* Peter Luschny, Feb 15 2022 *)
A271777[n_] := (-1)^(n + 1) Binomial[n + 1, 2] HypergeometricPFQ[{1 - n, n + 1, n + 2}, {2, 3}, 1]; Table[A271777[n], {n, 0, 22}] (* Peter Luschny, Feb 18 2022 *)

a(n) := sum((-1)^(n-k)*k/((n+1)^2+(k-1)*(n+1))*binomial(n+1,k+1)* binomial(n+k,k)^2,k,0,n);

a(n) = sum(k=0,n,((-1)^(n-k)*k/((n+1)^2+(k-1)*(n+1))*binomial(n+1, k+1)*binomial(n+k, k)^2)) \\ Felix Fröhlich, Jul 14 2016

a(n) ~ sqrt(17 - 38/sqrt(5)) * ((1+sqrt(5))/2)^(5*(n+1)) / (2*Pi*(n+1)^2). - Vaclav Kotesovec, Apr 14 2016. Equivalently, a(n) ~ n^(-2)*phi^(5*n + 1/2)/(2*Pi*5^(1/4)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021
Recurrence: a(n)*((n+1)*n*(n-1)*(5*n-6)) = ((n-2)*(n-3)*(5*n-1)*n*a(n-2) + (55*n^3 - 121*n^2 + 64*n - 12)*(n-1)*a(n-1)). - Vaclav Kotesovec, Apr 14 2016
From David Callan, Feb 15 2022: (Start)
a(n) = Sum_{k=1..n} binomial(2*n-1-k, n-k)*Narayana(n,k).
a(n) = Sum_{k=1..n} (-1)^(n-k)*binomial(n+k, 2*k)*binomial(n+k-1, k-1)* CatalanNumber(k) where CatalanNumber is A000108. (End)
a(n) = (-1)^(n+1)*binomial(n+1, 2)*hypergeom([1-n, n+1, n+2], [2, 3], 1). - Peter Luschny, Feb 18 2022

Offset set to 0 and then changed a(0) = 0 by Peter Luschny, Feb 18 2022

cp's OEIS Frontend

A362726 a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} A208675(k)*x^k/k ).

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A362727 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A208675(k)*x^k/k ).

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A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

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A108628 n-th term of the crystal ball sequence for A_{n+1} lattice for n >= 0.

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A352655 a(n) = (1/2)*(A005258(n) + A005258(n-1)).

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A208673 Number of words A(n,k), either empty or beginning with the first letter of the k-ary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A352654 a(0) = 0, a(n) = [x^n] P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.

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A271777 a(n) = Sum_{k=1..n} ((-1)^(n-k) * k / ((n+1)^2 + (k-1)(n+1))) binomial(n+1, k+1) * binomial(n+k, k)^2.