A108625 -id:A108625 - cp's OEIS frontend

A363871 a(n) = A108625(2n, 3n).

1, 37, 5321, 980407, 201186025, 43859522037, 9939874413899, 2314357836947571, 549694303511409641, 132569070434503802605, 32360243622138480889321, 7977001183875449759759807, 1982402220908671654519130731, 496031095735572731850517509727

Offset: 0

Peter Bala, Jun 27 2023

a(n) = B(2*n, 3*n, 2*n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that 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.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2) and A363870 (r = 1, s = 3).

G. C. Greubel, Table of n, a(n) for n = 0..400
Peter Bala, A recurrence for A363871

Cf. A005258, A099601, A108625, A363864 - A363870.

A363871:= func< n | (&+[Binomial(2*n,j)^2*Binomial(5*n-j,2*n): j in [0..2*n]]) >;
[A363871(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023

A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
seq(simplify(A108625(2*n, 3*n)), n = 0..20);

Table[HypergeometricPFQ[{-2*n,-3*n,2*n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)

def A363871(n): return sum(binomial(2*n,j)^2*binomial(5*n-j,2*n) for j in range(2*n+1))
[A363871(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

a(n) = Sum_{k = 0..2*n} binomial(2*n, k)^2 * binomial(5*n-k, 2*n).
a(n) = Sum_{k = 0..2*n} (-1)^k * binomial(2*n, k)*binomial(5*n-k, 2*n)^2.
a(n) = hypergeometric3F2([-2*n, -3*n, 2*n+1], [1, 1], 1).
a(n) = [x^(3*n)] 1/(1 - x)*Legendre_P(2*n, (1 + x)/(1 - x)).
a(n) ~ sqrt(1700 + 530*sqrt(10)) * (98729 + 31220*sqrt(10))^n / (120 * Pi * n * 3^(6*n)). - Vaclav Kotesovec, Feb 17 2024
a(n) = Sum_{k = 0..2*n} binomial(2*n, k) * binomial(3*n, k) * binomial(2*n+k, k). - Peter Bala, Feb 26 2024

A376458 a(n) = Sum_{k = 0..n} (-1)^(n+k)binomial(n, k)binomial(n+k, k)*A108625(n-1, n-k).

Original entry on oeis.org

1, 1, -7, 1, 569, -3749, -45151, 806737, 1052729, -130060889, 740060243, 16076432923, -238772815711, -1050791121197, 49401000432497, -171944622257999, -7658491447803847, 87632552103603679, 768037618172427023, -22023427875902878553, 19183786570616924819, 4030690809877385503081, -33792039667279104716677, -520860578851790657166869

Offset: 0

Peter Bala, Sep 23 2024

Compare with the following identity relating the sequence of Apéry numbers to the table of crystal ball sequences for the A_n lattices: A005259(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k), which can be verified by using the MulZeil procedure in Zeilberger's MultiZeilberger Maple package to find a recurrence for the double sum on the right-hand side of the identity.
The sequence of Apéry numbers A005259 satisfies the supercongruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies same supercongruences and in addition satisfies the stronger congruences a(p) == a(1) (mod p^5) for all primes p >= 7 (checked up to p = 199) and a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for all primes p >= 5 and integers r >= 2.

			Examples of supercongruences:
a(7) - a(1) = 806737 - 1 = (2^4)*3*(7^5) == 0 (mod 7^5).
a(11) - a(1) = 16076432923 - 1 =  2*3*(11^5)*127*131 == 0 (mod 11^5).
a(5^2) - a(5) = 22511570786292886382808751 - (-3749) = (2^2)*(3^2)*(5^9)*67*97*
7741*49223*129289 == 0 (mod 5^9).

Cf. A005259, A108625, A183204, A352653, A376459 - A376466.

A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, n-k), k = 0..n):
seq(a(n), n = 0..25);

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

P-recursive: (2*n - 3)*n^3*(n - 1)^2*(473*n^5 - 4988*n^4 + 20888*n^3 - 43462*n^2 + 45019*n - 18634)*a(n) = - 2*(n - 1)^2*(3784*n^9 - 51256*n^8 + 303801*n^7 - 1037327*n^6 + 2252744*n^5 - 3220636*n^4 + 3006247*n^3 - 1739455*n^2 + 555714*n - 75024)*a(n-1) - 2*(n - 2)*(2*n - 1)*(52030*n^9 - 756800*n^8 + 4787337*n^7 - 17271387*n^6 + 39143817*n^5 - 57806236*n^4 + 55708921*n^3 - 33926177*n^2 + 11955879*n - 1890360)*a(n-2) - 2*(n - 2)*(n - 3)^3*(2*n - 1)*(2*n - 3)*(473*n^5 - 2623*n^4 + 5666*n^3 - 5996*n^2 + 3172*n - 704)*a(n-3) with a(0) = 1, a(1) = -1 and a(2) = 7.

A108626 Antidiagonal sums of square array A108625, in which row n equals the crystal ball sequence for A_n lattice.

Original entry on oeis.org

1, 2, 5, 14, 41, 124, 383, 1200, 3799, 12122, 38919, 125578, 406865, 1322772, 4313155, 14099524, 46192483, 151628090, 498578411, 1641921014, 5414619739, 17878144968, 59097039545, 195548471268, 647665451911, 2146947613286

Offset: 0

Paul D. Hanna, Jun 12 2005

nonn

Limit a(n+1)/a(n) = 3.3829757679... = 1/r = 3 + r + r^2, where r is radius of convergence of A(x), which diverges at x=r.

Infinitely many recurrence relations of even order 2d can be built for this sequence; first define the following polynomial: P(d) = (1/2^d) * Sum_{i=0..floor(d/2)} binomial(d, 2*i) * (x^4 + 2*x^2 - 4*x + 1)^i * (x^2 + 2*x - 1)^(d - 2*i) then call c(d,k) the coefficient of term with power k in the polynomial P(d); then we have the relation: Sum_{k=0..2*d} c(d, 2*d-k)*a(n+k) = (-1)^d * Sum_{k=0..n} Sum_{i=0..k} binomial(n-k, d+i)*binomial(n-k, i)*binomial(n-i, k-i). - Thomas Baruchel, Jan 26 2015

			Log(A(x)) = 2*x + 6*x^2/2 + 20*x^3/3 + ... + A108627(n)*x^n/n + ...

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Thomas Baruchel and C. Elsner, On error sums formed by rational approximations with split denominators, arXiv preprint arXiv:1602.06445 [math.NT], 2016.
Sergi Elizalde, Symmetric peaks and symmetric valleys in Dyck paths, arXiv:2008.05669 [math.CO], 2020, p. 16.

Cf. A108625, A108627, A180091.

R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( 1/Sqrt(1-4*x+2*x^2+x^4) )); // G. C. Greubel, Oct 06 2023

a := n -> add(binomial(n,k)*hypergeom([-k,k-n,k-n], [1,-n], 1), k=0..n):
seq(simplify(a(n)), n=0..25); # Peter Luschny, Feb 13 2018

CoefficientList[Series[1/Sqrt[x^4+2*x^2-4*x+1], {x, 0, 50}], x] (* Vincenzo Librandi, Nov 08 2014 *)

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

{a(n)=polcoeff( sum(m=0, n, x^m * sum(k=0, m, binomial(m, k)^2 * x^k) / (1-x +x*O(x^n))^(m+1)) , n)}
for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Nov 08 2014

def A108626_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/sqrt(1-4*x+2*x^2+x^4) ).list()
A108626_list(40) # G. C. Greubel, Oct 06 2023

a(n) = Sum_{k=0..n} Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f.: 1 / sqrt(x^4 + 2*x^2 - 4*x + 1). - Thomas Baruchel, Nov 08 2014
G.f.: A(x) = exp( Sum_{n>=1} A108627(n)*x^n/n ), where A108627 has g.f.: 2*x*(1 - x - x^3)/((1-x)*(1 - 3*x - x^2 - x^3)).
a(n) = ( (5*n-3)*a(n-1) - (6*n-8)*a(n-2) + (2*n-4)*a(n-3) - (n-2)*a(n-4) + (n-3)*a(n-5) ) / n. - Thomas Baruchel, Nov 08 2014
a(n+2) - 2*a(n+1) - a(n) = 2*Sum_{k=0..n} Sum_{i=0..k} binomial(n-k+1,i-1)*binomial(n-k+1,i)*binomial(n-i+1,k-i) = Sum_{k=0..n} a(k)*A086581(n-k+1). - Thomas Baruchel, Nov 08 2014
G.f.: Sum_{n>=0} (2*n)!/n!^2 * x^(2*n) / ((1-x)*(1-2*x)^(3*n+1)). - Paul D. Hanna, Nov 08 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
Partial sums of A171155: a(n) = Sum_{i=0..n} A171155(n). - Thomas Baruchel, Nov 08 2014
Recurrence: n*a(n) = 2*(2*n-1)*a(n-1) - 2*(n-1)*a(n-2) - (n-2)*a(n-4). - Vaclav Kotesovec, Dec 20 2015
a(n) = Sum_{k=0..n} binomial(n,k)*hypergeometric3F2([-k,k-n,k-n], [1,-n], 1). - Peter Luschny, Feb 13 2018

A376466 a(n) = Sum_{k = 0..n} (-1)^(n+k)binomial(n, k)binomial(n+k, k)^2*A108625(n-1, k).

Original entry on oeis.org

1, 3, 127, 9435, 866751, 89591753, 9988439203, 1173951006987, 143456999185855, 18063466831218981, 2329136571942011877, 306174745758226208537, 40896708938016175140963, 5536767359542664588001285, 758259747093486125157272779, 104880152366856305370319427435, 14632959744552362547801104612799

Offset: 0

Peter Bala, Sep 25 2024

Cf. A005258(n) = Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(n+k, k)^2.
The sequence of Apéry numbers A005258 satisfies the pair of supercongruences
1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

Cf. A005258, A376458 - A376465.

A108625(n, k) := add( binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n-1, k), k = 0..n):
seq(a(n), n = 0..25);

Examples of supercongruences:
a(11) - a(1) = 306174745758226208537 - 3 = 2*(11^3)*17*79367*85245689663 == 0 (mod 11^3).
a(10) - a(0) = 2329136571942011877 - 1 = (2^2)*(11^3)*17011*25717400209 == 0 (mod 11^3).

A363867 a(n) = A108625(n,2*n).

Original entry on oeis.org

1, 5, 61, 923, 15421, 272755, 5006275, 94307855, 1811113021, 35301145037, 696227550811, 13863654392945, 278264498108611, 5622746346645953, 114268249446672151, 2333733620675302423, 47868774493665731645, 985608360056821004233, 20362035153323824192645

Offset: 0

Peter Bala, Jun 27 2023

a(n) = B(n,2*n,n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that 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.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2), A363870 (r = 1, s = 3) and A363871 (r = 2, s = 3).

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A005258, A099601, A108625, A363864 - A363871.

A363867:= func< n | (&+[Binomial(n,j)^2*Binomial(2*n+j,n): j in [0..n]]) >;
[A363867(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023

A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
seq(simplify(A108625(n, 2*n)), n = 0..18);

Table[HypergeometricPFQ[{-n,-2*n,n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)

def A363867(n): return sum(binomial(n,j)^2*binomial(2*n+j,n) for j in range(n+1))
[A363867(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(2*n+k, n).
a(n) = Sum_{k = 0..n} (-1)^(n+k)* binomial(n, k)*binomial(2*n+k, n)^2.
a(n) = hypergeom( [-n, -2*n, n+1], [1, 1], 1).
a(n) = [x^(2*n)] 1/(1 - x)*Legendre_P(n, (1 + x)/(1 - x)).
P-recursive: 4*(2*n - 1)^2*n^2*(85*n^2 - 235*n + 163)*a(n) = (29665*n^6 - 141345*n^5 + 264772*n^4 - 249181*n^3 + 124975*n^2 - 31902*n + 3276)*a(n-1) + 4*(2*n - 3)^2*(n-1)^2*(85*n^2 - 65*n + 13)*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*binomial(2*n, k). - Peter Bala, Feb 25 2024
a(n) ~ sqrt(13 + 53/sqrt(17)) * (349 + 85*sqrt(17))^n / (Pi * n * 2^(5*n + 5/2)). - Vaclav Kotesovec, Apr 26 2024

A363868 a(n) = A108625(3*n, n).

Original entry on oeis.org

1, 13, 505, 24691, 1337961, 76869013, 4586370139, 280973874215, 17552736006121, 1113134497824901, 71437216036404505, 4629194489296980715, 302391678415222922475, 19886936616891022422159, 1315438146193644502479255, 87445220568000089973356191, 5838332204000163260729138153

Offset: 0

Peter Bala, Jun 27 2023

a(n) = B(3*n, n, 3*n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that 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.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363869 (r = 3, s = 2), A363870 (r = 1, s = 3) and A363871(r = 2, s = 3).

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A005258, A099601, A108625, A363864 - A363871.

A363868:= func< n | (&+[Binomial(3*n,n-j)^2*Binomial(3*n+j,j): j in [0..n]]) >;
[A363868(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023

A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
seq(simplify(A108625(3*n, n)), n = 0..16);

Table[HypergeometricPFQ[{-3*n,-n,3*n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)

def A363868(n): return sum(binomial(3*n,n-j)^2*binomial(3*n+j, j) for j in range(n+1))
[A363868(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

a(n) = Sum_{k = 0..n} binomial(3*n, n-k)^2 * binomial(3*n+k, k).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(3*n, n-k) * binomial(3*n+k, k)^2.
a(n) = hypergeometric3F2( [-3*n, -n, 3*n+1], [1, 1], 1).
a(n) = [x^n] 1/(1 - x)*Legendre_P(3*n, (1 + x)/(1 - x)).
P-recursive: 3*(4797*n^4 - 26076*n^3 + 53055*n^2 - 47886*n + 16178)*(3*n - 1)^2*(3*n - 2)^2*n^2*a(n) = (82935333*n^10 - 699633963*n^9 + 2570641767*n^8 - 5402404662*n^7 + 7171181427*n^6 - 6264762171*n^5 + 3637752517*n^4 - 1382756780*n^3 + 328531700*n^2 - 44004160*n + 2529600)*a(n-1) + 3*(4797*n^4 - 6888*n^3 + 3609*n^2 - 816*n + 68)*(n - 1)^2*(3*n - 4)^2*(3*n - 5)^2*a(n-2) with a(0) = 1 and a(1) = 13.
a(n) ~ sqrt(17 + 61/sqrt(13)) * ((1921 + 533*sqrt(13))/54)^n / (6*Pi*sqrt(2)*n). - Vaclav Kotesovec, Feb 17 2024
a(n) = Sum_{k = 0..n} binomial(n, k) * binomial(3*n, k) * binomial(3*n+k, k). - Peter Bala, Feb 26 2024

A363869 a(n) = A108625(3n, 2n).

Original entry on oeis.org

1, 55, 12559, 3685123, 1205189519, 418856591055, 151353475289275, 56193989426243199, 21283943385478109071, 8185785098679048061837, 3186604888590691870779559, 1252744279186835597251089055, 496508748101370063304243706939, 198134918989716743103591120933103

Offset: 0

Peter Bala, Jun 27 2023

a(n) = B(3*n, 2*n, 3*n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that 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.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363870 (r = 1, s = 3) and A363871 (r = 2, s = 3).

Paolo Xausa, Table of n, a(n) for n = 0..350
Peter Bala, A recurrence for A363869

Cf. A005258, A099601, A108625, A363864 - A363871.

A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
seq(simplify(A108625(3*n, 2*n)), n = 0..20);

A363869[n_] := HypergeometricPFQ[{-2*n, -3*n, 3*n + 1}, {1, 1}, 1];
Array[A363869, 20, 0] (* Paolo Xausa, Feb 26 2024 *)

a(n) = Sum_{k = 0..2*n} binomial(3*n, k)^2 * binomial(5*n-k, 3*n).
a(n) = Sum_{k = 0..2*n} (-1)^k * binomial(3*n, k)*binomial(5*n-k, 3*n)^2.
a(n) = hypergeom( [-2*n, -3*n, 3*n+1], [1, 1], 1).
a(n) = [x^(2*n)] 1/(1 - x)*Legendre_P(3*n, (1 + x)/(1 - x)).
a(n) ~ 2^(4*n) * 3^(3*n) / (sqrt(5)*Pi*n). - Vaclav Kotesovec, Apr 27 2024

A363870 a(n) = A108625(n, 3*n).

Original entry on oeis.org

1, 7, 127, 2869, 71631, 1894007, 51978529, 1464209383, 42050906191, 1225778575021, 36156060825127, 1076772406867549, 32324178587781393, 976893529756053501, 29693248490460447747, 907027175886637081619, 27826656707376811715663, 856949305975908664414097

Offset: 0

Peter Bala, Jun 27 2023

a(n) = B(n, 3*n, n) in the notation of Straub, equation 24. It follows from Straub, Theorem 3.2, that 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.
More generally, for positive integers r and s the sequence {A108625(r*n, s*n) : n >= 0} satisfies the same supercongruences.
For other cases, see A099601 (r = 2, s = 1), A363867 (r = 1, s = 2), A363868 (r = 3, s = 1), A363869 (r = 3, s = 2) and A363871 (r = 2, s = 3).

G. C. Greubel, Table of n, a(n) for n = 0..500
Peter Bala, A recurrence for A363870

Cf. A005258, A099601, A108625, A363864 - A363871.

A363870:= func< n | (&+[Binomial(n,j)^2*Binomial(3*n+j,n): j in [0..n]]) >;
[A363870(n): n in [0..30]]; // G. C. Greubel, Oct 05 2023

A108625 := (n, k) -> hypergeom([-n, -k, n+1], [1, 1], 1):
seq(simplify(A108625(n, 3*n)), n = 0..20);

Table[HypergeometricPFQ[{-n,-3*n,n+1}, {1,1}, 1], {n,0,30}] (* G. C. Greubel, Oct 05 2023 *)

def A363870(n): return sum(binomial(n,j)^2*binomial(3*n+j,n) for j in range(n+1))
[A363870(n) for n in range(31)] # G. C. Greubel, Oct 05 2023

a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*n+k, n).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(3*n+k, n)^2.
a(n) = hypergeometric3F2( [-n, -3*n, n+1], [1, 1], 1).
a(n) = [x^(3*n)] 1/(1 - x)*Legendre_P(n, (1 + x)/(1 - x)).
a(n) ~ sqrt(25 + 151/sqrt(37)) * (11906 + 1961*sqrt(37))^n / (Pi * 2^(3/2) * n * 3^(6*n+1)). - Vaclav Kotesovec, Feb 17 2024
a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*binomial(3*n, k). - Peter Bala, Feb 25 2024

A364243 a(n) = A108625(2*n-1, n-1) for n >= 1.

Original entry on oeis.org

1, 13, 271, 6637, 176251, 4914427, 141573251, 4174790893, 125288929171, 3811804637263, 117248436333601, 3638993432201563, 113790712076898871, 3580847269415337487, 113299135244467189771, 3601766951734150461677, 114973519461796962202051

Offset: 1

Peter Bala, Jul 16 2023

The sequence of Apéry numbers A005258 forms the main diagonal of A108625, i.e., A005258(n) = A108625(n, n). The Apéry numbers satisfy the supercongruences A005258(n*p^r) == A005258(n^p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.

More generally, for positive integers r and s, the sequence defined by a(r,s;n) = A108625(r*n - 1, s*n - 1) may also satisfy the same supercongruences. This is the case r = 2, s = 1. Compare with the comments in A363867.

Cf. A005258, A108625, A363867, A364244.

seq(add(binomial(2*n-1,k)^2 * binomial(3*n-2-k,2*n-1), k = 0..n-1), n = 1..20);
# alternative program
seq( simplify(binomial(2*n-1,n-1)^2 * hypergeom([1 - n, 1 - n, 2*n], [1 + n, 1 + n], 1)), n = 1..20);

Table[Binomial[2*n-1, n-1]^2 * HypergeometricPFQ[{1 - n, 1 - n, 2*n}, {1 + n, 1 + n}, 1], {n, 1, 20}] (* Vaclav Kotesovec, Jul 16 2023 *)

a(n) = Sum_{k = 0..n-1} binomial(2*n-1,k)^2 * binomial(3*n-2-k,2*n-1) for n >= 1.
a(n) = Sum_{k = 0..n-1} (-1)^k * binomial(2*n-1,k) * binomial(3*n-2-k,2*n-1)^2 for n >= 1.
a(n) = binomial(2*n-1,n-1)^2 * hypergeom([1 - n, 1 - n, 2*n], [1 + n, 1 + n], 1).
a(n) = (-1)^(n+1) * binomial(2*n-1,n-1) * hypergeom([1 - n, 2*n, 2*n], [1, 1 + n], 1).
P-recursive: (n-1)^2*(2*n-1)^2*(48*n^2-162*n+137)*a(n) = (6528*n^6-48144*n^5+144800*n^4-226714*n^3+194349*n^2-86299*n+15515)*a(n-1) - (2*n-3)^2*(n-2)^2*(48*n^2-66*n+23)*a(n-2) with a(1) = 1 and a(2) = 13.
a(n) ~ (1 + sqrt(2))^(4*n - 3/2) / (2^(9/4)*Pi*n). - Vaclav Kotesovec, Jul 16 2023

A376459 a(n) = Sum_{k = 0..n} (-1)^(n+k)binomial(n, k)binomial(n+k, k)*A108625(n, n-k).

Original entry on oeis.org

1, -1, -17, 143, 751, -20251, 30871, 2584847, -21586193, -251907751, 5176221733, 5498864117, -913327142441, 5540080670669, 120825094592983, -1860921180719857, -8346832617144593, 401702184476719649, -1403893237226212151, -64680833271083055607, 743195619082337134501, 6754996433001423371159, -192371016736634220839987, 139058974519768723621493, 36163089652079749214625751, -298797649039016749340832751

Offset: 0

Peter Bala, Sep 24 2024

Compare with the following identity relating the sequence of Apéry numbers to the table of crystal ball sequences for the A_n lattices: A005259(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k), which can be verified by using the MulZeil procedure in Zeilberger's MultiZeilberger Maple package to find a recurrence for the double sum on the right-hand side of the identity.
The sequence of Apéry numbers A005259 satisfies the pair of supercongruences
1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
and
2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

			Examples of supercongruences:
(1a) a(11) - a(1) = 5498864117 - (-1) = 2*(3^3)*(11^3)*76507 == 0 (mod 11^3);
(1b) a(10) - a(0) = 5176221733 - 1 = (2^2)*(3^5)*(11^3)*4001 == 0 (mod 11^3).
(2a) a(5^2) - a(5) = -298797649039016749340832751 - (-20251) = -(2^2)*3*(5^6)*(11^2)*47*89*1683049*1870707593 == 0 (mod 5^6);
(2b) a(5^2 - 1) - a(5 - 1) = 36163089652079749214625751 - 751 = (2^3)*3*(5^6)*7*11*17*101*729412564491671 == 0 (mod 5^6).

Cf. A005259, A108625, A376458, A376460 - A376466.

A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, n-k), k = 0..n):
seq(a(n), n = 0..25);

P-recursive: n^3*(n - 1)*(946*n^2 - 3053*n + 2475)*a(n) = -2*(n - 1)*(3784*n^5 - 17888*n^4 + 31787*n^3 - 26726*n^2 + 11051*n - 1824)*a(n-1) - 2*(104060*n^6 - 752070*n^5 + 2212238*n^4 - 3374927*n^3 + 2802671*n^2 - 1196821*n + 205920)*a(n-2) - 2*(n - 2)^3*(2*n - 3)*(946*n^2 - 1161*n + 368)*a(n-3) with a(0) = 1, a(1) = -1 and a(2) = -17.

cp's OEIS Frontend

A363871 a(n) = A108625(2*n, 3*n).

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A376458 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, n-k).

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A108626 Antidiagonal sums of square array A108625, in which row n equals the crystal ball sequence for A_n lattice.

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A376466 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n-1, k).

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A363867 a(n) = A108625(n,2*n).

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A363868 a(n) = A108625(3*n, n).

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A363869 a(n) = A108625(3*n, 2*n).

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A363871 a(n) = A108625(2n, 3n).

A376458 a(n) = Sum_{k = 0..n} (-1)^(n+k)binomial(n, k)binomial(n+k, k)*A108625(n-1, n-k).

A376466 a(n) = Sum_{k = 0..n} (-1)^(n+k)binomial(n, k)binomial(n+k, k)^2*A108625(n-1, k).

A363869 a(n) = A108625(3n, 2n).

A376459 a(n) = Sum_{k = 0..n} (-1)^(n+k)binomial(n, k)binomial(n+k, k)*A108625(n, n-k).