cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A006480 De Bruijn's S(3,n): (3n)!/(n!)^3.

Original entry on oeis.org

1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000
Offset: 0

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Author

Keywords

Comments

Number of paths of length 3n in an n X n X n grid from (0,0,0) to (n,n,n), using steps (0,0,1), (0,1,0), and (1,0,0).
Appears in Ramanujan's theory of elliptic functions of signature 3.
S(s,n) = Sum_{k=0..2n} (-1)^(k+n) * binomial(2n, k)^s. The formula S(3,n) = (3n)!/(n!)^3 is due to Dixon (according to W. N. Bailey 1935). - Charles R Greathouse IV, Dec 28 2011
a(n) is the number of ballot results that end in a 3-way tie when 3n voters each cast two votes for two out of three candidates vying for 2 slots on a county board; in such a tie, each of the three candidates receives 2n votes. Note there are C(3n,2n) ways to choose the voters who cast a vote for the youngest candidate. The n voters who did note vote for the youngest candidate voted for the two older candidates. Then there are C(2n,n) ways to choose the other n voters who voted for both the youngest and the second youngest candidate. The remaining voters vote for the oldest candidate. Hence there are C(3n,2n)*C(2n,n)=(3n)!/(n!)^3 ballot results. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X+Y+1/(X*Y))^(3*n). - Mark van Hoeij, May 07 2013
For n > 2 a(n) is divisible by (n+2)*(n+1)^2, a(n) = (n+1)^2*(n+2)*A161581(n). - Alexander Adamchuk, Dec 27 2013
a(n) is the number of permutations of the multiset {1^n, 2^n, 3^n}, the number of ternary words of length 3*n with n of each letters. - Joerg Arndt, Feb 28 2016
Diagonal of the rational function 1/(1 - x - y - z). - Gheorghe Coserea, Jul 06 2016
At least two families of elliptic curves, x = 2*H1 = (p^2+q^2)*(1-q) and x = 2*H2 = p^2+q^2-3*p^2*q+q^3 (0Bradley Klee, Feb 25 2018
The ordinary generating function also determines periods along a family of tetrahedral-symmetric sphere curves ("du troisième ordre"). Compare links to Goursat "Étude des surfaces..." and "Proof Certificate". - Bradley Klee, Sep 28 2018

Examples

			G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...
		

References

  • L. A. Aizenberg and A. P. Yuzhakov, "Integral representations and residues in multidimensional complex analysis", American Mathematical Society, 1983, p. 194.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 174.
  • N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 3 of A187783.
Related to diagonal of rational functions: A268545-A268555. Elliptic Integrals: A002894, A113424, A000897. Factors: A005809, A000984. Integrals: A007004, A024486. Sphere Curves: A318245, A318495.

Programs

  • GAP
    List([0..20],n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018
    
  • Magma
    [Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011
    
  • Maple
    seq((3*n)!/(n!)^3, n=0..16); # Zerinvary Lajos, Jun 28 2007
  • Mathematica
    Sum [ (-1)^(k+n) Binomial[ 2n, k ]^3, {k, 0, 2n} ]
    a[ n_] := If[ n < 0, 0, (-1)^n HypergeometricPFQ[ {-2 n, -2 n, -2 n}, {1, 1}, 1]]; (* Michael Somos, Oct 22 2014 *)
    Table[Multinomial[n, n, n], {n, 0, 100}] (* Emanuele Munarini, Oct 25 2016 *)
    CoefficientList[Series[Hypergeometric2F1[1/3,2/3,1,27*x],{x,0,5}],x] (* Bradley Klee, Feb 28 2018 *)
    Table[(3n)!/(n!)^3,{n,0,20}] (* Harvey P. Dale, Mar 09 2025 *)
  • Maxima
    makelist(multinomial_coeff(n,n,n),n,0,24); /* Emanuele Munarini, Oct 25 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n!^3)}; /* Michael Somos, Dec 03 2002 */
    
  • PARI
    {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while( m<=n, m*=3; A = subst( (1 + 2*x) * subst(A, x, (x/3)^3), x, serreverse(x * (1 + x + x^2) / (1 + 2*x)^3 / 3 + O(x^m)))); polcoeff(A, n))}; /* Michael Somos, Dec 03 2002 */
    
  • Python
    from math import factorial
    def A006480(n): return factorial(3*n)//factorial(n)**3 # Chai Wah Wu, Oct 04 2022

Formula

Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 1/2 * sqrt(3) * 27^n / (Pi*n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Karol A. Penson, Nov 21 2001: (Start)
O.g.f.: hypergeom([1/3, 2/3], [1], 27*x).
E.g.f.: hypergeom([1/3, 2/3], [1, 1], 27*x).
Integral representation as n-th moment of a positive function on [0, 27]:
a(n) = int( x^n*(-1/24*(3*sqrt(3)*hypergeom([2/3, 2/3], [4/3], 1/27*x)* Gamma(2/3)^6*x^(1/3) - 8*hypergeom([1/3, 1/3], [2/3], 1/27*x)*Pi^3)/Pi^3 /x^(2/3)/Gamma(2/3)^3), x=0..27). This representation is unique. (End)
a(n) = Sum_{k=-n..n} (-1)^k*binomial(2*n, n+k)^3. - Benoit Cloitre, Mar 02 2005
a(n) = C(2n,n)*C(3n,n) = A104684(2n,n). - Paul Barry, Mar 14 2006
G.f. satisfies: A(x^3) = A( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x). - Paul D. Hanna, Oct 29 2010
D-finite with recurrence: n^2*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = (n+1)^2*(n+2)*A161581(n) for n>2. - Alexander Adamchuk, Dec 27 2013
0 = a(n)^2*(472392*a(n+1)^2 - 83106*a(n+1)*a(n+2) + 3600*a(n+2)^2) + a(n)*a(n+1)*(-8748*a(n+1)^2 + 1953*a(n+1)*a(n+2) - 120*a(n+2)^2) + a(n+1)^2*(36*a(n+1)^2 - 12*a(n+1)*a(n+2) + a(n+2)^2) for all n in Z. - Michael Somos, Oct 22 2014
0 = x*(27*x-1)*y'' + (54*x-1)*y' + 6*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
From Peter Bala, Jul 15 2016: (Start)
a(n) = 3*binomial(2*n - 1,n)*binomial(3*n - 1,n) = 3*[x^n] 1/(1 - x)^n * [x^n] 1/(1 - x)^(2*n) for n >= 1.
a(n) = binomial(2*n,n)*binomial(3*n,n) = ([x^n](1 + x)^(2*n)) *([x^n](1 + x)^(3*n)) = [x^n](F(x)^(6*n)), where F(x) = 1 + x + 2*x^2 + 14*x^3 + 127*x^4 + 1364*x^5 + 16219*x^6 + ... appears to have integer coefficients. Cf. A002894.
This sequence occurs as the right-hand side of several binomial sums:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)^3 = a(n) (Dixon's identity).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n,n - k)*binomial(3*n + k,k) = a(n) (Gould, Vol. 4, 6.86)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = a(n).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n + k,k)*binomial(3*n,n - k) = a(n).
Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = a(n).
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(4*n - k,2*n) = a(n) (see Gould, Vol.5, 9.23).
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = a(n) (Sprugnoli, Section 2.9, Table 10, p. 123). (End)
From Bradley Klee, Feb 28 2018: (Start)
a(n) = A005809(n)*A000984(n).
G.f.: F(x) = 1/(2*Pi) Integral_{z=0..2*Pi} 2F1(1/3,2/3; 1/2; 27*x*sin^2(z)) dz.
With G(x) = x*2F1(1/3,2/3; 2; 27*x): F(x) = d/dx G(x). (Cf. A007004) (End)
F(x)*G(1/27-x) + F(1/27-x)*G(x) = 1/(4*Pi*sqrt(3)). - Bradley Klee, Sep 29 2018
Sum_{n>=0} 1/a(n) = A091683. - Amiram Eldar, Nov 15 2020
From Peter Bala, Sep 20 2021: (Start)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)^2 * binomial(k,n). Cf. A001459.
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(3*n,2*n)*C(2*n,n) and apply Mestrovic, equation 39, p. 12). (End)
a(n) = 6*A060542(n). - R. J. Mathar, Jun 21 2023
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^3 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^3 * binomial(2*n, n+k)^3 = x*(x + n)*(x + 2*n)*a(n) (x arbitrary). Compare with Dixon's identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^3 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Aug 14 2023: (Start)
a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
Row 1 of A364509. (End)
From Peter Bala, Oct 10 2024: (Start)
The following hold for n >= 1:
a(n) = Sum_{k = 0.. 2*n} (-1)^(n+k) * k/n * binomial(2*n, k)^3 = 3/2 * Sum_{k = 0.. 2*n} (-1)^(n+k) * (k/n)^2 * binomial(2*n, k)^3.
a(n) = 3/2 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)^2*binomial(2*n-1, k).
a(n) = 3 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)*binomial(2*n-1, k)^2. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(2*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

Extensions

a(14)-a(16) from Eric W. Weisstein
Terms a(17) and beyond from T. D. Noe, Jun 29 2008

A364506 Square array read by ascending antidiagonals: T(n,k) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 40, 90, 20, 1, 350, 5880, 1680, 70, 1, 3528, 594594, 1101100, 34650, 252, 1, 38808, 75088728, 1299170600, 229265400, 756756, 924, 1, 453024, 10861066216, 2066315135040, 3164045050530, 50678855040, 17153136, 3432, 1, 5521230, 1721929279200, 3943172216808000
Offset: 0

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Author

Peter Bala, Jul 27 2023

Keywords

Comments

Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L we can define the factorial ratio sequence u_k(c, d) = (c_1*k)!*(c_2*k)!* ... *(c_K*k)!/ ( (d_1*k)!*(d_2*k)!* ... *(d_L*k)! ) and ask whether it is integral for all k >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
It is known that both row 0, the central binomial numbers, and row 1, the de Bruijn numbers, satisfy 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. We conjecture that all the row sequences of the table satisfy the same supercongruences.

Examples

			 Square array begins:
 n\k|  0     1         2              3                  4                 5
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1     2         6             20                 70               252 ...
  1 |  1     6        90           1680              34650            756756 ...
  2 |  1    40      5880        1101100          229265400       50678855040 ...
  3 |  1   350    594594     1299170600      3164045050530  8188909171581600 ...
  4 |  1  3528  75088728  2066315135040  63464046079757400  ...
  5 |  1 38808  ...
		

Crossrefs

A000984 (row 0), A006480 (row 1), A364507 (row 2), A364508 (row 3). Cf. A364303, A364509, A365025.

Programs

  • Maple
    # display as a square array
    T(n,k) := (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/((n*k)!^2 * ((n+1)*k)!^2):
    seq( print(seq(T(n,k), k = 0..10)), n = 0..10);
    # display as a sequence
    seq( seq(T(n-k,k), k = 0..n), n = 0..10);

Formula

T(n,k) = Sum_{i = -k..k} (-1)^i * binomial(2*k, k+i) * binomial(2*n*k, n*k+i)^2 (shows that the table entries are integers).
For n >= 1, T(n,k) = (-1)^k * binomial(2*n*k, (n+1)*k)^2 * hypergeom([-2*k, -(n+1)*k, -(n+1)*k], [1 + (n-1)*k, 1 + (n-1)*k], 1) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ) by Dixon's 3F2 summation theorem.
T(n,k) = (-1)^k * [x^((n + 1)*k)] ( (1 - x)^(2*(n+1)*k) * Legendre_P(2*n*k, (1 + x)/(1 - x)) ). - Peter Bala, Aug 15 2023

A364513 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k-1, (1 + x)/(1 - x)) for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, -2, 6, 1, 0, 0, -20, 1, 4, 0, 16, 70, 1, 10, 126, 0, 0, -252, 1, 18, 594, 4900, 0, -252, 924, 1, 28, 1716, 44200, 209950, 0, 0, -3432, 1, 40, 3900, 205920, 3640210, 9513504, 0, 4800, 12870, 1, 54, 7650, 685216, 27386100, 317678760, 447103440, 0, 0, -48620, 1, 70, 13566, 1847560, 133501500, 3861534768, 28782923400, 21558808128, 0, -100100, 184756
Offset: 0

Views

Author

Peter Bala, Jul 31 2023

Keywords

Comments

Compare with A364303.
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L), where c_1 + ... + c_K = d_1 + ... + d_L, we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431). Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row of the present table is an integral factorial ratio sequence of height 2. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.

Examples

			 Square array begins:
 n\k|  0    1       2         3            4              5
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1   -2       6       -20           70           -252   ... (see A000984)
  1 |  1   -2       0        16            0           -252   ...  A364514
  2 |  1    0       0         0            0              0
  3 |  1    4     126      4900       209950        9513504   ...  (1/3)*A352651
  4 |  1   10     594     44200      3640210      317678760   ...  A364515
  5 |  1   18    1716    205920     27386100     3861534768   ...  (3/5)*A352652
  6 |  1   28    3900    685216    133501500    27583083528   ...  A364516
  7 |  1   40    7650   1847560    494944450   140625140040   ...  A364517
		

Crossrefs

Cf. A000984 (row 0 unsigned), A276098, A295431, A352651 (3*row 3), A352652 ((5/3)*row 5), A364303, A364506, A364509, A364514 (row 1), A364515 (row 4), A364516 (row 6), A364517 (row 7).

Programs

  • Maple
    T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k-1, (1 + x)/(1 - x)), x, 11), x, k):
    # display as a square array
    seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
    # display as a sequence
    seq(seq(T(n-k, k), k = 0..n), n = 0..10);

Formula

T(n,k) = Sum_{i = 0..k} binomial(n*k-1, k-i)^2 * binomial((n-2)*k+i-2, i).
T(n,1) = 1 for all n and for n >= 2 and k >= 1, T(n,k) = binomial((k*n-1), k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 2)*k - 1 and b = -k.
For n >= 3 and k >= 1, T(n,k) = ((n*k - 1))! * ( ((n+2)*k - 1)/2 )! * ( ((n-2)*k - 1)/2 )! / ( k!^2 * ((n-2)*k - 1)! * ((n*k - 1)/2)!^2 ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.
For n >= 3 and k >= 1, T(n,k) = (n-2)/n * ((n+2)*k)!*(n*k/2)!^2 / ( ((n+2)*k/2)! * (n*k)! * ((n-2)*k/2)! * k!^2 ).
The central binomial numbers A000984, row 1 unsigned, satisfy the supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that each row sequence of the table satisfies the same supercongruences.

A364510 a(n) = binomial(4*n, n)^2.

Original entry on oeis.org

1, 16, 784, 48400, 3312400, 240374016, 18116083216, 1401950721600, 110634634890000, 8862957169158400, 718528370729238784, 58818762721626513424, 4853704694918904043024, 403242220875862752160000, 33694913171561404510440000, 2829611125043050701300998400
Offset: 0

Views

Author

Peter Bala, Jul 28 2023

Keywords

Crossrefs

Programs

  • Maple
    seq( binomial(4*n,n)^2, n = 0..15);
  • Mathematica
    A364510[n_]:=Binomial[4n,n]^2;Array[A364510,15,0] (* Paolo Xausa, Oct 05 2023 *)

Formula

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(4*n, 2*n+i).
Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.
a(n) = A005810(n)^2.
P-recursive: a(n) = 16 * ( (4*n - 1)*(4*n - 2)*(4*n - 3)/(3*n*(3*n - 1)*(3*n - 2)) )^2 * a(n-1) with a(0) = 1.
a(n) ~ c^n * 2/(3*Pi*n), where c = (2^16)/(3^6).
a(n) = [x^n] G(x)^(16*n), where the power series G(x) = 1 + x + 9*x^2 + 225*x^3 + 7525*x^4 + 295228*x^5 + 12787152*x^6 + 592477457*x^7 + 28827755219*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^16, where the power series F(x) = 1 + x + 25*x^2 + 1033*x^3 + 53077*x^4 + 3081944*x^5 + 193543624*x^6 + 12835533333*x^7 + 886092805699*x^8 + ... appears to have integer coefficients.
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.
a(n) = [x^(3*n)] ( (1 - x)^(6*n)*Legendre_P(2*n, (1 + x)/(1 - x)) ). - Peter Bala, Aug 14 2023

A364511 a(n) = (6*n)!*(5*n)!*(2*n)! / ((4*n)!^2 * (3*n)! * n!^2).

Original entry on oeis.org

1, 50, 8910, 2011100, 503909070, 133954543800, 36992598142500, 10491379251679040, 3034472729231379150, 891028813210575018980, 264787855164104281785160, 79455812929030151249454000, 24035311050907120054564683300, 7320107028326385998504601648000
Offset: 0

Views

Author

Peter Bala, Jul 28 2023

Keywords

Comments

Row 4 of A364509.

Crossrefs

Programs

  • Maple
    seq((6*n)!*(5*n)!*(2*n)! / ((4*n)!^2 * (3*n)! * n!^2), n = 0..15);
  • Mathematica
    A364511[n_]:=(6n)!(5n)!(2n)!/((4n)!^2(3n)!n!^2);Array[A364511,15,0] (* Paolo Xausa, Oct 05 2023 *)

Formula

a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(6*n, 3*n+i).
Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.
a(n) = (-1)^n*binomial(6*n,2*n) * hypergeom([-2*n, -2*n, -4*n], [1, 2*n+1], 1).
P-recursive: a(n) = (15/4)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*(6*n-1)*(6*n-5)/((4*n-1)^2*(4*n-3)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(5)/(4*Pi*n), where c = (3^3)*(5^5)/(2^8).
a(n) = [x^n] G(x)^(10*n), where the power series G(x) = 1 + 5*x + 208*x^2 + 19960*x^3 + 2580710*x^4 + 390721786*x^5 + 65243160516*x^6 + 11646611942100*x^7 + 2182248792056787*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^10, where the power series F(x) = 1 + 5*x + 458*x^2 + 69285*x^3 + 13037740*x^4 + 2773287786*x^5 + 638122182196*x^6 + 155077758079485*x^7 + 39234250338228617*x^8 + ... appears to have integer coefficients.
Conjecture: 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 [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since a(n) = binomial(6*n, 3*n)*binomial(5*n, n)* binomial(2*n, n)/binomial(4*n, n)].
a(n) = (-1)^n * [x^(4*n)] (1 - x)^(8*n)*Legendre_P(2*n, (1 + x)/(1 - x)). - Peter Bala, Aug 14 2023

A365025 Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ) for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 126, 300, 1, 1, 1716, 79380, 11440, 1, 1, 24310, 20612592, 65523780, 485100, 1, 1, 352716, 5318784900, 328206021000, 60634147860, 21841260, 1, 1, 5200300, 1368494343216, 1552041334596844, 5876083665270000, 59774707082376, 1022041020, 1
Offset: 0

Views

Author

Peter Bala, Aug 17 2023

Keywords

Comments

Fractional factorials are defined in terms of the gamma function; for example, ((2*n+1/2)*k)! = Gamma(1 + (2*n+1/2)*k).
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L we can define the factorial ratio sequence u_k(c, d) = (c_1*k)!*(c_2*k)!* ... *(c_K*k)!/ ( (d_1*k)!*(d_2*k)!* ... *(d_L*k)! ) and ask whether it is integral for all k >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
It is usually assumed that the c's and d's are integers but here we allow for some of the c's and d's to be half-integers. See A276098 for further examples of this type.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
Conjecture: each row sequence of the table 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.

Examples

			 Square array begins:
 n\k|  0      1               2                    3                      4
  - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1      1               1                    1                      1  ...
  1 |  1     10             300                11440                 485100  ...
  2 |  1    126           79380             65523780            60634147860  ...
  3 |  1   1716        20612592         328206021000       5876083665270000  ...
  4 |  1  24310      5318784900     1552041334596844  510031828417402714500  ...
  5 |  1 352716   1368494343216  7108360304262169344 ...
		

Crossrefs

Cf. A275652 (row 1), A365026 (row 2), A365027 (row 3).

Programs

  • Maple
    # display as a square array
    T(n, k) := (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ):
    seq( print(seq(simplify(T(n, k)), k = 0..10)), n = 0..10);
    # display as a sequence
    seq( seq(simplify(T(n-k, k)), k = 0..n), n = 0..10);
  • Python
    from itertools import count, islice
    from math import factorial
    from sympy import factorial2
    def A365025_T(n,k): return int(factorial2(k)*factorial(r:=((m:=n<<1)+1)*k)*factorial2(((m<<1)+1)*k)//((factorial(n*k)*factorial2(r))**2*factorial(k)))
    def A365025_gen(): # generator of terms
        for n in count(0):
            yield from (A365025_T(n-k,k) for k in range(n+1))
    A365025_list = list(islice(A365025_gen(),20)) # Chai Wah Wu, Aug 24 2023

Formula

T(n,k) = Sum_{j = 0..n*k} binomial((2*n+1)*k, n*k-j)^2 * binomial(k+j-1, j).
T(n,k) = binomial((2*n+1)*k,n*k)^2 * hypergeom([k, -n*k, -n*k], [1 + (n+1)*k, 1 + (n+1)*k], 1) = (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ) by Dixon's 3F2 summation theorem.
T(n,k) = [x^(n*k)] ( (1 - x)^(2*n*k) * Legendre_P((2*n+1)*k, (1 + x)/(1 - x)) ).
T(n,k) = k!!*((2*n+1)*k)!*((4*n+1)*k)!!/(k!*((n*k)!*((2*n+1)*k)!!)^2). - Chai Wah Wu, Aug 24 2023

A364518 Square array read by ascending antidiagonals: T(n,k) = [x^(2*k)] ( (1 + x)^(n+2)/(1 - x)^(n-2) )^k for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 6, -10, -20, 1, 16, 70, 0, 70, 1, 30, 630, 924, 198, -252, 1, 48, 2310, 28672, 12870, 0, 924, 1, 70, 6006, 204204, 1385670, 184756, -4420, -3432, 1, 96, 12870, 860160, 19122246, 69206016, 2704156, 0, 12870, 1, 126, 24310, 2704156, 130378950, 1848483780, 3528923580, 40116600, 104006, -48620
Offset: 0

Views

Author

Peter Bala, Aug 07 2023

Keywords

Comments

Compare with A364303 and A364519.
Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L), where c_1 + ... + c_K = d_1 + ... + d_L, we can define the factorial ratio sequence u_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431). Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row of the present table is an integral factorial ratio sequence of height 1. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.
It is known that the unsigned version of row 0 (the central binomial numbers A000984) and row 2 satisfy 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. We conjecture that all the row sequences of the table satisfy the same supercongruences.

Examples

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252   ... see A000984
  1 |  1   0    -10        0         198             0   ... see A211419
  2 |  1   6     70      924       12870        184756   ... A001448
  3 |  1  16    630    28672     1385670      69206016   ... A091496
  4 |  1  30   2310   204204    19122246    1848483780   ... A061162
  5 |  1  48   6006   860160   130378950   20392706048   ... A276098
  6 |  1  70  12870  2704156   601080390  137846528820   ... A001448 bisected
  7 |  1  96  24310  7028736  2149374150  678057476096   ... A276099
		

Crossrefs

Cf. A000984 (row 0 unsigned), A211419 (row 1 unsigned without 0's), A001448 (row 2), A091496 (row 3), A061162 (row 4), A276098 (row 5), A001448 bisected (row 6), A276099 (row 7).

Programs

  • Maple
    T(n,k) = add( binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j), j = 0..2*k):
    # display as a square array
    seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
    # display as a sequence
    seq(seq(T(n-k, k), k = 0..n), n = 0..10);
  • PARI
    T(n,k) = sum(j = 0, 2*k, binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j));
    lista(nn) = for( n=0, nn, for (k=0, n, print1(T(n-k, k), ", "))); \\ Michel Marcus, Aug 13 2023

Formula

T(n,k) = Sum_{j = 0..2*k} binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j).
T(2,k) = binomial(4*k,2*k).
For n >= 3, T(n,k) = binomial(n*k-1,2*k) * hypergeom([-(n+2)*k, -2*k], [1 - n*k], -1) except when (n,k) = (3,1).
For n >= 2, T(n,k) = ((n+2)*k)!*((n-2)*k/2)!/(((n+2)*k/2)!*((n-2)*k)!*(2*k)!) by Kummer's Theorem.
T(n,k) = [x^k] (1 - x)^(2*k) * Chebyshev_T(n*k, (1 + x)/(1 - x)).
T(n,k) = Sum_{j = 0..k} binomial(2*n*k, 2*j)*binomial((n-1)*k-j-1, k-j).
For n >= 3, T(n,k) = binomial((n-1)*k-1,k) * hypergeom([-n*k, -k, -n*k + 1/2], [1 - (n-1)*k, 1/2], 1).
The row generating functions are algebraic functions over the field of rational functions Q(x).
Showing 1-7 of 7 results.