A108628 -id:A108628 - 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)

A103882 a(n) = Sum_{i=0..n} C(n+1,i)C(n-1,i-1)C(2n-i,n).

Original entry on oeis.org

1, 2, 12, 92, 780, 7002, 65226, 623576, 6077196, 60110030, 601585512, 6078578508, 61908797418, 634756203018, 6545498596110, 67830161708592, 705951252118284, 7375213677918294, 77310179609631564, 812839595630249540, 8569327862277434280, 90562666977432643862

Offset: 0

Ralf Stephan, Feb 20 2005

Number of permutations of n copies of 1..3 with all adjacent differences <= 1 in absolute value. - R. H. Hardin, May 06 2010 [Cf. A177316. - Peter Bala, Jan 14 2020]

Alois P. Heinz, Table of n, a(n) for n = 0..950 (terms n=1..94 from R. H. Hardin)
A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Cf. A005258, A156554, A177316, A352654.
Equals A103881(n, n).
Row n=3 of A331562.

[1] cat [&+[Binomial(n+1, i)*Binomial(n-1, i-1) * Binomial(2*n-i, n): i in [0..n]]:n in  [1..21]]; // Marius A. Burtea, Jan 19 2020

[&+[Binomial(n, k)^2*Binomial(n+k-1, k): k in [0..n]]:n in  [0..21]]; // Marius A. Burtea, Jan 19 2020

a:= proc(n) option remember; `if`(n<2, n+1,
      ((n-1)*(55*n^3-143*n^2+102*n-24)*a(n-1)+
      n*(5*n-3)*(n-2)^2*a(n-2))/((n-1)*(5*n-8)*n^2))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Jun 29 2015
# Alternative:
a := n -> hypergeom([-n, -n, n], [1, 1], 1):
seq(simplify(a(n)), n=0..21); # Peter Luschny, Jan 19 2020

Drop[Table[Sum[Sum[Multinomial[r, g, n + 1 - r - g] Binomial[n - 1,n - r] Binomial[n - 1, n - g], {g, 1, n}], {r, 1, n}], {n, 0, 18}], 1] (* Geoffrey Critzer, Jun 29 2015 *)
Table[Sum[Binomial[n+1,k]Binomial[n-1,k-1]Binomial[2n-k,n],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jun 19 2021 *)

a(n) = polcoef(pollegendre(n, (1 + x)/(1 - x)) + O(x^(n+1)), n); \\ Michel Marcus, Dec 20 2020

def A103882(n):
    if n == 0: return 1
    m, g = 1, 0
    for k in range(n+1):
        g += m*n//(n+k)
        m *= (n+k+1)*(n-k)**2
        m //= (k+1)**3
    return g # Chai Wah Wu, Oct 04 2022

def A103882(n): return hypergeometric([-n,-n,n], [1,1], 1).simplify()
[A103882(n) for n in range(31)] # G. C. Greubel, May 24 2023

a(n) = (A005258(n-1) + 3*A005258(n))/5 (Apéry numbers). - Mark van Hoeij, Jul 13 2010
n^2*(n-1)*(5*n-8)*a(n) = (n-1)*(55*n^3-143*n^2+102*n-24)*a(n-1) + n*(n-2)^2*(5*n-3)*a(n-2). - Alois P. Heinz, Jun 29 2015
a(n) ~ phi^(5*n + 3/2) / (2*Pi*5^(1/4)*n), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jul 21 2019
From Peter Bala, Jan 14 2020: (Start)
a(n) = Sum_{k = 0..n} C(n,k)^2*C(n+k-1,k). Cf. A005258.
For any prime p >= 5, a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k (follows from known supercongruences satisfied by the Apéry numbers A005258 - see Straub, Example 3.4). (End)
a(n) = hypergeometric([-n, -n, n], [1, 1], 1). - Peter Luschny, Jan 19 2020
From Peter Bala, Dec 19 2020: (Start)
a(n) = Sum_{k = 1..n} C(n,k)*C(n+k,k)*C(n-1,k-1) for n >= 1.
a(n) = [x^n] P(n, (1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A156554. (End)
a(n) = Sum_{k = 0..n} binomial(2*n-k-1,n-k)*binomial(n,k)^2. Cf. A108628. - Peter Bala, Mar 24 2022
From Peter Bala, Apr 15 2022: (Start)
a(-n) = (-1)^n*A352654(n).
a(n) = [x^n*y^n*z^(n-1)] 1/(1 - x - y - z + x*z + y*z - x*y*z) for n >= 1.
a(n) = B(n,n,n-1) in the notation of Straub, see equation 24.
a(n) = [x^n*y^n*z^(n-1)] (x + y + z)^n*(x + y)^n*(y + z)^(n-1) for n >= 1. (End)
D-finite with recurrence 9*n^2*a(n) -3*(31*n^2-27*n+6)*a(n-1) -2*(37*n^2-138*n+108)*a(n-2) -(n-3)*(17*n-56)*a(n-3) -(n-4)^2*a(n-4) = 0. - R. J. Mathar, Aug 01 2022
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n-1, n-k)*binomial(n+k, k)*binomial(n+k-1, k). - Peter Bala, Aug 13 2023
a(n) = Sum_{k = 0..n} (-1)^k * binomial(n+1, k)*binomial(2*n-k, n-k)^2. - Peter Bala, Oct 05 2024

a(0)=1 prepended by Alois P. Heinz, Jun 29 2015

A208675 Number of words, either empty or beginning with the first letter of the ternary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet.

Original entry on oeis.org

1, 1, 5, 37, 309, 2751, 25493, 242845, 2360501, 23301307, 232834755, 2349638259, 23905438725, 244889453043, 2523373849701, 26132595017037, 271826326839477, 2838429951771795, 29740725671232119, 312573076392760183, 3294144659048391059, 34802392680979707121

Offset: 0

Alois P. Heinz, Feb 29 2012

Also the number of (3*n-1)-step walks on 3-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.

			a(2) = 5 = |{aabbcc, aabcbc, aabccb, ababcc, abccba}|.
a(3) = 37 = |{aaabbbccc, aaabbcbcc, aaabbccbc, aaabbcccb, aaabcbbcc, aaabcbcbc, aaabcbccb, aaabccbbc, aaabccbcb, aaabcccbb, aababbccc, aababcbcc, aababccbc, aababcccb, aabbabccc, aabbcccba, aabcbabcc, aabcbccba, aabccbabc, aabccbcba, aabcccbab, aabcccbba, abaabbccc, abaabcbcc, abaabccbc, abaabcccb, abababccc, ababcccba, abbaabccc, abbcccbaa, abcbaabcc, abcbccbaa, abccbaabc, abccbcbaa, abcccbaab, abcccbaba, abcccbbaa}|.

Alois P. Heinz, Table of n, a(n) for n = 0..960
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.

Column k=3 of A208673.

Cf. A000108, A001622, A005258, A108625, A108628, A271777.

A208675:= func< n | (&+[Binomial(n,j)*Binomial(n-1,j)*Binomial(n+j-1,j): j in [0..2*n]]) >;
[A208675(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023

a:= n-> add(binomial(n-1, k)^2 *binomial(2*n-1-k, n-k), k=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, Jun 26 2012

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

def A208675(n): return sum(binomial(n,j)*binomial(n-1,j)*binomial(n+j-1,j) for j in range(n+1))
[A208675(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

From Michael Somos, Jun 03 2012: (Start)
a(n) = A108625(n-1, n).
a(n) = Hypergeometric3F2([1-n, -n, n], [1, 1], 1).
(n+1)^2 * (1 -4*n +5*n^2) * a(n+1) = (5 -5*n -26*n^2 +11*n^3 +55*n^4) * a(n) + (n-1)^2 * (2 +6*n +5*n^2) * a(n-1). (End)
a(n) ~ sqrt((5-sqrt(5))/10)/(2*Pi*n) * ((1+sqrt(5))/2)^(5*n). - Vaclav Kotesovec, Dec 06 2012. Equivalently, a(n) ~ phi^(5*n - 1/2) / (2 * 5^(1/4) * Pi * n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 15*x^3 + 94*x^4 + 668*x^5 + 5144*x^6 + 41884*x^7 + 355307*x^8 + ... appears to have integer coefficients. Cf. A108628. - Peter Bala, Jan 12 2016
From Peter Bala, Apr 05 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 = 0..n} binomial(-n,k)*binomial(-n-1,k)*binomial(-n+k-1,k).
a(-n) = Sum_{k = 0..n} (-1)^k* binomial(n+k-1,k)*binomial(n+k,k)*binomial(n,k)
a(-n) = (-1)^n*A108628(n-1), for n >= 1.
a(n) = Sum_{k = 1..n} binomial(n,k)*binomial(n-1,k-1)*binomial(n+k-1,k-1) for n >= 1.
Equivalently, a(n) = [(x^n)*(y*z)^(n-1)] (x + y + z)^n*(x + y)^(n-1)*(y + z)^(n-1) for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n-1,k)*binomial(2*n-k-1,n-k)^2.
a(n) = (1/5)*(A005258(n) + 2*A005258(n-1)) for n >= 1.
a(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Compare with A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).
a(n) = B(n,n-1,n-1) in the notation of Straub, equation 24. Hence
a(n) = [(x^n)*(y*z)^(n-1)] 1/(1 - x - y - z + x*z + y*z - x*y*z) for n >= 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:
1) a(n) = [(x*y)^n*z^(n-1)] 1/(1 - x - y - z + x*y + x*y*z) for n >= 1.
2) a(n) = - [(x*z)^(n-1)*(y^n)] 1/(1 + y + z + x*y + y*z + x*z + x*y*z) for n >= 1.
3) a(n) = [x^(n-1)*(y*z)^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) for n >= 1. (End)
From Peter Bala, Mar 17 2023: (Start)
For n >= 1:
a(n) = Sum_{k = 0..n} ((n-k)/(n+k))*binomial(n,k)^2*binomial(n+k,k).
a(n) = Sum_{k = 0..n} (-1)^(n+k-1) * ((n-k)/(n+k)) * binomial(n,k) * binomial(n+k,k)^2. (End)

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.

A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)binomial(n-1,k)binomial(n+k,k)*binomial(n-1+k,k).

Original entry on oeis.org

0, 1, 13, 253, 5741, 142001, 3713305, 100961365, 2825369965, 80843126905, 2354354542013, 69555586698533, 2079506849058809, 62797822901091409, 1912714261589863633, 58691012753161717253, 1812592783819686728045, 56298976785759622597385, 1757493098495181029912485

Offset: 0

Peter Bala, Apr 06 2022

Compare with A005259(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2.
Main superdiagonal (or main subdiagonal) of the symmetric square array A143007.
This is the sequence (A(n,n-1,n,n-1)) in the notation of Straub 2014, where it is proved that the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
More generally, for nonnegative integers A, B and C with A >= 2 let S(n;A,B,C) = Sum_{k = 1..n} binomial(n,k)^A * binomial(n+k-1,k-1)^B * binomial(2*k,n)^C. Then we conjecture the sequence (S(n;A,B,C))n>=1 satisfies the same supercongruences. This is the case A = B = 2, C = 0. Compare with Osburn et al., Theorem 1.2.

			Examples of supercongruences:
a(11) - a(1) = 69555586698533 - 1 = (2^2)*(11^3)*1321*2521*3923 == 0 (mod 11^3).
a(2*7) - a(2) = 1912714261589863633 - 13 = (2^2)*(3^2)*5*(7^3)*776221* 39911503 == 0 (mod 7^3).

Robert Osburn, Brundaban Sahu and Armin Straub, Supercongruences for sporadic sequences, arXiv:1312.2195 [math.NT], 2013-2014.
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.
Eric Weisstein's World of Mathematics, Strehl identities

Cf. A002002, A005259, A108628, A143007, A177316.

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

a[n_] := n^2 HypergeometricPFQ[{1 - n, 1 - n, 1 + n, 1 + n}, {1, 2, 2}, 1];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Apr 17 2022 *)

a(n) = sum(k=1, n, binomial(n,k)^2*binomial(n+k-1,k-1)^2); \\ Michel Marcus, Apr 19 2022

def A352653(n):
    if n == 0: return 0
    m, g = 1, 0
    for k in range(n+1):
        g += m*n**2//(n+k)**2
        m *= ((n+k+1)*(n-k))**2
        m //= (k+1)**4
    return g>>1 # Chai Wah Wu, Oct 03 2022

The sequence can be extended to negative values of n:
a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+k,k)*binomial(-n-1,k)*binomial(-n-1+k,k) = a(n), since binomial(-m,k) = (-1)^k*binomial(m+k-1,k) for nonnegative k.
a(n) = A177316(n)/2 for n >= 1.
a(n) = Sum_{k = 1..n} binomial(n,k)^2*binomial(n+k-1,k-1)^2.
a(n) = (1/2)*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k-1,k)^2 for n >= 1.
a(n) = hypergeom([n, n+1, -n, -n+1], [1, 1, 1], 1) for n >= 1.
a(n) = (1/2)*hypergeom([n, n, -n, -n], [1, 1, 1], 1) for n >= 1.
a(n) = (1/4)*binomial(2*n,n)^2*hypergeom([-n+1, -n+1, -n, -n], [1, -2*n+1, -2*n+1], 1) for n >= 1
a(n) = [x^n*y^(n-1)*z^n*t^(n-1)] 1/((1 - x - y)(1 - z - t) - x*y*z*t).
a(n) = [x^n] 1/(1 - x) * P(n-1,(1 + x)/(1 - x))^2 = [x^(n-1)] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2 for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Cf. A005259(n) = [x^n] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2.
(n + 1)^3*(2*n - 1)(3*n^2 - 3*n + 1)*a(n+1) = 2*(102*n^6 - 68*n^4 + 21*n^2 - 3)*a(n) - (n - 1)^3*(2*n + 1)*(3*n^2 + 3*n + 1)*a(n-1) with a(0) = 0 and a(1) = 1.
The Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p and all positive integers n and r. It follows that the expansion of exp( Sum_{n >= 1} a(n)/n*x^n ) = 1 + x + 7*x^2 + 91*x^3 + 1544*x^4 + 30448*x^5 + 661506*x^6 + 15377010*x^7 + ... has integer coefficients.
a(n) = n^2*hypergeom([1 - n, 1 - n, 1 + n, 1 + n], [1, 2, 2], 1). - Peter Luschny, Apr 17 2022
From Peter Bala, Mar 18 2023: (Start)
a(n) = Sum_{k = 0..n-1} (n-k)/(n+k) * binomial(n,k)^2 * binomial(n+k,k)^2.
a(n) = 1/6*(A005259(n) + A005259(n-1)). (End)
a(n) ~ (1 + sqrt(2))^(4*n) / (2^(9/4) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2023
a(n) = (1/3)*Sum_{k = 0..n} binomial(n,k)*binomial(n+k-1,k)*f(k) for n >= 1, where f(n) = A000172(n) denotes the n-th Franel number. Cf. A005259(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*f(k) (the second Strehl identity). - Peter Bala, Jun 26 2023
a(n) = Sum_{0 <= j <= k <= n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)* binomial(n+j-1, j)^2*binomial(n-1, k-j) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, k). Cf. Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k) = A005259(n). - Peter Bala, Sep 22 2024

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

cp's OEIS Frontend

A362728 a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} A108628(k-1)*x^k/k ).

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A362729 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A108628(k-1)*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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A103882 a(n) = Sum_{i=0..n} C(n+1,i)*C(n-1,i-1)*C(2n-i,n).

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A208675 Number of words, either empty or beginning with the first letter of the ternary alphabet, where each letter of the alphabet occurs n times and letters of neighboring word positions are equal or neighbors in the alphabet.

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

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A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)*binomial(n-1,k)*binomial(n+k,k)*binomial(n-1+k,k).

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A103882 a(n) = Sum_{i=0..n} C(n+1,i)C(n-1,i-1)C(2n-i,n).

A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)binomial(n-1,k)binomial(n+k,k)*binomial(n-1+k,k).