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 1-6 of 6 results.

A275652 a(n) = binomial(3*n,3*n/2)*binomial(2*n,n)*binomial(5*n/2,n/2)/binomial(n,n/2).

Original entry on oeis.org

1, 10, 300, 11440, 485100, 21841260, 1022041020, 49128552000, 2408829328620, 119918393838100, 6042249840712800, 307438844121252480, 15770112362658517500, 814459593645444166560, 42308586942403276440000, 2208850973597860123741440, 115825519836558228435979500
Offset: 0

Views

Author

Peter Bala, Aug 04 2016

Keywords

Comments

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)* binomial(3*n+k,3*n-k)*binomial(2*k,k)*binomial(2*n-k,n) = binomial(3*n,3*n/2)*binomial(2*n,n)*binomial(5*n/2,n/2)/binomial(n,n/2).
We also have Sum_{k = 0..3*n} (-1)^k*binomial(3*n+k,3*n-k)* binomial(2*k,k)*binomial(2*n-k,n) = binomial(3*n,3*n/2)*binomial(2*n,n)* binomial(5*n/2,n/2) /binomial(n,n/2).
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n+k,2*n-k)* binomial(2*k,k)*binomial(2*n-k,n) = binomial(2*n,n)^2 = A002894(n). See also A275653, A275654 and A275655.

Crossrefs

Cf. A002894, A245086, A275653, A275654, A275655, A276098, A276100, A276101, A276102, A352651, A352652, A365025.

Programs

  • Maple
    seq(simplify(factorial(3*n)*factorial(n/2)*factorial(5*n/2)/(factorial(n)^3*factorial(3*n/2)^2)), n = 0 .. 20);
  • Mathematica
    Table[Binomial[3 n, 3 n/2] Binomial[2 n, n] Binomial[5 n/2, n/2] / Binomial[n, n/2], {n, 0, 16}] (* Michael De Vlieger, Aug 07 2016 *)
  • PARI
    a(n) = sum(k = 0, n, binomial(2*n-k-1,n-k)*binomial(3*n,k)^2); \\ Michel Marcus, Apr 21 2022
  • Python
    from math import factorial
from sympy import factorial2
def A275652(n): return int(factorial(3*n)*factorial2(5*n)*factorial2(n)//factorial2(3*n)**2//factorial(n)**3) # Chai Wah Wu, Aug 08 2023

Formula

a(n) = (3*n)!*(5*n/2)!*(n/2)!/((3*n/2)!^2*n!^3).
Recurrence: a(n) = 5*(3*n - 1)*(3*n - 5)*(5*n - 2)*(5*n - 4)*(5*n - 6)*(5*n - 8)/(n^2*(n - 1)^2*(3*n - 2)*(3*n - 4)) * a(n-2).
a(n) = [x^n] G(x)^(5*n), where G(x) = 1 + 2*x + 12*x^2 + 184*x^3 + 3811*x^4 + 92796*x^5 + 2497358*x^6 + ... appears to have integer coefficients.
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^5, where F(x) = 1 + 2*x + 32*x^2 + 824*x^3 + 26291*x^4 + 947506*x^5 + 36934522*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(5/3)*5^(5*n/2)/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 22 2022: (Start)
For n >= 1, a(n) = (5/3)*binomial(m*n,2*n)*binomial(m*n/2,2*n)*binomial(2*n,n)^2/ binomial(m*n/2,n)^2 at m = -1. See A352651 for the case m = 1.
a(n) = Sum_{k = 0..n} binomial(2*n-k-1,n-k)*binomial(3*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(3*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A245086.
a(p) == a(1) (mod p^3) for prime p >= 5.
Conjecture: 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. (End)
Row 1 of A365025. - Peter Bala, Aug 18 2023

A364303 Square array read by ascending antidiagonals: T(n,k) = [x^k] (1 - x)^(2*k) * Legendre_P(n*k, (1 + x)/(1 - x)) for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 4, -6, -20, 1, 10, 36, 0, 70, 1, 18, 300, 400, 90, -252, 1, 28, 1050, 11440, 4900, 0, 924, 1, 40, 2646, 77616, 485100, 63504, -1680, -3432, 1, 54, 5544, 316540, 6370650, 21841260, 853776, 0, 12870, 1, 70, 10296, 972400, 42031990, 554822268, 1022041020, 11778624, 34650, -48620
Offset: 0

Views

Author

Peter Bala, Jul 19 2023

Keywords

Comments

The first row of the table is a signed version of the central binomial coefficients A000984. The central binomial coefficients 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 (see Meštrović, equation 39). We conjecture that each row sequence of the table satisfies the same supercongruences.

Examples

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252    ...  (-1)^k*A000984(k)
  1 |  1   0     -6        0          90             0    ...  A245086
  2 |  1   4     36      400        4900         63504    ...  A002894
  3 |  1  10    300    11440      485100      21841260    ...  A275652
  4 |  1  18   1050    77616     6370650     554822268    ...  A275653
  5 |  1  28   2646   316540    42031990    5921058528    ...  A275654
  6 |  1  40   5544   972400   189290920   39089615040    ...  A275655
  7 |  1  54  10296  2484000   665091000  188907932304    ...  A364304
  8 |  1  70  17550  5567380  1960044750  732012601320    ...  A364305

Links

Crossrefs

Cf. A000984 (row 0 unsigned), A245086 (row 1), A002894 (row 2), A275652 (row 3), A275653 (row 4), A275654 (row 5), A275655 (row 6), A364304 (row 7), A364305 (row 8).
Cf. A364113, A364298

Programs

  • Maple
    T(n,k) := coeff(series( (1 - x)^(2*k) * LegendreP(n*k, (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, k-i)^2 * binomial((n-2)*k+i-1, i).
For n >= 2, T(n,k) = binomial((n-1)*k, k)^2 * hypergeom([a, b, b], [1 + a - b, 1 + a - b], 1), where a = (n - 3)*k and b = -k.
For n >= 3, T(n,k) = ((n - 1)*k)! * ((n + 1)*k/2)! * ((n - 3)*k/2)! / ( ((n - 1)*k/2)!^2 * k!^2 * ((n - 3)*k)! ) by Dixon's 3F2 summation theorem, where fractional factorials are defined in terms of the gamma function.

A352652 a(n) = ( binomial(7*n,2*n)*binomial(7*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(7*n/2,n)^2.

Original entry on oeis.org

1, 30, 2860, 343200, 45643500, 6435891280, 942422020540, 141696569678400, 21724714133822700, 3381208130986900500, 532553441617598475360, 84695057996350934903680, 13578009523892192555221500, 2191530567314796197691108600, 355765014009052303028935320000
Offset: 0

Views

Author

Peter Bala, Apr 03 2022

Keywords

Comments

We write x! as shorthand for Gamma(x+1) and binomial(x,y) as shorthand for x!/(y!*(x-y)!) = Gamma(x+1)/(Gamma(y+1)*Gamma(x-y+1)).
Given two sequences of numbers 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. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2. Usually, it is 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 rational numbers. See A276098 and the cross references for further examples of factorial ratios of this type.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. The case n = k = 1 is easily proved.

Examples

			Examples of supercongruences:
a(11) - a(1) = 84695057996350934903680 - 30 = 2*(5^2)*(11^3)*23*593* 3671*5693*4464799 == 0 (mod 11^3)
a(2*7) - a(2) = 355765014009052303028935320000 - 2860 = (2^2)*5*(7^3)*11* 269*3307*375101*14129010228023 == 0 (mod 7^3)

Links

Crossrefs

Cf. A275654, A276098, A276100, A276101, A276102, A352651.

Programs

  • Maple
    a := n -> if n = 0 then 1 elif n = 1 then 30 else
7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) *a(n-2) end if:
seq(a(n), n = 0..20);
  • Python
    from math import factorial
from sympy import factorial2
def A352652(n): return int(factorial(7*n)*factorial2(5*n)**2//factorial(5*n)//factorial2(7*n)//factorial2(3*n)//factorial(n)**2) # Chai Wah Wu, Aug 08 2023

Formula

a(n) = (5/3)*Sum_{k = 0..n} (-1)^(n+k)*binomial(7*n,n-k)*binomial(5*n+k-1,k)^2 for n >= 1 (this formula shows 3*a(n) is integral; how to show a(n) is integral?).
a(n) = (5/3)*Sum_{k = 0..n} binomial(4*n-k-2,n-k)*binomial(5*n-1,k)^2 for n >= 1.
a(n) = (7*n)!*(5*n/2)!^2/((5*n)!*(7*n/2)!*(3*n/2)!*n!^2!).
a(n) = (5/3) * [x^n] ( (1 - x)^(2*n) * P(5*n-1,(1 + x)/(1 - x)) ) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
a(n) = (5/3)*(-1)^n*binomial(7*n,n)*hypergeom([-n, 5*n, 5*n], [1, 6*n+1], 1) for n >= 1.
a(n) ~ sqrt(15)/Pi * 7^(7*n/2)/3^(3*n/2) * ( 1/(6*n) - 29/(945*n^2) + 841/(297675*n^3) + O(1/n^4) ).
a(n) = 7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) * a(n-2) with a(0) = 1 and a(1) = 30.
a(n)*A275654(n) = (7*n)!/(n!^4*(3*n)!) = A071549(n)/A006480(n).
a(p) == 30 (mod p^3) for all primes p >= 5.

A275653 a(n) = binomial(4*n,2*n)*binomial(3*n,2*n).

Original entry on oeis.org

1, 18, 1050, 77616, 6370650, 554822268, 50199951984, 4664758248000, 442077195513690, 42533571002422500, 4141601026094832300, 407220411993767798400, 40363606408574136870000, 4028061310168832261158176, 404311537318239680601595200, 40785601782042745410592271616
Offset: 0

Views

Author

Peter Bala, Aug 04 2016

Keywords

Comments

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)* binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
We also have Sum_{k = 0..4*n} (-1)^(n+k)*binomial(4*n + k,4*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(4*n,2*n)* binomial(3*n,2*n).
Compare with the identities
Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)* binomial(2*n,n) = A275655(n)
Sum_{k = 0..n} (-1)^(n+k)*binomial(8*n + k,8*n - k)* binomial(2*k,k)*binomial(2*n - k,n) = binomial(8*n,4*n)* binomial(5*n,2*n)*binomial(2*n,n)/binomial(6*n,3*n).
See also A275652, A275654 and A275655.

Crossrefs

Cf. A001448, A002894, A005809, A275652, A275654, A275655.

Programs

  • Maple
    seq((4*n)!*(3*n)!/(n!*(2*n)!^3), n = 0..20);
  • Mathematica
    Table[Binomial[4 n, 2 n] Binomial[3 n, 2 n], {n, 0, 15}] (* Michael De Vlieger, Aug 07 2016 *)

Formula

a(n) = (4*n)!*(3*n)!/(n!*(2*n)!^3).
a(n) = A001448(n) * A005809(n).
Recurrence: a(n) = 3*(3*n - 1)*(3*n - 2)*(4*n - 1)*(4*n - 3)/(n^2*(2*n - 1)^2) * a(n-1).
a(n) = [x^n] ((1 + x)^2/(1 - x))^(2*n) * [x^n] (1 + x)^(3*n) = [x^n] G(x)^(6*n) where G(x) = 1 + 3*x + 38*x^2 + 1150*x^3 + 47099^x^4 + 2264968*x^5 + 120311611*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^6, where F(x) = 1 + 3*x + 92*x^2 + 4579*x^3 + 282605*x^4 + 19698991*x^5 + 1484923315*x^6 + ... appears to have integer coefficients.
a(n) ~ sqrt(3/2)*108^n/(2*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(3*n-k-1,n-k)*binomial(4*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(4*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
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. (End)

A275655 a(n) = binomial(6*n,3*n)*binomial(2*n,n).

Original entry on oeis.org

1, 40, 5544, 972400, 189290920, 39089615040, 8385425017200, 1847301025078080, 415026659401497000, 94660194875011205440, 21850091031597537252544, 5092815839064962373499680, 1196622940864849837505171824, 283073284848591452381449360000
Offset: 0

Views

Author

Peter Bala, Aug 04 2016

Keywords

Comments

Right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k) *binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
We also note that Sum_{k = 0..6*n} (-1)^(n+k)*binomial(6*n + k,6*n - k)*binomial(2*k,k)*binomial(2*n - k,n) = binomial(6*n,3*n)*binomial(2*n,n).
Compare with Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k, k)*binomial(2*n - k,n) = binomial(2*n,n)^2 = A002894(n). See also A275652, A275653 and A275654.

Crossrefs

Cf. A000984, A002894, A066802, A275652, A275653, A275654.

Programs

  • Maple
    seq((6*n)!*(2*n)!/((3*n)!*n!)^2, n = 0..20);
  • Mathematica
    Table[Binomial[6 n, 3 n] Binomial[2 n, n], {n, 0, 13}] (* Michael De Vlieger, Aug 07 2016 *)

Formula

a(n) = (6*n)!*(2*n)!/((3*n)!*n!)^2.
a(n) = A066802(n) * A000984(n).
Recurrence: a(n) = 16*(2*n - 1)^2*(6*n - 1)*(6*n - 5)/(n^2*(3*n - 1)*(3*n - 2)) * a(n-1).
a(n) = [x^(3*n)] (1 + x)^(6*n) * [x^n] (1 + x)^(2*n) = [x^n] G(x)^(8*n) where G(x) = 1 + 5*x + 159*x^2 + 11690*x^3 + 1160817*x^4 + 135123516*x^5 + 17357714116*x^6 + ... appears to have integer coefficients.
exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^8, where F(x) = 1 + 5*x + 359*x^2 + 42270*x^3 + 6182313*x^4 + 1021669966*x^5 + 182605696304*x^6 + ... appears to have integer coefficients.
a(n) ~ 256^n/(sqrt(3)*Pi*n). - Ilya Gutkovskiy, Aug 07 2016
From Peter Bala, Mar 23 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(5*n-k-1,n-k)*binomial(6*n,k)^2.
a(n) = [x^n] (1 - x)^(2*n) * P(6*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A275652.
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. (End)

A364304 a(n) = (7*n)!*(9*n/2)!*(5*n/2)!/((5*n)!*(7*n/2)!^2*n!^2).

Original entry on oeis.org

1, 54, 10296, 2484000, 665091000, 188907932304, 55737530929080, 16888537352985408, 5218680924762089400, 1637124203403474142500, 519752205290081232622296, 166620892958456148158454144, 53846423260084127389865311800
Offset: 0

Views

Author

Peter Bala, Jul 21 2023

Keywords

Comments

Fractional factorials are defined in terms of the gamma function; for example, (9*n/2)! = gamma(1 + 9*n/2).

Links

Crossrefs

Row 7 of A364303.
Cf. A275652, A275654.

Programs

  • Maple
    seq( simplify((7*n)!*(9*n/2)!*(5*n/2)!/((5*n)!*(7*n/2)!^2*n!^2)), n = 0..12);
  • Mathematica
    A364304[n_]:=(7n)!(9n/2)!(5n/2)!/((5n)!(7n/2)!^2n!^2);Array[A364304,15,0] (* Paolo Xausa, Oct 06 2023 *)
  • Python
    from math import factorial
from sympy import factorial2
def A364304(n): return int(factorial(7*n)*factorial2(9*n)*factorial2(5*n)//factorial(5*n)//factorial2(7*n)**2//factorial(n)**2) # Chai Wah Wu, Aug 10 2023

Formula

a(n) = Sum_{k = 0..n} binomial(7*n, n - k)^2 * binomial(5*n + k - 1, k).
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(7*n + k, 7*n - k)*binomial(2*k, k)*binomial(2*n - k, n).
Conjecture: a(n) = Sum_{k = 0..7*n} (-1)^k * binomial(7*n + k, 7*n - k)* binomial(2*k, k)*binomial(2*n - k, n)
a(n) = [x^n] (1 - x)^(2*n) * P(7*n, (1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial.
a(n) = [x^n] G(x)^(9*n), where G(x) = 1 + 6*x + 266*x^2 + 27104*x^3 + 3726380*x^4 + 600232416*x^5 + 106662768380*x^6 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^9, where F(x) = 1 + 6*x + 590*x^2 + 95468*x^3 + 19200692*x^4 + 4364084760*x^5 + 1072849548644*x^6 + ... appears to have integer coefficients.
a(p) == a(1) (mod p^3).
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.
P-recursive: a(n) = 9*(9*n-2)*(9*n-4)*(9*n-6)*(9*n-8)*(9*n-10)*(9*n-12)*(9*n-14)*(9*n-16)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/((7*n-2)*(7*n-4)*(7*n-6)*(7*n-8)*(7*n-10)*(7*n-12)*(5*n-1)*(5*n-3)*(5*n-5)*(5*n-7)*(5*n-9)*n^2*(n-1)) * a(n-2) with a(0) = 1 and a(1) = 54.
a(n) ~ c^n * 3*sqrt(7)/(14*Pi*n), where c = (3^9)/(5^3) * sqrt(5).
Showing 1-6 of 6 results.

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