A103882 -id:A103882 - cp's OEIS frontend

A103881 Square array T(n,k) (n >= 1, k >= 0) read by antidiagonals: coordination sequence for root lattice A_n.

1, 1, 2, 1, 6, 2, 1, 12, 12, 2, 1, 20, 42, 18, 2, 1, 30, 110, 92, 24, 2, 1, 42, 240, 340, 162, 30, 2, 1, 56, 462, 1010, 780, 252, 36, 2, 1, 72, 812, 2562, 2970, 1500, 362, 42, 2, 1, 90, 1332, 5768, 9492, 7002, 2570, 492, 48, 2, 1, 110, 2070, 11832, 26474, 27174, 14240, 4060, 642, 54, 2, 1, 132, 3080, 22530, 66222, 91112, 65226, 26070, 6040, 812, 60, 2

Offset: 1

Ralf Stephan, Feb 20 2005

T(n,k) is the number of integer sequences of length n+1 with sum zero and sum of absolute values 2k. - R. H. Hardin, Feb 23 2009

			Array begins:
  1,   2,     2,      2,       2,        2,         2,          2, ... A040000;
  1,   6,    12,     18,      24,       30,        36,         42, ... A008458;
  1,  12,    42,     92,     162,      252,       362,        492, ... A005901;
  1,  20,   110,    340,     780,     1500,      2570,       4060, ... A008383;
  1,  30,   240,   1010,    2970,     7002,     14240,      26070, ... A008385;
  1,  42,   462,   2562,    9492,    27174,     65226,     137886, ... A008387;
  1,  56,   812,   5768,   26474,    91112,    256508,     623576, ... A008389;
  1,  72,  1332,  11832,   66222,   271224,    889716,    2476296, ... A008391;
  1,  90,  2070,  22530,  151560,   731502,   2777370,    8809110, ... A008393;
  1, 110,  3080,  40370,  322190,  1815506,   7925720,   28512110, ... A008395;
  1, 132,  4422,  68772,  643632,  4197468,  20934474,   85014204, ... A035837;
  1, 156,  6162, 112268, 1219374,  9129276,  51697802,  235895244, ... A035838;
  1, 182,  8372, 176722, 2206932, 18827718, 120353324,  614266354, ... A035839;
  1, 210, 11130, 269570, 3838590, 37060506, 265953170, 1511679210, ... A035840;
  ...
Antidiagonals:
  1;
  1,  2;
  1,  6,    2;
  1, 12,   12,    2;
  1, 20,   42,   18,    2;
  1, 30,  110,   92,   24,    2;
  1, 42,  240,  340,  162,   30,    2;
  1, 56,  462, 1010,  780,  252,   36,   2;
  1, 72,  812, 2562, 2970, 1500,  362,  42,  2;
  1, 90, 1332, 5768, 9492, 7002, 2570, 492, 48,  2;

Muniru A Asiru, Table of n, a(n) for n = 1..5050 (antidiagonals 1 to 100, flattened)
M. Baake and U. Grimm, Coordination sequences for root lattices and related graphs, arXiv:cond-mat/9706122 [cond-mat.stat-mech], 1997.
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
Arun Padakandla, P. R. Kumar, and Wojciech Szpankowski, On the Discrete Geometry of Differential Privacy via Ehrhart Theory, November 2017.
Arun Padakandla, P. R. Kumar, and Wojciech Szpankowski, Preserving Privacy and Fidelity via Ehrhart Theory, July 2017.
Joan Serra-Sagrista, Enumeration of lattice points in l_1 norm, Inf. Proc. Lett. 76 (1-2) (2000) 39-44.

Rows include A040000, A008458, A005901, A008383, A008385, A008387, A008389, A008391, A008393, A008395, A035837, A035838, A035839, A035840, A035841 - A035876.
Columns include A002376, A001621.
Main diagonal is in A103882.
Cf. A047085, A103884, A103903, A103998, A143007.

T:=Flat(List([1..12],n->Concatenation([1],List([1..n-1],k->Sum([1..n],i->Binomial(n-k+1,i)*Binomial(k-1,i-1)*Binomial(n-i,k)))))); # Muniru A Asiru, Oct 14 2018

A103881:= func< n,k | k le 0 select 1 else (&+[Binomial(n-k+1, j)*Binomial(k-1, j-1)*Binomial(n-j, k): j in [1..n-k]]) >;
[A103881(n,k): k in [0..n-1], n in [1..15]]; // G. C. Greubel, Oct 16 2018; May 24 2023

T:=proc(n,k) option remember; local i;
if k=0 then 1 else
add( binomial(n+1,i)*binomial(k-1,i-1)*binomial(n-i+k,k),i=1..n); fi;
end:
g:=n->[seq(T(n-i,i),i=0..n-1)]:
for n from 1 to 14 do lprint(op(g(n))); od:

T[n_, k_]:= (n+1)*(n+k-1)!*HypergeometricPFQ[{1-k,1-n,-n}, {2,-n-k+1}, 1]/(k!*(n-1)!); T[, 0]=1; Flatten[Table[T[n-k, k], {n,12}, {k,0,n-1}]] (* _Jean-François Alcover, Dec 27 2012 *)

A103881(n,k) = if(k==0, 1, sum(j=1, n-k, binomial(n-k+1, j)*binomial(k-1, j-1)*binomial(n-j, k)));
for(n=1, 15, for(k=0, n-1, print1(A103881(n,k), ", "))) \\ G. C. Greubel, Oct 16 2018; May 24 2023

def A103881(n,k): return 1 if k==0 else (n-k+1)*binomial(n-1,k)*hypergeometric([k-n,1+k-n,1-k], [2,1-n], 1).simplify()
flatten([[A103881(n,k) for k in range(n)] for n in range(1,16)]) # G. C. Greubel, May 24 2023

T(n,k) = Sum_{i=1..n} C(n+1, i)*C(k-1, i-1)*C(n-i+k, k), T(n,0)=1.
G.f. of n-th row: (Sum_{i=0..n} C(n, i)^2*x^i)/(1-x)^n.
From G. C. Greubel, May 24 2023: (Start)
T(n, k) = Sum_{j=0..n} binomial(n,j)^2 * binomial(n+k-j-1, n-1) (array).
T(n, k) = (n+1)*binomial(n+k-1,k)*hypergeometric([-n,1-n,1-k], [2,1-n-k], 1), with T(n, k) = 1 (array).
t(n, k) = (n-k+1)*binomial(n-1,k)*hypergeometric([k-n,1+k-n,1-k], [2,1-n], 1), with t(n, 0) = 1 (antidiagonals).
Sum_{k=0..n-1} t(n, k) = A047085(n). (End)
From Peter Bala, Jul 09 2023: (Start)
T(n,k) = [x^k] Legendre_P(n, (1 + x)/(1 - x)).
(n+1)*T(n+1,k) = (n+1)*T(n+1,k-1) + (2*n+1)*(T(n,k) + T(n,k-1)) - n*(T(n-1,k) - T(n-1,k-1)). (End)

Corrected by N. J. A. Sloane, Dec 15 2012, at the suggestion of Manuel Blum

A156554 The number of integer sequences of length d = 2n+1 such that the sum of the terms is 0 and the sum of the absolute values of the terms is d-1.

Original entry on oeis.org

1, 6, 110, 2562, 66222, 1815506, 51697802, 1511679210, 45076309166, 1364497268946, 41800229045610, 1292986222651646, 40317756506959050, 1265712901796074842, 39965073938276694002, 1268208750951634765562, 40419340092267053380782, 1293151592990764737265490

Offset: 0

W. Edwin Clark, Feb 09 2009

Let b(n) = S(d,n) be the coordination sequence of the lattice A_d. Then this sequence is a(n) = S(2n,n). See Conway-Sloane. The sequence is defined by Couveignes et al.

			The a(1) = 6 sequences are (1,-1,0), (-1,1,0), (1,0,-1), (-1,0,1), (0,1,-1) and (0,-1,1).

R. H. Hardin and Colin Barker, Table of n, a(n) for n = 0..300 (terms up to n=26 from R. H. Hardin)
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
J.-M. Couveignes, T. Ezome, and R. Lercier, Elliptic periods and primality proving, arXiv:0810.2853 [math.NT], 2008-2009.

a(n) = A103881(2n, n), A103882.

S:=proc(d,n) add(binomial(d,k)^2*binomial(n-k+d-1,d-1),k=0..d); end proc; a:=n->S(2*n,n);

Table[ Binomial[-1 + 3 n, -1 + 2 n] HypergeometricPFQ[{-2 n, -2 n, -n}, {1, 1 - 3 n}, 1], {n, 0, 10}]  (* Eric W. Weisstein, Feb 10 2009 *)

S(d, n) = sum(k=0, d, binomial(d,k)^2*binomial(n-k+d-1, d-1));
concat(1, vector(20, n, S(2*n,n))) \\ Colin Barker, Dec 24 2015

a(n) = S(2n,n) where S(d,n) = Sum_{k=0..d} C(d,k)^2*C(n-k+d-1,d-1) from formula (22) in Conway-Sloane.
a(n) ~ (1 + sqrt(2))^(4*n + 1/2) / (2^(5/4) * Pi * n). - Vaclav Kotesovec, Apr 10 2018
From Peter Bala, Dec 19 2020: (Start)
a(n) = Sum_{k = 0..n} C(2*n,n-k)^2 * C(2*n+k-1,k).
a(n) = Sum_{k = 1..n} C(2*n, k)*C(2*n+k, k)*C(n-1,k-1) for n >= 1.
a(n) = [x^n] P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A103882.
a(n) = C(2*n,n)^2 * hypergeom([-n, -n, 2*n], [n+1, n+1], 1).
n^2*(2*n - 1)^2*(24*n^3 - 105*n^2 + 152*n - 73)*a(n) = (3264*n^7 - 20808*n^6 + 53900*n^5 - 73159*n^4 + 55963*n^3 - 24107*n^2 + 5436*n - 504)*a(n - 1) - (2*n - 1)*(2*n - 3)*(n - 2)^2*(24*n^3 - 33*n^2 + 14*n - 2)*a(n - 2).
Conjectural: 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.
More generally, if r and s are positive integers, we conjecture that the same supercongruences hold for the sequence defined by [x^(r*n)] P(s*n,(1 + x)/(1 - x)). (End)
Even more generally, we conjecture that the same supercongruences hold for the sequence defined by [x^(r*n)] (1 + x)^(A*n) * (1 - x)^(B*n) * P(s*n,(1 + x)/(1 - x)), where A and B are integers. - Peter Bala, Mar 17 2023
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(k+1)*binomial(2*n-1, k)*binomial(n+k, k)* binomial(2*n+k-1, k) for n >= 1. - Peter Bala, Sep 25 2024

Formula incorrectly copied from A143699 removed by R. J. Mathar, Mar 11 2010

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

Original entry on oeis.org

1, 2, 16, 146, 1408, 14002, 142000, 1459810, 15158272, 158611106, 1669752016, 17664712562, 187641279616, 2000029880786, 21380213588848, 229129634462146, 2460955893981184, 26482855453375042, 285475524009208720, 3082024598888203090, 33319523640218177408

Offset: 0

Ralf Stephan, Feb 20 2005

From Peter Bala, Mar 01 2020: (Start)
The recurrence given below can be rewritten in the form
(2*n+1)*(2*n+2)*P(2,n)*a(n+1) - (2*n-1)*(2*n-2)*P(2,-n)*a(n-1) = Q(2,n^2)*a(n), where the polynomial Q(2,n) = 4*(55*n^2 - 34*n + 3) and the polynomial P(2,n) = 5*n^2 - 5*n + 1 satisfies the symmetry condition P(2,n) = P(2,1-n) and has real zeros.
More generally, for fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form ( Product_{k = 1..2*m} (2*m*n + k) ) * P(2*m,n)*b(n+1) + (-1)^m*( Product_{k = 1..2*m} (2*m*n - k) ) * P(2*m,-n)*b(n-1) = Q(2*m,n^2)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial P(2*m,n) = P(2*m,1-n) and has real zeros in the interval [0, 1]. The 4*m zeros of the polynomial Q(2*m,n^2) seem to belong to the interval [-1, 1] and 4*m - 2 of these zeros appear to be approximated by the rational numbers +- k/(3*m), where 1 <= k <= 3*m - 2, k not a multiple of 3. (End)

G. C. Greubel, Table of n, a(n) for n = 0..950 [a(0) = 1 inserted by _Georg Fischer_, Apr 03 2020]
Peter Bala, Notes on A103885
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A002003, A006318, A027307, A103882, A103884, A123164, A144097, A156894, A266213, A370102.

A103885:= func< n | n eq 0 select 1 else (&+[ Binomial(n, k)*Binomial(2*n+k-1, n-1): k in [0..n]]) >;
[A103885(n): n in [0..40]]; // G. C. Greubel, Oct 27 2024

a := n -> `if`(n=0, 1, 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Dec 30 2019
# Alternative (after Peter Bala ):
gf := n -> ( (1 + x)/(1 - x) )^n: ser := n -> series(gf(n), x, 40):
seq(coeff(ser(n), x, 2*n), n=0..17); # Peter Luschny, Mar 20 2020

Prepend[Table[Sum[2^i Binomial[n, i] Binomial[2n-1, i-1], {i, 1, 2n}], {n,1,20}], 1] (* Vaclav Kotesovec, Jul 01 2015 *)

a(n) = if (n==0, 1, sum(i=0, n, 2^i * binomial(n, i) * binomial(2*n-1, i-1))); \\ Michel Marcus, Mar 21 2020

def A103885(n): return 1 if n==0 else sum(binomial(n, k)*binomial(2*n+k-1, n-1) for k in range(n+1))
[A103885(n) for n in range(41)] # G. C. Greubel, Oct 27 2024

a(n) = Sum_{i=0..n} 2^i * binomial(n,i) * binomial(2*n-1,i-1). [Original definition, with summation range {i=1..n}.]
a(n) = A103884(n, n).
G.f.: A(x) = x*B(x)'/B(x), where B(x) is g.f. of A027307. - Vladimir Kruchinin, Jun 30 2015
From Vaclav Kotesovec, Jul 01 2015: (Start)
Recurrence: n*(2*n-1)*(5*n^2 - 15*n + 11)*a(n) = 2*(55*n^4 - 220*n^3 + 296*n^2 - 152*n + 24)*a(n-1) + (n-2)*(2*n-3)*(5*n^2 - 5*n + 1)*a(n-2).
a(n) ~ ((11 + 5*sqrt(5))/2)^n / (2 * 5^(1/4) * sqrt(Pi*n)). (End)
a(n) = [x^n] (1/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ...))))))^n, a continued fraction. - Ilya Gutkovskiy, Sep 29 2017
a(n) = 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2) for n >= 1. - Peter Luschny, Dec 30 2019
From Peter Bala, Mar 01 2020: (Start)
a(n) = Sum_{k = 0..n} C(n, k)*C(2*n+k-1, n-1), with a(0) = 1.
a(n) = Sum_{k = 0..n} C(2*n, 2*k)*C(2*n-k-1, n-1), with a(0) = 1.
a(n) = (1/2)*Sum_{k = 0..n} C(2*n, n-k)*C(2*n+k-1, k). Cf. A156894.
a(n) = [x^n] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
a(n) = (1/2) * [x^(n)] ( (1 + x)/(1 - x) )^(2*n). Cf. A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n.
Conjecture: a(n) = - [x^n] G(x)^(-n), where G(x) = 1 + 2*x + 14*x^2 + 134*x^3 + 1482*x^4 + ... is the o.g.f. of A144097.
a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
From Peter Bala, Sep 22 2021: (Start)
a(n) = Sum_{k = 0..n} 4^k*binomial(n+k-1,n)*binomial(n,k)^2 / binomial(2*k,k).
Equivalently, a(n) = [x^n] T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A103882. (End)
For n>0, a(n) = (1/3) * [x^n] (1/S(-x))^(3*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318. Cf. A370102. - Peter Bala, Jul 29 2024

a(0) = 1 added and new name by Peter Bala, Mar 01 2020

A331562 Number A(n,k) of sequences with k copies each of 1,2,...,n avoiding absolute differences between adjacent elements larger than one; 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, 2, 1, 1, 1, 6, 2, 1, 1, 1, 20, 12, 2, 1, 1, 1, 70, 92, 26, 2, 1, 1, 1, 252, 780, 506, 48, 2, 1, 1, 1, 924, 7002, 11482, 2288, 86, 2, 1, 1, 1, 3432, 65226, 284002, 135040, 10010, 148, 2, 1, 1, 1, 12870, 623576, 7426610, 8956752, 1543862, 41618, 250, 2, 1

Offset: 0

Alois P. Heinz, Jan 20 2020

All columns are linear recurrences with constant coefficients and for k > 0 the order of the recurrence is bounded by 3*k-1. For k up to at least 17 this upper bound is exact. - Andrew Howroyd, May 16 2020

Row 2, the sequence of central binomial numbers A000984, satisfies the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r (see Meštrović, equation 39). This is also known to be true for row 3 (A103882) and row 4 (A177316). We conjecture that each row sequence of the table satisfies the same congruences. - Peter Bala, Oct 26 2024.

			A(2,2) = 6: 1122, 1212, 1221, 2112, 2121, 2211.
A(3,2) = 12: 112233, 112323, 112332, 121233, 123321, 211233, 233211, 321123, 323211, 332112, 332121, 332211.
A(2,3) = 20: 111222, 112122, 112212, 112221, 121122, 121212, 121221, 122112, 122121, 122211, 211122, 211212, 211221, 212112, 212121, 212211, 221112, 221121, 221211, 222111.
A(3,3) = 92: 111222333, 111223233, 111223323, 111223332, ..., 333221112, 333221121, 333221211, 333222111.
Square array A(n,k) begins:
  1, 1,  1,     1,       1,         1,           1, ...
  1, 1,  1,     1,       1,         1,           1, ...
  1, 2,  6,    20,      70,       252,         924, ...
  1, 2, 12,    92,     780,      7002,       65226, ...
  1, 2, 26,   506,   11482,    284002,     7426610, ...
  1, 2, 48,  2288,  135040,   8956752,   640160976, ...
  1, 2, 86, 10010, 1543862, 276285002, 54331653686, ...

Andrew Howroyd, Antidiagonals n = 0..50, flattened (antidiagonals 0..15 from Alois P. Heinz)
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).

Columns k=0-9 give: A000012, A130130 (for n>0), A177282, A177291, A177298, A177301, A177304, A177307, A177310, A177313.
Rows n=0+1, 2-9 give: A000012, A000984, A103882, A177316, A177317, A177318, A177319, A177320, A177321.
Main diagonal gives A331623.
Cf. A269129, A275784.

b:= proc(l, q) option remember; (n-> `if`(n<2, 1, add(
     `if`(l[j]=1, `if`(j in [1, n], b(subsop(j=[][], l),
     `if`(j=1, 0, n)), 0), b(subsop(j=l[j]-1, l), j)), j=
     `if`(q<0, 1..n, max(1, q-1)..min(n, q+1)))))(nops(l))
    end:
A:= (n, k)-> `if`(k=0, 1, b([k$n], -1)):
seq(seq(A(n, d-n), n=0..d), d=0..10);

b[l_, q_] := b[l, q] = With[{n = Length[l]}, If[n < 2, 1, Sum[
      If[l[[j]] == 1, If[j == 1 || j == n, b[ReplacePart[l, j -> Nothing],
      If[j == 1, 0, n]], 0], b[ReplacePart[l, j -> l[[j]] - 1], j]], {j,
      If[q < 0, Range[n], Range[Max[1, q - 1], Min[n, q + 1]]]}]]];
A[n_, k_] := If[k == 0, 1, b[Table[k, {n}], -1]];
Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Jan 03 2021, after Alois P. Heinz *)

step(m,R)={my(M=matrix(3, m+1, q, p, q--; p--; sum(j=0, m-p-q, sum(i=max(p+j-#R+1, 2*p+q+j-m), p, R[1+q, 1+p+j-i] * binomial(p,i) * binomial(p+q+j-i-1, j) * binomial(m-1, 2*p+q+j-i-1))))); M[3,]+=2*M[2,]+M[1,]; M[2,]+=M[1,]; M}
AdjPathsBySig(sig)={if(#sig<1, 1, my(R=[1;1;1]); for(i=1, #sig-1, R=step(sig[i], R)); my(m=sig[#sig]); sum(i=1, min(m, #R), binomial(m-1, i-1)*R[3,i]))}
T(n,k) = {if(k==0, 1, AdjPathsBySig(vector(n,i,k)))} \\ Andrew Howroyd, May 16 2020

A245086 Central values of the n-th discrete Chebyshev polynomials of order 2n.

Original entry on oeis.org

1, 0, -6, 0, 90, 0, -1680, 0, 34650, 0, -756756, 0, 17153136, 0, -399072960, 0, 9465511770, 0, -227873431500, 0, 5550996791340, 0, -136526995463040, 0, 3384731762521200, 0, -84478098072866400, 0, 2120572665910728000, 0, -53494979785374631680, 0

Offset: 0

Nikita Gogin, Jul 11 2014

In the general case the n-th discrete Chebyshev polynomial of order N is D(N,n;x) = Sum_{i = 0..n} (-1)^i*C(n,i)*C(N-x,n-i)*C(x,i). For N = 2*n , x = n, one gets a(n) = D(2n,n;n) = Sum_{i = 0..n} (-1)^i*C(n,i)^3 that equals (due to Dixon's formula) 0 for odd n and (-1)^m*(3m)!/(m!)^3 for n = 2*m. (Riordan, 1968) So, a(2*m) = (-1)^m*A006480(m).

John Riordan, Combinatorial Identities, John Willey&Sons Inc., 1968.

Nikita Gogin and Mika Hirvensalo, On the Generating Function of Discrete Chebyshev Polynomials, TUCS Technical Reports 819, Turku Centre for Computer Science, 2007.
Nikita Gogin and Mika Hirvensalo, Recurrent Constructions of the MacWilliams and Chebyshev Matrices, Fundamenta Informaticae, vol.116, issue 1-4, January 2012, pages 93-110.
H. W. Gould, Tables of Combinatorial Identities, Edited by J. Quaintance.
Darij Grinberg, Introduction to Modern Algebra (UMN Spring 2019 Math 4281 Notes), University of Minnesota (2019).
Y. Sun and X. Wang, A new proof of a curious identity, Math Gazette, Vol. 91, No. 520 (Mar 2007) 105-107.
Wikipedia, Dixon's identity

Cf. A000172, A002894, A005260, A006480, A103882, A183204, A273630, A364303.

Table[Coefficient[Simplify[JacobiP[n,0,-(2*n+1),(1+t^2)/(1-t^2)]*(1-t^2)^n],t,n],{n,0,20}]

from math import factorial
def A245086(n): return 0 if n&1 else (-1 if (m:=n>>1)&1 else 1)*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022

a(n) is a coefficient at t^n in (1-t^2)^n*P(0,-(2*n+1);n;(1+t^2)/(1-t^2)), where P(a,b;k;x) is the k-th Jacobi polynomial (Gogin and Hirvensalo, 2007).
G.f.: Hypergeometric2F1[1/3,2/3,1,-27*x^2].
a(2*m+1) = 0, a(2*m) = (-1)^m*A006480(m).
From Peter Bala, Aug 04 2016: (Start)
a(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(n + k,n) (Sun and Wang).
a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k, n - k)*binomial(2*k, k)*binomial(2*n - k, n) (Gould, Vol.5, 9.23).
a(n) = -1/(n + 1)^3 * A273630(n+1). (End)
From Peter Bala, Mar 22 2022: (Start)
a(n) = - (3*(3*n-2)*(3*n-4)/n^2)*a(n-2).
a(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A002894(n) = binomial(2*n,n)^2 = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)). Cf. A103882. (End)
From Peter Bala, Jul 23 2023: (Start)
a(n) = [x^n] G(x)^(3*n), where the power series G(x) = 1 - x^2 + 2*x^4 - 14*x^6 + 127*x^8 - 1364*x^10 + ... appears to have integer coefficients.
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^3, where the power series F(x) = 1 - x^2  + 8*x^4 - 101*x^6 + 1569*x^8 - 27445*x^10 + ..., appears to have integer coefficients. See A229452.
Row 1 of A364303. (End)
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(n+k, k)^2 * binomial(3*n+1, n-k). Cf. A183204.- Peter Bala, Sep 20 2024

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.

A177316 Number of permutations of n copies of 1..4 with all adjacent differences <= 1 in absolute value.

Original entry on oeis.org

1, 2, 26, 506, 11482, 284002, 7426610, 201922730, 5650739930, 161686253810, 4708709084026, 139111173397066, 4159013698117618, 125595645802182818, 3825428523179727266, 117382025506323434506, 3625185567639373456090, 112597953571519245194770

Offset: 0

R. H. Hardin, May 06 2010

nonn

See A103882 and A177317 through A177328 for the number of permutations of n copies of 1..k (for different values of k) with adjacent differences restricted in size. We conjecture that all these sequences satisfy the congruences A(n*p^k) == A(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k and any prime p >= 5. - Peter Bala, Jan 16 2020

Alois P. Heinz, Table of n, a(n) for n = 0..656 (terms n=1..31 from R. H. Hardin)

Cf. A005259, A103882, A177317 - A177328, A352653.

Row n=4 of A331562.

a:= proc(n) option remember; `if`(n<3, [1, 2, 26][n+1],
       (3*((105*n^4-356*n^3+402*n^2-208*n+43)*a(n-1)
      -(105*n^4-904*n^3+2868*n^2-3932*n+1930)*a(n-2))
      +(9*n-11)*(n-3)^3*a(n-3))/((9*n-16)*n^3))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, Jan 22 2020
A177316 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1):
seq(simplify(A177316(n)), n = 0..17); # Peter Luschny, Mar 27 2023

a[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1];
Table[a[n], {n, 0, 17}] (* Jean-François Alcover, May 28 2023, after Peter Luschny *)

def A177316(n):
    if n == 0: return 1
    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 # Chai Wah Wu, Oct 03 2022

From Peter Bala, Jan 14 2020: (Start)
Conjecture: a(n) = (1/3)*( A005259(n) + A005259(n-1) ).
Equivalently, a(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k-1,k)^2. Cf. A103882. If true, then the sequence satisfies the recurrence a(n) = (2*(102*n^6 - 612*n^5 + 1462*n^4 - 1768*n^3 + 1143*n^2 - 382*n+52) * a(n-1) - (2*n-1)*(3*n^2 - 3*n+1) * (n-2)^3 * a(n-2)) / (n^3*(2*n - 3) * (3*n^2 - 9*n+7)) and the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k and any prime p >= 5. [added Apr 18 2022: assuming the recurrence given in the Maple program below is correct then these conjectures are true.] (End)
a(n) = 2*A352653(n) for n >= 1. - Peter Bala, Apr 18 2022
a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1). - Peter Luschny, Mar 27 2023
a(n) ~ (1 + sqrt(2))^(4*n) / (2^(5/4) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2023

a(0)=1 prepended by Alois P. Heinz, Jan 20 2020

A361715 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k-1,k).

Original entry on oeis.org

0, 1, 9, 82, 745, 6876, 64764, 621860, 6070761, 60085720, 601493134, 6078225792, 61907445340, 634751002718, 6545478537810, 67830084149832, 705950951578089, 7375212511115184, 77310175072063914, 812839577957617640, 8569327793354169870, 90562666708303706642, 959212007563384494522, 10180245921386807485152

Offset: 0

Peter Bala, Mar 23 2023

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.
Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k), which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

Cf. A005258, A103882, A361712, A361713, A361714, A361717.

seq( add( binomial(n,k)^2*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
#faster alternative program
P(n) := 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532:
Q(n) := (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936):
R(n) := (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320):
a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 elif n = 2 then 9 else (Q(n)*a(n-1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3))/((n - 1)*(n - 2)*n^2*P(n)) end if; end:
seq(a(n), n = 0..25);
# Alternative:
A361715 := n -> hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n):
seq(simplify(A361715(n)), n = 0..23); # Peter Luschny, Mar 27 2023

Table[Sum[Binomial[n,k]^2 Binomial[n+k-1,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Nov 01 2023 *)

a(n) = A103882(n) - binomial(2*n-1,n) = (3*A005258(n) + A005258(n-1))/5 - binomial(2*n-1,n) for n >= 1.
a(n) ~ sqrt(sqrt(5)/10 + 1/4)*(5*sqrt(5)/2 + 11/2)^n/(Pi*n)
P-recursive:
(n - 1)*(n - 2)*n^2*P(n)*a(n) = Q(n)*a(n - 1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 9 and where
P(n) = 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532,
Q(n) = (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936) and
R(n) = (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320).
a(n) = hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n). This is another way to write the first formula. - Peter Luschny, Mar 27 2023

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

A103881 Square array T(n,k) (n >= 1, k >= 0) read by antidiagonals: coordination sequence for root lattice A_n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

SageMath

Formula

Extensions

A156554 The number of integer sequences of length d = 2n+1 such that the sum of the terms is 0 and the sum of the absolute values of the terms is d-1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

Extensions

A331562 Number A(n,k) of sequences with k copies each of 1,2,...,n avoiding absolute differences between adjacent elements larger than one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A245086 Central values of the n-th discrete Chebyshev polynomials of order 2n.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Python

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma