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 1-10 of 13 results. Next

A256890 Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.

Original entry on oeis.org

1, 2, 2, 4, 12, 4, 8, 52, 52, 8, 16, 196, 416, 196, 16, 32, 684, 2644, 2644, 684, 32, 64, 2276, 14680, 26440, 14680, 2276, 64, 128, 7340, 74652, 220280, 220280, 74652, 7340, 128, 256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172, 256, 512, 72076, 1637860, 10978444, 27227908, 27227908, 10978444, 1637860, 72076, 512
Offset: 0

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Author

Dale Gerdemann, Apr 12 2015

Keywords

Comments

Related triangles may be found by varying the function f(x). If f(x) is a linear function, it can be parameterized as f(x) = a*x + b. With different values for a and b, the following triangles are obtained:
a\b 1.......2.......3.......4.......5.......6
The row sums of these, and similarly constructed number triangles, are shown in the following table:
a\b 1.......2.......3.......4.......5.......6.......7.......8.......9
The formula can be further generalized to: t(n,m) = f(m+s)*t(n-1,m) + f(n-s)*t(n,m-1), where f(x) = a*x + b. The following table specifies triangles with nonzero values for s (given after the slash).
a\b 0 1 2 3
-2 A130595/1
-1
0
With the absolute value, f(x) = |x|, one obtains A038221/3, A038234/4, A038247/5, A038260/6, A038273/7, A038286/8, A038299/9 (with value for s after the slash).
If f(x) = A000045(x) (Fibonacci) and s = 1, the result is A010048 (Fibonomial).
In the notation of Carlitz and Scoville, this is the triangle of generalized Eulerian numbers A(r, s | alpha, beta) with alpha = beta = 2. Also the array A(2,1,4) in the notation of Hwang et al. (see page 31). - Peter Bala, Dec 27 2019

Examples

			Array, t(n, k), begins as:
   1,    2,      4,        8,        16,         32,          64, ...;
   2,   12,     52,      196,       684,       2276,        7340, ...;
   4,   52,    416,     2644,     14680,      74652,      357328, ...;
   8,  196,   2644,    26440,    220280,    1623964,    10978444, ...;
  16,  684,  14680,   220280,   2643360,   27227908,   251195000, ...;
  32, 2276,  74652,  1623964,  27227908,  381190712,  4677894984, ...;
  64, 7340, 357328, 10978444, 251195000, 4677894984, 74846319744, ...;
Triangle, T(n, k), begins as:
    1;
    2,     2;
    4,    12,      4;
    8,    52,     52,       8;
   16,   196,    416,     196,      16;
   32,   684,   2644,    2644,     684,      32;
   64,  2276,  14680,   26440,   14680,    2276,     64;
  128,  7340,  74652,  220280,  220280,   74652,   7340,   128;
  256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172,   256;
		

Crossrefs

Programs

  • Magma
    A256890:= func< n,k | (&+[(-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n: j in [0..k]]) >;
    [A256890(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Oct 18 2022
    
  • Mathematica
    Table[Sum[(-1)^(k-j)*Binomial[j+3, j] Binomial[n+4, k-j] (j+2)^n, {j,0,k}], {n,0, 9}, {k,0,n}]//Flatten (* Michael De Vlieger, Dec 27 2019 *)
  • PARI
    t(n,m) = if ((n<0) || (m<0), 0, if ((n==0) && (m==0), 1, (m+2)*t(n-1, m) + (n+2)*t(n, m-1)));
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n-k, k), ", ");); print(););} \\ Michel Marcus, Apr 14 2015
    
  • SageMath
    def A256890(n,k): return sum((-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n for j in range(k+1))
    flatten([[A256890(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Oct 18 2022

Formula

T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0 else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.
Sum_{k=0..n} T(n, k) = A001715(n).
T(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(j+3,j)*binomial(n+4,k-j)*(j+2)^n. - Peter Bala, Dec 27 2019
Modified rule of Pascal: T(0,0) = 1, T(n,k) = 0 if k < 0 or k > n else T(n,k) = f(n-k) * T(n-1,k-1) + f(k) * T(n-1,k), where f(x) = x + 2. - Georg Fischer, Nov 11 2021
From G. C. Greubel, Oct 18 2022: (Start)
T(n, n-k) = T(n, k).
T(n, 0) = A000079(n). (End)

A090802 Triangle read by rows: a(n,k) = number of k-length walks in the Hasse diagram of a Boolean algebra of order n.

Original entry on oeis.org

1, 2, 1, 4, 4, 2, 8, 12, 12, 6, 16, 32, 48, 48, 24, 32, 80, 160, 240, 240, 120, 64, 192, 480, 960, 1440, 1440, 720, 128, 448, 1344, 3360, 6720, 10080, 10080, 5040, 256, 1024, 3584, 10752, 26880, 53760, 80640, 80640, 40320
Offset: 0

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Author

Ross La Haye, Feb 10 2004

Keywords

Comments

Row sums = A010842(n); Row sums from column 1 on = A066534(n) = n*A010842(n-1) = A010842(n) - 2^n.
a(n,k) = n! = k! = A000142(n) for n = k; a(n,n-1) = 2*n! = A052849(n) for n > 1; a(n,n-2) = 2*n! = A052849(n) for n > 2; a(n,n-3) = (4/3)*n! = A082569(n) for n > 3; a(n,n-1)/a(2,1) = n!/2! = A001710(n) for n > 1; a(n,n-2)/ a(3,1) = n!/3! = A001715(n) for n > 2; a(n,n-3)/a(4,1) = n!/4! = A001720(n) for n > 3.
a(2k, k) = A052714(k+1). a(2k-1, k) = A034910(k).
a(n,0) = A000079(n); a(n,1) = A001787(n) = row sums of A003506; a(n,2) = A001815(n) = 2!*A001788(n-1); a(n,3) = A052771(n) = 3!*A001789(n); a(n,4) = A052796(n) = 4!*A003472(n); ceiling[a(n,1) / 2] = A057711(n); a(n,5) = 5!*A054849(n).
In a class of n students, the number of committees (of any size) that contain an ordered k-sized subcommittee is a(n,k). - Ross La Haye, Apr 17 2006
Antidiagonal sums [1,2,5,12,30,76,198,528,1448,4080,...] appear to be binomial transform of A000522 interleaved with itself, i.e., 1,1,2,2,5,5,16,16,65,65,... - Ross La Haye, Sep 09 2006
Let P(A) be the power set of an n-element set A. Then a(n,k) = the number of ways to add k elements of A to each element x of P(A) where the k elements are not elements of x and order of addition is important. - Ross La Haye, Nov 19 2007
The derivatives of x^n evaluated at x=2. - T. D. Noe, Apr 21 2011

Examples

			{1};
{2, 1};
{4, 4, 2};
{8, 12, 12, 6};
{16, 32, 48, 48, 24};
{32, 80, 160, 240, 240, 120};
{64, 192, 480, 960, 1440, 1440, 720};
{128, 448, 1344, 3360, 6720, 10080, 10080, 5040};
{256, 1024, 3584, 10752, 26880, 53760, 80640, 80640, 40320}
a(5,3) = 240 because P(5,3) = 60, 2^(5-3) = 4 and 60 * 4 = 240.
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[n!/(n-k)! * 2^(n-k), {n, 0, 8}, {k, 0, n}]] (* Ross La Haye, Feb 10 2004 *)

Formula

a(n, k) = 0 for n < k. a(n, k) = k!*C(n, k)*2^(n-k) = P(n, k)*2^(n-k) = (2n)!!/((n-k)!*2^k) = k!*A038207(n, k) = A068424*2^(n-k) = Sum[C(n, m)*P(n-m, k), {m, 0, n-k}] = Sum[C(n, n-m)*P(n-m, k), {m, 0, n-k}] = n!*Sum[1/(m!*(n-m-k)!), {m, 0, n-k}] = k!*Sum[C(n, m)*C(n-m, k), {m, 0, n-k}] = k!*Sum[C(n, n-m)*C(n-m, k), {m, 0, n-k}] = k!*C(n, k)*Sum[C(n-k, n-m-k), {m, 0, n-k}] = k!*C(n, k)*Sum[C(n-k, m), {m, 0, n-k}] for n >= k.
a(n, k) = 0 for n < k. a(n, k) = n*a(n-1, k-1) for n >= k >= 1.
E.g.f. (by columns): exp(2x)*x^k.

Extensions

More terms from Ray Chandler, Feb 26 2004
Entry revised by Ross La Haye, Aug 18 2006

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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Keywords

Crossrefs

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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Author

Keywords

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)

A034911 One fifth of octo-factorial numbers.

Original entry on oeis.org

1, 13, 273, 7917, 292929, 13181805, 698635665, 42616775565, 2940557513985, 226422928576845, 19245948929031825, 1789873250399959725, 180777198290395932225, 19704714613653156612525, 2305451609797419323665425, 288181451224677415458178125, 38328133012882096255937690625
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (8*n-3)*Self(n-1): n in [1..40]]; // G. C. Greubel, Oct 20 2022
    
  • Mathematica
    With[{nn=20},CoefficientList[Series[(-1+(1-8x)^(-5/8))/5,{x,0,nn}],x] Range[0,nn]!] (* or *) FoldList[Times,Range[5,200,8]]/5 (* Harvey P. Dale, May 25 2016 *)
  • SageMath
    [8^n*rising_factorial(5/8,n)/5 for n in range(1,40)] # G. C. Greubel, Oct 20 2022

Formula

5*a(n) = (8*n-3)(!^8) = Product_{j=1..n} 8*j-3.
E.g.f.: (-1+(1-8*x)^(-5/8))/5.
G.f.: x/(1-13*x/(1-8*x/(1-21*x/(1-16*x/(1-29*x/(1-24*x/(1-37*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
D-finite with recurrence: a(n) = (8*n-3)*a(n-1). - R. J. Mathar, Jan 28 2020
a(n) = (1/5)* 8^n * Pochhammer(n, 5/8). - G. C. Greubel, Oct 20 2022
From Amiram Eldar, Dec 20 2022: (Start)
a(n) = A147625(n+1)/5.
Sum_{n>=1} 1/a(n) = 5*(e/8^3)^(1/8)*(Gamma(5/8) - Gamma(5/8, 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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Author

Keywords

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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Author

Keywords

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)

A203516 a(n) = Product_{1 <= i < j <= n} 2*(i+j-1).

Original entry on oeis.org

1, 4, 192, 184320, 4954521600, 4794391461888000, 204135216112950312960000, 451965950843675288237663846400000, 60040562704967329457107799785403842560000000, 542366306792798635131534558788357929673196306432000000000
Offset: 1

Views

Author

Clark Kimberling, Jan 03 2012

Keywords

Comments

Each term divides its successor, as in A034910.
See A093883 for a guide to related sequences.

Crossrefs

Programs

  • Magma
    [2^Binomial(n,2)*(&*[Factorial(2*k)/Factorial(k): k in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Feb 19 2024
    
  • Maple
    a:= n-> mul(mul(2*(i+j-1), i=1..j-1), j=2..n):
    seq(a(n), n=1..12);  # Alois P. Heinz, Jul 23 2017
  • Mathematica
    f[j_] := 2 j - 1; z = 15;
    v[n_] := Product[Product[f[k] + f[j], {j, 1, k - 1}], {k, 2, n}]
    d[n_] := Product[(i - 1)!, {i, 1, n}]    (* A000178 *)
    Table[v[n], {n, 1, z}]                   (* A203516 *)
    Table[v[n + 1]/(4 v[n]), {n, 1, z - 1}]  (* A034910 *)
    Table[v[n]/d[n], {n, 1, 20}]             (* A203517 *)
    Table[2^(-1/24 - 3*n/2 + 3*n^2/2) * Glaisher^(3/2) * Pi^(1/4 - n/2) * BarnesG[1/2 + n]/E^(1/8), {n, 1, 10}] (* Vaclav Kotesovec, Sep 01 2023 *)
  • PARI
    a(n) = my(pd=1); for(j=1, n, for(i=1, j-1, pd=pd*2*(i+j-1))); pd \\ Felix Fröhlich, Jul 23 2017
    
  • SageMath
    [2^binomial(n,2)*product(factorial(2*k)/factorial(k) for k in range(n)) for n in range(1,21)] # G. C. Greubel, Feb 19 2024

Formula

a(n) ~ sqrt(A) * 2^(-7/24 - n + 3*n^2/2) * exp(-1/24 + n/2 - 3*n^2/4) * n^(1/24 - n/2 + n^2/2), where A = A074962 is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Sep 01 2023
From G. C. Greubel, Feb 19 2024: (Start)
a(n) = BarnesG(n+1)*A203517(n).
a(n) = 2^binomial(n,2) * Product_{j=1..n-1} (2j)!/j!. (End)

Extensions

Name edited by Alois P. Heinz, Jul 23 2017

A254619 a(n) = 4^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 24, 960, 53760, 3870720, 340623360, 35424829440, 4250979532800, 578133216460800, 87876248902041600, 14763209815542988800, 2716430606059909939200, 543286121211981987840000, 117349802181788109373440000
Offset: 0

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Author

Peter Bala, Feb 03 2015

Keywords

Crossrefs

Programs

  • Maple
    seq(4^n*(2*n + 1)!/n!, n = 0..13);
  • Mathematica
    Table[4^n (2n+1)!/n!,{n,0,20}] (* Harvey P. Dale, Oct 02 2021 *)

Formula

E.g.f.: 1/(1 - 16*x)^(3/2) = 1 + 24*x + 960*x^2/2! + 53760*x^3/3! + ....
Recurrence equation: a(n) = 8*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = (20*n + 6)*a(n-1) - 16*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 24.
Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*4^k) beginning [1, 26, 1052, 59032, 4251984, 374204832, 38917967808, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). From this observation we can obtain the continued fraction expansion
log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k) = 1 + 2/(24 - 16*3^2/(46 - 16*5^2/(66 - ... - 16*(2*n - 1)^2/((20*n + 6) - ... )))).
Alternative 2nd order recurrence equation: a(n) = (12*n + 10)*a(n-1) + 16*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 24.
Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*4^k) beginning [1, 22, 892, 49832, 3589584, 315853152, 32849393088, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion 2*arctan(1/2) = Sum {k >= 0} (-1)^k/((2*k + 1)*4^k) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ... )))). Cf. A254381 and A254620.

A254381 a(n) = 3^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 18, 540, 22680, 1224720, 80831520, 6304858560, 567437270400, 57878601580800, 6598160580211200, 831368233106611200, 114728816168712345600, 17209322425306851840000, 2787910232899709998080000, 485096380524549539665920000, 90227926777566214377861120000
Offset: 0

Views

Author

Peter Bala, Feb 04 2015

Keywords

Crossrefs

Programs

  • Maple
    seq(3^n*(2*n + 1)!/n!, n = 0..13);
  • Mathematica
    Table[3^n(2n + 1)!/n!, {n, 0, 19}] (* Alonso del Arte, Feb 04 2015 *)

Formula

E.g.f.: 1/(1 - 12*x)^(3/2) = 1 + 18*x + 540*x^2/2! + 22680*x^3/3! + ....
Recurrence equation: a(n) = 6*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = 8*(n + 1)*a(n-1) + 12*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 18.
Define a sequence b(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*3^k) beginning [1, 16, 492, 20544, 1111056, 73299456, 5718022848, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion Pi/(2*sqrt(3)) = Sum {k >= 0} (-1)^k/( (2*k + 1)*3^k ) = 1 - 2/(18 + 12*3^2/(24 + 12*5^2/(32 + ... + 12*(2*n - 1)^2/((8*n + 8) + ... )))). Cf. A254619 and A254620.
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