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 1-3 of 3 results.

A168467 a(n) = Product_{k=0..n} ((2*k+2)*(2*k+3))^(n-k).

Original entry on oeis.org

1, 6, 720, 3628800, 1316818944000, 52563198423859200000, 327312129899898454671360000000, 428017682605583614976547335700480000000000, 152240508705590071980086429193304853792686080000000000000
Offset: 0

Views

Author

Paul Barry, Nov 26 2009

Keywords

Comments

Hankel transform of A000698(n+1).
The sequence 1,1,6,720,... with general term Product_{k=0..n, ((2k+1)(2k+0^k))^(n-k)} is the Hankel transform of A112934. - Paul Barry, Dec 04 2009
a(n) is also the determinant of the n X n matrix M(i,j) = i^(2*j)*sinh(2*j*arccsch(i))/(2*sqrt(i^2+1)), with i and j from 1 to n, which is the same matrix generated by sequences of length n by the linear recurrences with kernel { 2*(k^2 + z), -k^4 }, and initial conditions { 1, 2*(k^2 + z) }, with k from 1 to n, and z = 2. Regardless of the value of z, for every n, the determinant of the n X n matrix of polynomials generated gives always a(n) as result. - Federico Provvedi, Feb 01 2021

Examples

			From _Federico Provvedi_, Apr 01 2021: (Start)
From both formulas in the comment above and in particular with z=2 from the linear recurrences, the determinant of the 5 X 5 matrix: ( (1,6,35,204,1189), (1,12,128,1344,14080),(1,22,403,7084,123205), (1,36,1040,28224,749824), (1,54,2291,89964,3426181) ) = 1316818944000 = a(5).
For a generic z, the determinant doesn't change as shown in this example, where the determinant of the 3 X 3 square matrix:
( ( 1, 2*(z+1), (2*z + 1)*(2*z+3)  ),
  ( 1, 2*(z+4), 4*(z+6)*(z+2)      ),
  ( 1, 2*(z+9), (2*z + 9)(2*z + 27)) ) = 720 = a(3). (End)

Links

Crossrefs

Cf. A000178, A000698, A098694, A112934, A306635, A001109, A099157, A246645, A202768, A000142.

Programs

  • Mathematica
    Table[2^(n^2 + 2*n + 23/24) Glaisher^(3/2) Pi^(-n/2 - 3/4) BarnesG[n + 2] BarnesG[n + 5/2]/E^(1/8), {n, 0, 10}] (* Vladimir Reshetnikov, Sep 06 2016 *)
Table[Product[((2k+2)(2k+3))^(n-k),{k,0,n}],{n,0,10}] (* Harvey P. Dale, Dec 26 2019 *)
Table[Det@Table[LinearRecurrence[{2*k^2,-k^4},{1, 2*k^2},n], {k, 1, n}], {n,1,20}] (* Federico Provvedi, Feb 01 2021 *)
Det@Expand@Array[(#1^(2 #2))/(4 Sqrt[1 + #1^2])((Sqrt[1+1/#1^2]+1/#1)^(2 #2)-(Sqrt[1+1/#1^2]-1/#1)^(2 #2))&,{#,#}]&/@Range[20] (* Federico Provvedi, Apr 01 2021 *)
  • Python
    from math import prod
def A168467(n): return prod(((m:=k+1<<1)*(m+1))**(n-k) for k in range(1,n+1))*3**n<Chai Wah Wu, Nov 26 2023

Formula

G.f.: Q(0)/(2*x) - 1/x, where Q(k) = 1 + 1/(1 - (2*k+1)!*x/((2*k+1)!*x + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Sep 17 2013
a(n) = Product_{k=1..n} (2*k+1)!. - Vladimir Reshetnikov, Sep 06 2016
a(n) ~ A^(-1/2) * 2^(n^2 + 3*n + 53/24) * exp((-3/2)*n^2 + (-5/2)*n + 1/24) * n^(n^2 + (5/2)*n + 35/24) * Pi^((n+1)/2), where A = A074962 is the Glaisher-Kinkelin constant. - Vladimir Reshetnikov, Sep 06 2016
a(n) = A000178(2*n + 1) / A098694(n). - Vaclav Kotesovec, Oct 28 2017
a(n) = A202768(n)*A000142(n). - Federico Provvedi, Feb 01 2021
For n > 0, a(n) = n * (2*n+1) * sqrt(BarnesG(2*n)) * Gamma(2*n)^2 / (sqrt(Gamma(n)) * 2^((n-3)/2)). - Vaclav Kotesovec, Nov 27 2024

A099158 a(n) = 5^(n-1) * U(n-1, 7/5) where U is the Chebyshev polynomial of the second kind.

Original entry on oeis.org

0, 1, 14, 171, 2044, 24341, 289674, 3446911, 41014904, 488035881, 5807129734, 69098919251, 822206626164, 9783419785021, 116412711336194, 1385192464081191, 16482376713731824, 196123462390215761
Offset: 0

Views

Author

Paul Barry, Oct 01 2004

Keywords

Links

Crossrefs

Cf. A099141, A099157.

Programs

  • Magma
    [n le 2 select n-1 else 14*Self(n-1) -25*Self(n-2): n in [1..30]]; // G. C. Greubel, Jul 20 2023
  • Mathematica
    LinearRecurrence[{14,-25}, {0,1}, 40] (* G. C. Greubel, Jul 20 2023 *)
  • PARI
    a(n) = 5^(n-1)*polchebyshev(n-1, 2, 7/5); \\ Michel Marcus, Sep 08 2019
  • SageMath
    A099158=BinaryRecurrenceSequence(14,-25,0,1)
[A099158(n) for n in range(41)] # G. C. Greubel, Jul 20 2023

Formula

G.f.: x/(1 - 14*x + 25*x^2).
E.g.f.: exp(7*x)*sinh(2*sqrt(6)*x)/sqrt(6).
a(n) = 14*a(n-1) - 25*a(n-2).
a(n) = sqrt(6)*(sqrt(6)+1)^(2*n)/24 - sqrt(6)*(sqrt(6)-1)^(2*n)/24.
a(n) = Sum_{k=0..n} binomial(2n, 2k+1)*6^k/2.
a(n) = 5^(n-1)*U(n-1, 7/5), where U is the Chebyshev polynomial of the second kind.

A337129 Triangular array read by rows: T(n,0) = 2^n, T(n,k) = Sum_{i=n-k..n, j=0..i-n+k, i<>n or j<>k} T(i,j) for k > 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 16, 8, 12, 32, 84, 16, 24, 64, 168, 440, 32, 48, 128, 336, 880, 2304, 64, 96, 256, 672, 1760, 4608, 12064, 128, 192, 512, 1344, 3520, 9216, 24128, 63168, 256, 384, 1024, 2688, 7040, 18432, 48256, 126336, 330752, 512, 768, 2048, 5376, 14080, 36864, 96512, 252672, 661504, 1731840
Offset: 0

Views

Author

Oboifeng Dira, Sep 14 2020

Keywords

Examples

			The triangle  T(n,k) begins:
   n\k  0    1    2    3    4    5
   0:   1
   1:   2    3
   2:   4    6    16
   3:   8    12   32  84
   4:   16   24   64  168  440
   5:   32   48   128 336  880  2304
   ...
T(3,2) = ((3+sqrt(5))^3-(3-sqrt(5))^3)*(2)/(4*sqrt(5)) = (64*sqrt(5))/(2*sqrt(5)) = 32.

Links

Crossrefs

Cf. A000079 (1st column), A069429 (diagonal), A018903 (row sums), A001906, A004171.

Programs

  • Maple
    T := proc (n, k) if k = 0 and 0 <= n then 2^n elif 1 <= k and k <= n then round((((3+sqrt(5))^(k+1)-(3-sqrt(5))^(k+1))*(2^(n-k))/(4*sqrt(5)))) else 0 end if end proc:seq(print(seq(T(n, k), k=0..n)), n=0..9);
  • Mathematica
    T[n_, 0] := 2^n;
T[n_, n_] := 2^(n-1) Fibonacci[2n+2];
T[n_, k_] /; 0Jean-François Alcover, Nov 13 2020 *)
  • PARI
    T(n,k) = if (k == 0, 2^n, my(w=quadgen(5, 'w)); ((2*w+2)^(k+1)-(4-2*w)^(k+1))*(2^(n-k))/(4*(2*w-1))); \\ Michel Marcus, Sep 14 2020
  • PARI
    Row(n)={Vecrev(polcoef((1-x*y)*(1-2*x*y)/((1-6*x*y+4*x^2*y^2)*(1-2*x)) + O(x*x^n), n))} \\ Andrew Howroyd, Sep 23 2020

Formula

T(n,0) = 2^n.
T(n,k) = ((3+sqrt(5))^(k+1)-(3-sqrt(5))^(k+1))*(2^(n-k))/(4*sqrt(5)) for 1<=k<=n.
T(n+1,n) = 2*T(n,n).
T(n+m,n) = 2^m*T(n,n), for m>=1.
T(n,n) = A069429(n) = 2^(n-1)*A001906(n+1) for n>=1.
T(2*n,n) = (1/2)*A099157(n+1) = A004171(n-1)*A001906(n+1) for n>=1.
G.f.: (1 - x*y)*(1 - 2*x*y)/((1 - 6*x*y + 4*x^2*y^2)*(1 - 2*x)). - Andrew Howroyd, Sep 23 2020
Showing 1-3 of 3 results.

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