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.

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A347854 a(n) = (6*n)!/((3*n)!*(2*n)!) * (n/2)!/(3*n/2)!.

Original entry on oeis.org

1, 40, 4620, 622336, 89237148, 13236695040, 2005604901300, 308350245273600, 47913489552349980, 7505566011722039296, 1183237138556438547120, 187495217080545878999040, 29836408028165719837829700, 4764790302634058161217077248
Offset: 0

Views

Author

Peter Bala, Sep 16 2021

Keywords

Comments

Fractional factorials are defined using the Gamma function; for example, (n/2)! := Gamma(1 + n/2).
The sequence defined by u(n) = (12*n)!*n!/((6*n)!*(4*n)!*(3*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 1). See A295431. It is known that u(n) is integral and satisfies the congruences u(n*p) == u(n) ( mod p^3 ) for prime p >= 5 and any positive integer n (Zudilin, Section 5); the o.g.f. Sum_{n >= 0} u(n)*x^n is algebraic over Q(x) (Rodriguez-Villegas).
Here we are essentially considering the sequence ( u(n/2) )n>=0. The sequence is conjectured to be integral.

Examples

			a(11) - a(1) = 187495217080545878999040 - 40 = (2^3)*(5^3)*(11^3)*140867931690868429 == 0 (mod 11^3).
		

Crossrefs

Programs

  • Maple
    seq(binomial(6*n,2*n)*binomial(4*n,n)/binomial(3*n/2,n), n = 0..13);
  • Python
    from math import factorial
    from sympy import factorial2
    def A347854(n): return int((factorial(6*n)*factorial2(n)<Chai Wah Wu, Aug 10 2023

Formula

a(n) = binomial(6*n,2*n)*binomial(4*n,n)/binomial(3*n/2,n).
a(2*n) = A295431(n).
a(2*n) = 24*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)/( n*(2*n - 1)*(3*n - 1)*(3*n - 2) )*a(2*n-2);
a(2*n+1) = 96*(12*n + 1)*(12*n - 1)*(12*n + 5)*(12*n - 5)/( n*(2*n + 1)*(6*n + 1)*(6*n - 1) )*a(2*n-1).
Asymptotics: a(n) ~ 32^n/sqrt(6*Pi*n) * 3^(3*n/2) as n -> infinity.
O.g.f.: A(x) = hypergeom([1/12, 5/12, 7/12, 11/12], [1/3, 1/2, 2/3], 27648*x^2) + 40*x*hypergeom([11/12, 13/12, 7/12, 17/12], [3/2, 5/6, 7/6], 27648*x^2) is conjectured to be algebraic over Q(x).
Conjectural: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k.
From Karol A. Penson, Feb 17 2023: (Start)
An integral representation of a(n) as the n-th power moment of the weight function W(x) is given by a(n) = Integral_{x=0..96*sqrt(3)} x^n*W(x), where W(x) = W_1(x) + W_2(x) + W_3(x) + W_4(x) and the functions W_n(x) are:
W_1(x) = sqrt(2)*3^(3/4)*hypergeom([1/12, 5/12, 7/12, 3/4], [1/6, 1/2, 2/3], x^2/27648)*Gamma(3/4)/(18*sqrt(Pi)*x^(5/6)*Gamma(2/3)*Gamma(7/12)).
W_2(x) = sqrt(2)*cos((5*Pi)/12)*Gamma(2/3)*csc(Pi/12)*Gamma(3/4)*3^(1/4)* hypergeom([5/12, 3/4, 11/12, 13/12], [1/2, 5/6, 4/3], x^2/27648)/(2304*Pi^(3/2)* Gamma(11/12)*x^(1/6)).
W_3(x) = cos((5*Pi)/12)*3^(1/4)*Gamma(11/12)*x^(1/6)*hypergeom([7/12, 11/12, 13/12, 5/4], [2/3, 7/6, 3/2], x^2/27648)/(3456*sqrt(Pi)*Gamma(2/3)*Gamma(3/4)).
W_4(x) = 7*sin((5*Pi)/12)*Gamma(2/3)*Gamma(7/12)*3^(3/4)*x^(5/6)*hypergeom([11/12, 5/4,17/12, 19/12], [4/3, 3/2, 11/6], x^2/27648))/(1327104*Pi^(3/2)*Gamma(3/4)).
The function W(x) is positive on the support x = (0..96*sqrt(3)) and is singular at both endpoints of the support. The function W(x) is unique as it is the solution of the Hausdorff moment problem. (End)

A347858 a(n) = (9*n)!/((3*n)!*(2*n)!) * (n/2)!/(9*n/2)!.

Original entry on oeis.org

1, 512, 1021020, 2399141888, 6016814703900, 15626259253952512, 41477110789150966020, 111745115394167694950400, 304331361887290342345862940, 835666006020766806513664655360, 2309513382640863232775760738593520, 6416034331756986890806503962421755904
Offset: 0

Views

Author

Peter Bala, Sep 19 2021

Keywords

Comments

Fractional factorials are defined using the Gamma function; for example, (9*n/2)! := Gamma(1 + 9*n/2).
The sequence defined by u(n) = (18*n)!*(n)!/((4*n)!*(6*n)!*(9*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 7). See A295437. Here we are essentially considering the sequence (u(n/2))n>=0. The sequence is conjectured to be integral.

Examples

			Congruence: a(11) - a(1) = 6416034331756986890806503962421755904 - 512 = (2^9)*(11^3)*7207*1306363809854375553476366323 == 0 (mod 11^3).
		

Crossrefs

Programs

  • Maple
    seq( (9*n)!/((3*n)!*(2*n)!) * GAMMA(1+n/2)/GAMMA(1+9*n/2), n = 0..11);
  • Python
    from math import factorial
    from sympy import factorial2
    def A347858(n): return int((factorial(9*n)*factorial2(n)<<(n<<2))//(factorial(3*n)*factorial(n<<1)*factorial2(9*n))) # Chai Wah Wu, Aug 10 2023

Formula

a(n) = binomial(9*n,3*n)*binomial(6*n,2*n)/binomial(9*n/2,4*n).
a(2*n) = A295437(n).
a(2*n) = 72*(18*n-1)*(18*n-5)*(18*n-7)*(18*n-11)*(18*n-13)*(18*n-17)/(n*(2*n-1)*(3*n-1)*(3*n-2)*(4*n-1)*(4*n-3))*a(2*n-2);
a(2*n+1) = 18432*(81*n^2-1)*(81*n^2-4)*(81*n^2-16)/(n*(2*n+1)*(16*n^2-1)*(36*n^2-1))*a(2*n-1).
Asymptotics: a(n) ~ 1/(2*sqrt(3*Pi*n)) * 2916^n as n -> infinity.
O.g.f.: A(x) = hypergeom([1/18, 5/18, 7/18, 11/18, 13/18, 17/18], [1/4, 1/3, 1/2, 2/3, 3/4], (2^4)*(3^12)*x^2) + 512*x*hypergeom([5/9, 7/9, 8/9, 10/9, 11/9, 13/9], [3/4, 5/6, 7/6, 5/4, 3/2], (2^4)*(3^12)*x^2) is conjectured to be algebraic over Q(x).
Conjectural congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k.

A347855 a(n) = (4*n)!/((2*n)!*(n)!) * (n/3)!/(4*n/3)!.

Original entry on oeis.org

1, 9, 189, 4620, 120285, 3241134, 89237148, 2493521172, 70429218525, 2005604901300, 57481750139814, 1656023714623980, 47913489552349980, 1391243084942932620, 40519970408738302020, 1183237138556438547120
Offset: 0

Views

Author

Peter Bala, Sep 17 2021

Keywords

Comments

Fractional factorials are defined using the Gamma function; for example, (n/3)! := Gamma(1 + n/3). The sequence defined by u(n) = (12*n)!*n! / ((6*n)!*(4*n)!*(3*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 1). See A295431. Here we are essentially considering the sequence (u(n/3))n>=0. The sequence is conjectured to be integral.

Examples

			Congruence: a(11) - a(1) = 1656023714623980 - 9 = (3^2)*7*(11^3)*17* 1161713471 == 0 (mod 11^3).
		

Crossrefs

Programs

  • Maple
    seq( (4*n)!/((2*n)!*(n)!) * GAMMA(1+n/3)/GAMMA(1+4*n/3), n = 0..15);
  • Mathematica
    Table[Binomial[4n,2n] Binomial[2n,n]/Binomial[4 n/3,n],{n,0,20}] (* Harvey P. Dale, Apr 09 2022 *)

Formula

a(n) = binomial(4*n,2*n)*binomial(2*n,n)/binomial(4*n/3,n).
a(3*n) = A295431(n).
D-finite with recurrence -n*(n-1)*(n-2)*(2*n-3)*a(n) + 216*(4*n-11)*(4*n-1)*(4*n-5)*(4*n-7)*a(n-3).
Asymptotics: a(n) ~ 1/(2*sqrt(Pi*n))*2^(10*n/3)*3^n as n -> infinity.
O.g.f.: A(x) = hypergeom([11/12, 7/12, 5/12, 1/12], [2/3, 1/2, 1/3], 27648*x^3) + 9*x*hypergeom([11/12, 5/4, 5/12, 3/4], [5/6, 4/3, 2/3], 27648*x^3) + 189*x^2*hypergeom([19/12, 13/12, 5/4, 3/4], [7/6, 5/3, 4/3], 27648*x^3) is conjectured to be algebraic over Q(x).
Conjectural congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k.

A364173 a(n) = (9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!).

Original entry on oeis.org

1, 128, 43758, 17039360, 7012604550, 2976412336128, 1288415796384780, 565399665327996928, 250622090889055155270, 111950839825145979207680, 50312973039218473430585508, 22723567527558510746926055424, 10304958075870392958137083227804
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295440, defined by A295440(n) = (18*n)!*(4*n)!*(3*n)! / ((9*n)!*(8*n)!*(6*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 10). Here we are essentially considering the sequence {A295440(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).
This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((9*n)!*(2*n)!*(3*n/2)!/((9*n/2)!*(4*n)!*(3*n)!*n!)) , n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(4*Pi*n), where c = (3^7)/(2^3) * sqrt(3) = 473.4993895191418....
a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(n*(n - 1)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 128.

A364172 a(n) = (6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!).

Original entry on oeis.org

1, 45, 6237, 1021020, 178719453, 32427545670, 6016814703900, 1133540594837892, 215925912619400925, 41477110789150966020, 8019784929635201045862, 1558875476359831844951100, 304331361887290342345862940, 59629409730107012112361325820
Offset: 0

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Author

Peter Bala, Jul 12 2023

Keywords

Comments

Fractional factorials are defined in terms of the gamma function; for example, (n/3)! := Gamma(n/3 + 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. For a list of the 52 sporadic integral factorial ratio sequences see A295431.
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 rational numbers.
A295437, defined by A295437(n) = (18*n)!*n! / ((9*n)!*(6*n)!*(4*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 7). Here we are essentially considering the sequence {A295437(n/3) : n >= 0}. This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((6*n)!*(n/3)!/((3*n)!*(2*n)!*(4*n/3)!)), n = 0..15);
  • Mathematica
    Table[Product[36*(6*k - 5)*(6*k - 1)/(k*(3*k + n)), {k, 1, n}], {n, 0, 20}] (*  Vaclav Kotesovec, Jul 13 2023 *)

Formula

a(n) ~ 2^(4*n/3 - 3/2) * 3^(4*n) / sqrt(Pi*n). - Vaclav Kotesovec, Jul 13 2023
a(n) = 5832*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(n*(n - 1)*(n - 2)*(2*n - 3)*(4*n - 3)*(4*n - 9))*a(n-3) for n >= 3 with a(0) = 1, a(1) = 45 and a(2) = 6237.

A364183 a(n) = (12*n)!*(2*n)!*(n/2)!/((6*n)!*(4*n)!*(7*n/2)!*n!).

Original entry on oeis.org

1, 4224, 76488984, 1626105446400, 36856530424884600, 864687003650148532224, 20728451893251973782071160, 504292670666772382512278667264, 12401082728528113445556802226795640, 307453669544695584297743425538327838720, 7671567513095586883562392061857092727662984
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295479, defined by A295479(n) = (24*n)!*(4*n)!*n! / ((12*n)!*(8*n)!*(7*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 49). Here we are essentially considering the sequence {A295479(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (7*n/2)! := Gamma(1 + 7*n/2).
This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((12*n)!*(2*n)!*(n/2)!/((6*n)!*(4*n)!*(7*n/2)!*n!)), n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(14*Pi*n), where c = (2^15)*(3^6)/(7^4) * sqrt(7).
a(n) = 1327104*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)*(12*n - 13)*(12*n - 17)*(12*n - 19)*(12*n - 23)/(7*n*(n - 1)*(7*n - 2)*(7*n - 4)*(7*n - 6)*(7*n - 8)*(7*n - 10)*(7*n - 12))*a(n-2) with a(0) = 1 and a(1) = 4224.

A364174 a(n) = (9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!).

Original entry on oeis.org

1, 48, 4862, 549120, 65132550, 7945986048, 987291797996, 124259864002560, 15789207515217990, 2021092963752345600, 260227401685879140612, 33665720694993527504896, 4372592850984736084611996, 569819472537519480058675200, 74468439316740019538310543000
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295442, defined by A295442(n) = (18*n)!*(5*n)!*(3*n)!/((10*n)!*(9*n)!*(6*n)!*n!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 12). Here we are essentially considering the sequence {A295442(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (3*n/2)! := Gamma(1 + 3*n/2).
This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((9*n)!*(5*n/2)!*(3*n/2)!/((5*n)!*(9*n/2)!*(3*n)!*(n/2)!)), n = 0..15)

Formula

a(n) ~ c^n * 1/sqrt(2*Pi*n), where c = 2*(3^7)/(5^3) * sqrt(15) = 135.5234332504899....
a(n) = 108*(9*n - 1)*(9*n - 5)*(9*n - 7)*(9*n - 11)*(9*n - 13)*(9*n - 17)/(5*n*(n - 1)*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9))*a(n-2) for n >= 2 with a(0) = 1 and a(1) = 48.

A364175 a(n) = (6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!).

Original entry on oeis.org

1, 36, 3564, 408408, 49697388, 6249195036, 802241960520, 104466877291260, 13746018177013356, 1823169705017624880, 243331037661693468564, 32641262295291161362656, 4396944340992842923469640, 594371374049863341847620936, 80586283761263090599592845140
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295445, defined by A295445(n) = (18*n)!*(2*n)! / ((9*n)!*(6*n)!*(5*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 15). Here we are essentially considering the sequence {A295445(n/3) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (2*n/3)! := Gamma(1 + 2*n/3).
This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((6*n)!*(2*n/3)!/((3*n)!*(2*n)!*(5*n/3)!)), n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(5*Pi*n) where c = (1296/25)*20^(1/3) = 140.7154092442799....
a(n) = 93312*(2*n - 3)*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11)*(6*n - 13)*(6*n - 17)/(5*n*(n - 1)*(n - 2)*(5*n - 3)*(5*n - 6)*(5*n - 9)*(5*n - 12))*a(n-3) with a(0) = 1, a(1) = 36 and a(2) = 3564.

A364176 a(n) = (15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(6*n)!*(5*n)!*n!).

Original entry on oeis.org

1, 7168, 168043980, 4488240824320, 126694219977836700, 3688258943632086663168, 109504706026534324525391988, 3295939064766794222800490987520, 100204869963549181630558779565943580, 3070025447039504554088467623457608171520, 94632263448378916462441320194245442445186480
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295456, defined by A295456(n) = (30*n)!*(5*n)!*(4*n)! / ((15*n)!*(12*n)!*(10*n)!*(2*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 26). Here we are essentially considering the sequence {A295456(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (5*n/2)! := Gamma(1 + 5*n/2).
This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(6*n)!*(5*n)!*n!)), n = 0..15)

Formula

a(n) ~ c^n * 1/sqrt(6*Pi*n), where c = 18750*sqrt(3).
a(n) = 4800*(15*n - 1)*(15*n - 7)*(15*n - 11)*(15*n - 13)*(15*n - 17)*(15*n - 19)*(15*n - 23)*(15*n - 29)/(n*(n - 1)*(3*n - 2)*(3*n - 4)*(6*n - 1)*(6*n - 5)*(6*n - 7)*(6*n - 11))*a(n-2) with a(0) = 1 and a(1) = 7168.

A364177 a(n) = (15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!).

Original entry on oeis.org

1, 35840, 5545451340, 991901222174720, 188242272043069768860, 36901030731039027064995840, 7383354803839076831124554790900, 1498315221854950975184507333477662720, 307213802011837003346320048243705086348060
Offset: 0

Views

Author

Peter Bala, Jul 13 2023

Keywords

Comments

A295458, defined by A295458(n) = (30*n)!*(5*n)!*(4*n)! / ((15*n)!*(10*n)!*(8*n)!*(6*n)!), is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 28). Here we are essentially considering the sequence {A295458(n/2) : n >= 0}. Fractional factorials are defined in terms of the gamma function; for example, (5*n/2)! := Gamma(1 + 5*n/2).
This sequence is only conjecturally an integer sequence.
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.

Crossrefs

Programs

  • Maple
    seq( simplify((15*n)!*(5*n/2)!*(2*n)!/((15*n/2)!*(5*n)!*(4*n)!*(3*n)!)), n = 0..15);

Formula

a(n) ~ c^n * 1/sqrt(12*Pi*n), where c = (3^4)*(5^5) * sqrt(3)/2.
a(n) = 43200*(15*n - 1)*(15*n - 7)*(15*n - 11)*(15*n - 13)*(15*n - 17)*(15*n - 19)*(15*n - 23)*(15*n - 29)/(n*(n - 1)*(3*n - 2)*(3*n - 4)*(4*n - 1)*(4*n - 3)*(4*n - 5)*(4*n - 7))*a(n-2) with a(0) = 1 and a(1) = 35840.
Showing 1-10 of 16 results. Next