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 1-5 of 5 results.

A034910 One quarter of octo-factorial numbers.

Original entry on oeis.org

1, 12, 240, 6720, 241920, 10644480, 553512960, 33210777600, 2258332876800, 171633298636800, 14417197085491200, 1326382131865190400, 132638213186519040000, 14324927024144056320000, 1661691534800710533120000, 206049750315288106106880000
Offset: 1

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Comments

A034910 occurs in connection with the Vandermonde permanent of (1,3,5,7,9,...); see the Mathematica section of A203516. - Clark Kimberling, Jan 03 2012

Examples

			G.f. = x + 12*x^2 + 240*x^3 + 6720*x^4 + 241920*x^5 + 10644480*x^6 + ...
		

Crossrefs

Programs

  • Magma
    [n le 2 select 12^(n-1) else (7*n-3)*Self(n-1) +4*(n-1)*(2*n-3)*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 20 2022
    
  • Maple
    [seq((2*n)!/(n)!*2^(n-2), n=1..14)]; # Zerinvary Lajos, Sep 25 2006
  • Mathematica
    s=1;lst={s};Do[s+=n*s;AppendTo[lst, s], {n, 11, 5!, 8}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *)
    a[ n_] := Pochhammer[ 1/2, n] 8^n / 4; (* Michael Somos, Feb 04 2015 *)
  • PARI
    {a(n) = if( n==1, 1, n>1, a(n-1) * (8*n - 4), a(n+1) / (8*n + 4))}; /* Michael Somos, Feb 04 2015 */
    
  • SageMath
    [2^(3*n-2)*rising_factorial(1/2, n) for n in range(1,40)] # G. C. Greubel, Oct 20 2022

Formula

4*a(n) = (8*n-4)(!^8) = Product_{j=1..n} (8*j-4) = 4^n*A001147(n) = 2^n*(2*n)!/n!, A001147(n) = (2*n-1)!!.
E.g.f. (-1+(1-8*x)^(-1/2))/4.
a(n) = A090802(2n-1, n). - Ross La Haye, Oct 18 2005
a(n) = ((2*n)!/n!)*2^(n-2). - Zerinvary Lajos, Sep 25 2006
G.f.: x/(1-12*x/(1-8*x/(1-20*x/(1-16*x/(1-28*x/(1-24*x/(1-36*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2011
From Peter Bala, Feb 01 2015: (Start)
Recurrence equation: a(n) = (7*n - 3)*a(n-1) + 4*(n - 1)*(2*n - 3)*a(n-2).
The sequence b(n) := a(n)* Sum_{k = 0..n-1} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) beginning [1, 11, 222, 6210, 223584, ...] satisfies the same recurrence. This leads to the finite continued fraction expansion b(n)/a(n) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/(7*n - 3) )))) for n >= 3.
Letting n tend to infinity gives the continued fraction expansion Sum_{k>=0} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (4/3)*log(2) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/((7*n - 3) + ... ))))). (End)
From Peter Bala, Feb 03 2015: (Start)
This sequence satisfies several other second order recurrence equations leading to some continued fraction expansions.
1) a(n) = (9*n + 4)*a(n-1) - 4*n*(2*n - 1)*a(n-2).
This recurrence is also satisfied by the (integer) sequence c(n) := a(n)*Sum_{k = 0..n} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (8/sqrt(7))*arctan(sqrt(7)/7) = (8/sqrt(7))*A195699 = 1 + 1/(12 - 24/(22 - 60/(31 - ... - 4*n*(2*n - 1)/((9*n + 4) - ... )))).
2) a(n) = (12*n + 2)*a(n-1) - 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence d(n) := a(n)*Sum_{k = 0..n} 1/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( (2*k + 1)*2^k ) = (1/sqrt(2))*log(3 + 2*sqrt(2)) = 1 + 2/(12 - 8*3^2/(26 - 8*5^2/(38 - ... - 8*(2*n - 1)^2/((12*n + 2) - ... )))). Cf. A002391.
3) a(n) = (4*n + 6)*a(n-1) + 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence e(n) := a(n)*Sum_{k = 0..n} (-1)^k/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} (-1)^k/( (2*k + 1)*2^k ) = (1/sqrt(2))*arctan(sqrt(2)/2) = 1 - 2/(12 + 8*3^2/(14 + 8*5^2/(18 + ... + 8*(2*n - 1)^2/((4*n + 6) + ... )))). Cf. A073000. (End)
a(n) = (-1)^n / (16*a(-n)) for all n in Z. - Michael Somos, Feb 04 2015
From Amiram Eldar, Jan 08 2022: (Start)
Sum_{n>=1} 1/a(n) = e^(1/8)*sqrt(2*Pi)*erf(1/(2*sqrt(2))), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = e^(-1/8)*sqrt(2*Pi)*erfi(1/(2*sqrt(2))), where erfi is the imaginary error function. (End)

A034909 One third of octo-factorial numbers.

Original entry on oeis.org

1, 11, 209, 5643, 197505, 8492715, 433128465, 25554579435, 1712156822145, 128411761660875, 10658176217852625, 969894035824588875, 96019509546634298625, 10274087521489869952875, 1181520064971335044580625, 145326967991474210483416875, 19037832806883121573327610625
Offset: 1

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Formula

3*a(n) = (8*n-5)(!^8) = product(8*j-5, j=1..n).
E.g.f.: (-1+(1-8*x)^(-3/8))/3.
G.f.: x/(1-11x/(1-8x/(1-19x/(1-16x/(1-27x/(1-24x/(1-35x/(1-32x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
From Amiram Eldar, Dec 20 2022: (Start)
a(n) = A144756(n)/3.
Sum_{n>=1} 1/a(n) = 3*(e/8^5)^(1/8)*(Gamma(3/8) - Gamma(3/8, 1/8)). (End)

A034912 One sixth of octo-factorial numbers.

Original entry on oeis.org

1, 14, 308, 9240, 351120, 16151520, 872182080, 54075288960, 3785270227200, 295251077721600, 25391592684057600, 2386809712301414400, 243454590654744268800, 26780004972021869568000, 3160040586698580609024000, 398165113924021156737024000, 53354125265818835002761216000
Offset: 1

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Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (8*n-2)*Self(n-1): n in [1..40]]; // G. C. Greubel, Oct 20 2022
    
  • Maple
    f:= proc(n) option remember; procname(n-1)*(8*n-2) end proc:
    f(1):= 1:
    map(f,[$1..20]); # Robert Israel, Mar 20 2018
  • Mathematica
    Table[8^n*Pochhammer[3/4,n]/6, {n,40}] (* G. C. Greubel, Oct 20 2022 *)
  • SageMath
    [8^n*rising_factorial(3/4,n)/6 for n in range(1,40)] # G. C. Greubel, Oct 20 2022

Formula

6*a(n) = (8*n-2)(!^8) = Product_{j=1..n} (8*j - 2) = 2^n*3*A034176(n), where 3*A034176(n) = (4*n-1)(!^4) = Product_{j=1..n} (4*j - 1).
E.g.f.: (-1+(1-8*x)^(-3/4))/6.
G.f.: x/(1-14*x/(1-8*x/(1-22*x/(1-16*x/(1-30*x/(1-24*x/(1-38*x/(1-32*x/(1-...(continued fraction). - Philippe Deléham, Jan 07 2012
From G. C. Greubel, Oct 20 2022: (Start)
a(n) = (1/6) * 8^n * Pochhammer(n, 3/4).
a(n) = 2*(4*n - 1)*a(n-1). (End)
From Amiram Eldar, Dec 20 2022: (Start)
a(n) = A147626(n+1)/6.
Sum_{n>=1} 1/a(n) = 6*(e/8^2)^(1/8)*(Gamma(3/4) - Gamma(3/4, 1/8)). (End)

A034975 One seventh of octo-factorial numbers.

Original entry on oeis.org

1, 15, 345, 10695, 417105, 19603935, 1078216425, 67927634775, 4822862069025, 381006103452975, 33147531000408825, 3149015445038838375, 324348590839000352625, 36002693583129039141375, 4284320536392355657823625, 544108708121829168543600375, 73454675596446937753386050625
Offset: 1

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Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (8*n-1)*Self(n-1): n in [1..40]]; // G. C. Greubel, Oct 21 2022
    
  • Mathematica
    Table[8^n*Pochhammer[7/8, n]/7, {n, 40}] (* G. C. Greubel, Oct 21 2022 *)
  • SageMath
    [8^n*rising_factorial(7/8,n)/7 for n in range(1,40)] # G. C. Greubel, Oct 21 2022

Formula

7*a(n) = (8*n-1)!^8 = Product_{j=1..n} (8*j-1) = (8*n)!/((2*n)!*2^(6*n)*3^2*5 * A045755(n)*A007696(n)*A034909(n)*A034911(n)*A034176(n)).
E.g.f.: (-1+(1-8*x)^(-7/8))/7.
G.f.: x/(1-15*x/(1-8*x/(1-23*x/(1-16*x/(1-31*x/(1-24*x/(1-39*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
a(n) = (1/7) * 8^n * Pochhammer(n, 7/8). - G. C. Greubel, Oct 21 2022
From Amiram Eldar, Dec 20 2022: (Start)
a(n) = A049210(n)/7.
Sum_{n>=1} 1/a(n) = 7*(e/8)^(1/8)*(Gamma(7/8) - Gamma(7/8, 1/8)). (End)

A034976 One eighth of octo-factorial numbers.

Original entry on oeis.org

1, 16, 384, 12288, 491520, 23592960, 1321205760, 84557168640, 6088116142080, 487049291366400, 42860337640243200, 4114592413463347200, 427917611000188108800, 47926772432021068185600, 5751212691842528182272000, 736155224555843607330816000
Offset: 1

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Crossrefs

Programs

  • Magma
    [8^(n-1)*Factorial(n): n in [1..40]]; // G. C. Greubel, Oct 20 2022
    
  • Mathematica
    Table[8^(n-1)*n!, {n,40}] (* G. C. Greubel, Oct 20 2022 *)
  • SageMath
    [8^(n-1)*factorial(n) for n in range(1,40)] # G. C. Greubel, Oct 20 2022

Formula

8*a(n) = (8*n)!^8 = Product_{j=1..n} 8*j = 8^n*n!.
E.g.f.: (-1+(1-8*x)^(-1))/8.
G.f.: x/(1-16*x/(1-8*x/(1-24*x/(1-16*x/(1-32*x/(1-24*x/(1-40*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
From Amiram Eldar, Jan 08 2022: (Start)
Sum_{n>=1} 1/a(n) = 8*(exp(1/8)-1).
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*(1-exp(-1/8)). (End)
Showing 1-5 of 5 results.