A099601 -id:A099601 - 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)

A050983 de Bruijn's S(4,n).

Original entry on oeis.org

1, 14, 786, 61340, 5562130, 549676764, 57440496036, 6242164112184, 698300344311570, 79881547652046140, 9301427008157320036, 1098786921802152516024, 131361675994216221116836, 15863471168011822803270200, 1932252897656224864335299400

Offset: 0

Eric W. Weisstein

a(n) is divisible by (n+1). Prime p divides a(p-1). Prime p>2 divides all a(n) from a((p+1)/2) to a(p-1). - Alexander Adamchuk, Jul 05 2006

N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Seiichi Manyama, Table of n, a(n) for n = 0..471
T. Amdeberhan, De Bruijn's sequence is odd iff n = 2m - 1: Part I, MathOverflow 383035.
Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A000984, A006480, A050984, A099601, A177322, A227357, A249332.

Sum[ (-1)^(k+n)Binomial[ 2n, k ]^4, {k, 0, 2n} ]
RecurrenceTable[{a[0] == 1, a[1] == 14, 4 (n + 1) (2 n + 1)^3 (48 n^2 + 162 n + 137) a[n] + (n + 2)^3 (2 n + 3) (48 n^2 + 66 n + 23) a[n + 2] == 2 (4 (n + 1)^2 (2 n + 3)^2 (408 n^2 + 969 n + 431) - (n + 1) (2 n + 3) (69 n + 31) + 57 n + 92) a[n + 1]}, a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 26 2016 *)

a(n)=sum(k=0,2*n,(-1)^(k+n)*binomial(2*n,k)^4) \\ Charles R Greathouse IV, Dec 28 2011

a(n) = Sum_{k=-n..+n} (-1)^k*C(2*n,n+k)^4. - Benoit Cloitre, Mar 02 2005
a(n) = (-1)^n * HypergeometricPFQ[ {-2n, -2n, -2n, -2n}, {1, 1, 1}, -1]. - Alexander Adamchuk, Jul 05 2006
E.g.f.: Sum(n>=0,I^n*x^n/n!^4) * Sum(n>=0,(-I)^n*x^n/n!^4) = Sum(n>=0,a(n)*x^(2*n)/n!^4) where I^2=-1. - Paul D. Hanna, Dec 21 2011
a(n) ~ 0.125 k^(8n+3)/(Pi*n)^(3/2) where k = 2 cos(Pi/8) = A179260. This formula is due to de Bruijn 1958. - Charles R Greathouse IV, Dec 28 2011
Recurrence: a(0) = 1, a(1) = 14, 4 * (n + 1) * (2*n + 1)^3 * (48*n^2 + 162*n + 137) * a(n) + (n + 2)^3 * (2*n + 3) * (48*n^2 + 66*n + 23) * a(n+2) = 2 * (4 * (n + 1)^2 * (2*n + 3)^2 * (408*n^2 + 969*n + 431) - (n + 1) * (2*n + 3) * (69*n + 31) + 57*n + 92) * a(n+1). - Vladimir Reshetnikov, Sep 26 2016
From Peter Bala, Nov 02 2024; (Start)
a(n) = 1/n * Sum_{k = 0..2*n} (-1)^(n+k) * k * binomial(2*n, k)^4 for n >= 1.
a(n) = binomial(2*n, n) * Sum_{k = 0..n} binomial(2*n, n+k)^2 * binomial(2*n+k,k) = binomial(2*n, n) * Sum_{k = 0..n} (-1)^(n+k) * binomial(2*n, n+k) * binomial(2*n+k, k)^2. (End)

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

Original entry on oeis.org

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

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

A339946 a(n) = [x^n] 1/Legendre_P(n,(1 - x)/(1 + x)).

Original entry on oeis.org

2, 24, 812, 52920, 5635002, 889789866, 195289709624, 56872979140536, 21222308525755790, 9874215185197183524, 5604584032515576621372, 3811820779676364251891562, 3060364611485092496329558842, 2863915888926428097267223280790, 3090075825959616714726175633059312

Offset: 1

Peter Bala, Dec 23 2020

We conjecture that the following supercongruences hold 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.
It is known that other sequences, which are related to the Legendre polynomials in a similar manner to this one, satisfy these congruences. Examples include A103882 and the two kinds of Apéry numbers A005258 and A005259: it can be shown that A103882(n) = [x^n] Legendre_P(n,(1 + x)/(1 - x)), while A005258(n) = [x^n] 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)) and A005259(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^2.
Calculation suggests that, for the present sequence, we have stronger congruences for prime p >= 7, namely a(p) == a(1) ( mod p^5 ) (checked up to p = 61).
a(p) == 2 (mod p^5) verified for primes p with 7 <= p <= 401. - Robert Israel, Dec 29 2020
For m a positive integer, define a_m(n) = [x^(m*n)] 1/Legendre_P(n,(1 - x)/(1 + x)). We conjecture that the supercongruence a_m(p) == a_m(1) ( mod p^5 ) holds for all primes p >= 7. - Peter Bala, Mar 10 2022

Robert Israel, Table of n, a(n) for n = 1..220

Cf. A005258, A005259, A099601, A103882.

with(orthopoly):
a:= n->coeftayl(1/P(n,(1-x)/(1+x)), x = 0, n):
seq(a(n), n = 1..20);

A099608 Table of crystal ball sequences for A_n lattices read by antidiagonals.

Original entry on oeis.org

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

Offset: 0

Michael Somos, Oct 25 2004

Rows (n=0..10) are A000012, A005408, A003215, A005902, A008384, A008386, A008388, A008390, A008392, A008394, A008396.

Row reverse of A108625. - Peter Bala, Jul 18 2008

T(n, n)=A005258(n), T(2n, n)=A099601(n).

Cf. A108625.

T(n,k)=sum(j=0,min(n,k),binomial(n+j,j)*binomial(n,j)*binomial(k,j))

T(n, k)=hypergeom([n+1, -n, -k], [1, 1], 1).

A187056 G.f.: A(x,y,z) = Sum_{n>=0} ((2n)!/n!^2)[Sum_{k=0..2n} T(n,k)z^k]x^(2n)y^n/(1-x-xy)^(4n+1) where A(x,y,x+xy) = Sum_{n>=0, k=0..n} C(n,k)^4x^ny^k at z = x+xy; this is the triangle of coefficients T(n,k), read by rows.

Original entry on oeis.org

1, 7, 4, 1, 131, 176, 96, 16, 1, 3067, 6588, 5895, 2416, 477, 36, 1, 79459, 235456, 298816, 197824, 73120, 14656, 1504, 64, 1, 2181257, 8252300, 13668975, 12563200, 6966400, 2373504, 490700, 58400, 3675, 100, 1, 62165039, 286326288, 587324232, 692965040, 516541455, 252283968, 81432456, 17138304, 2276145, 179440, 7632, 144, 1

Offset: 0

Paul D. Hanna, Mar 02 2011

			G.f.: A(x,y,z) = 1/(1-x-x*y)
+ 2*(7 + 4*z + z^2)*x^2*y/(1-x-x*y)^5
+ 6*(131 + 176*z + 96*z^2 + 16*z^3 + z^4)*x^4*y^2/(1-x-x*y)^9
+ 20*(3067 + 6588*z + 5895*z^2 + 2416*z^3 + 477*z^4 + 36*z^5 + z^6)*x^6*y^3/(1-x-x*y)^13 +...
G.f. at z = x+xy yields: A(x,y,x+xy) = 1 + (1 + y)*x
+ (1 + 16*y + y^2)*x^2
+ (1 + 81*y + 81*y^2 + y^3)*x^3
+ (1 + 256*y + 1296*y^2 + 256*y^3 + y^4)*x^4
+ (1 + 625*y + 10000*y^2 + 10000*y^3 + 625*y^4 + y^5)*x^5 +...
which is a series involving binomial coefficients to the 4th power.
...
This triangle of coefficients T(n,k) of z^k, k=0..2n, begins:
[1];
[7, 4, 1];
[131, 176, 96, 16, 1];
[3067, 6588, 5895, 2416, 477, 36, 1];
[79459, 235456, 298816, 197824, 73120, 14656, 1504, 64, 1];
[2181257, 8252300, 13668975, 12563200, 6966400, 2373504, 490700, 58400, 3675, 100, 1];
[62165039, 286326288, 587324232, 692965040, 516541455, 252283968, 81432456, 17138304, 2276145, 179440, 7632, 144, 1];
[1818812387, 9876304172, 24205612067, 34939683632, 32837525567, 21029302364, 9356637759, 2899564224, 619135629, 88879924, 8237341, 461776, 14161, 196, 1];
[54257991011, 339398092544, 968547444480, 1655445817088, 1881608595776, 1496188189440, 853911382016, 353544477440, 106191762336, 22927328512, 3492995968, 364541184, 24932320, 1044736, 24192, 256, 1];
...

Cf. A000897, A099601.

Row sums equal A000897(n) = (4n)!/((2n)!*n!^2).

Column 0 equals A099601(n) = quotient of de Bruijn sums S(4,n)/S(2,n).

cp's OEIS Frontend

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

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A050983 de Bruijn's S(4,n).

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

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

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

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

A187056 G.f.: A(x,y,z) = Sum_{n>=0} ((2n)!/n!^2)[Sum_{k=0..2n} T(n,k)z^k]x^(2n)y^n/(1-x-xy)^(4n+1) where A(x,y,x+xy) = Sum_{n>=0, k=0..n} C(n,k)^4x^ny^k at z = x+xy; this is the triangle of coefficients T(n,k), read by rows.