A093123 -id:A093123 - cp's OEIS frontend

A093129 Binomial transform of Fibonacci(2n-1) (A001519).

1, 2, 5, 15, 50, 175, 625, 2250, 8125, 29375, 106250, 384375, 1390625, 5031250, 18203125, 65859375, 238281250, 862109375, 3119140625, 11285156250, 40830078125, 147724609375, 534472656250, 1933740234375, 6996337890625

Offset: 0

Paul Barry, Mar 23 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-5).

Cf. A000032, A000045, A001519, A020876, A030191, A093123.

a:=[1,2];; for n in [3..30] do a[n]:=5*(a[n-1]-a[n-2]); od; a; # G. C. Greubel, Dec 27 2019

I:=[1,2]; [n le 2 select I[n] else 5*(Self(n-1) - Self(n-2)): n in [1..30]]; // G. C. Greubel, Dec 27 2019

a:= n-> (<<0|1>, <-5|5>>^n. <<1,2>>)[1,1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 29 2015

LinearRecurrence[{5, -5}, {1, 2}, 25] (* Jean-François Alcover, May 11 2019 *)
Table[If[EvenQ[n], 5^(n/2)*Fibonacci[n-1], 5^((n-1)/2)*LucasL[n-1]], {n,0,30}] (* G. C. Greubel, Dec 27 2019 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-5*x+5*x^2)) \\ G. C. Greubel, Dec 27 2019

[lucas_number2(n,5,5) for n in range(-1,25)] # Zerinvary Lajos, Jul 08 2008

G.f.: (1-3*x)/(1-5*x+5*x^2).
a(n) = (5-sqrt(5))*((5+sqrt(5))/2)^n/10 + (5+sqrt(5))*((5-sqrt(5))/2)^n/10.
a(n) = A093123(n)/2^n.
a(n) = A020876(n-1). - R. J. Mathar, Sep 05 2008
a(n) = A030191(n) - 3*A030191(n-1). - R. J. Mathar, Jun 29 2012
a(2*n) = 5^n*Fibonacci(2*n-1), a(2*n+1) = 5^n*Lucas(2*n). - G. C. Greubel, Dec 27 2019
E.g.f.: (1/10)*exp((1/2)*(5-sqrt(5))*x)*(5 + sqrt(5) + (5 - sqrt(5))*exp(sqrt(5)*x)). - Stefano Spezia, Dec 28 2019

A247239 Array a(n,m) = ((n+2)/2)^mSum_{k=1..n+1} 1/sin(kPi/(n+2))^(2m), n>=0, k>=0, read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 10, 8, 1, 5, 20, 36, 16, 1, 6, 35, 120, 136, 32, 1, 7, 56, 329, 800, 528, 64, 1, 8, 84, 784, 3611, 5600, 2080, 128, 1, 9, 120, 1680, 13328, 42065, 40000, 8256, 256, 1, 10, 165, 3312, 42048, 241472, 499955, 288000, 32896, 512, 1

Offset: 0

Jean-François Alcover, Nov 28 2014

Unexpectedly, it is conjectured (proof wanted) that the expression ((n+2)/2)^m * Sum_{k=1..n+1} 1/sin(k*Pi/(n+2))^(2m), n>=0, k>=0, always gives an integer.

For example, a(3,1) = (5/2)*(1/sin(Pi/5)^2 + 1/sin((2*Pi)/5)^2 + 1/sin((3*Pi)/5)^2 + 1/sin((4*Pi)/5)^2) = (5/2)*(2/(5/8 - sqrt(5)/8) + 2/(5/8 + sqrt(5)/8)), which simplifies to 20.

			Array a(n,m) begins:
  1,  1,   1,    1,     1,      1,       1,        1, ... 1 (A000012)
  2,  4,   8,   16,    32,     64,     128,      256, ... 2^(m+1) (A000079)
  3, 10,  36,  136,   528,   2080,    8256,    32896, ... A007582
  4, 20, 120,  800,  5600,  40000,  288000,  2080000, ... A093123
  5, 35, 329, 3611, 42065, 499955, 5980889, 71698571, ... not in the OEIS
  ...
1st column is n+1 (A000027).
2nd column is A000292.
3rd column is not in the OEIS.

I. M. Gessel, Generating functions and generalized Dedekind sums, Electron. J. Combin.4 (1997), no. 2, Research Paper 11, 17 pp.
Les Mathematiques, Somme des 1/sin^2, Sketch of a proof [in French].
MilesB, How to prove this sum involving powers of cosec is an integer?, MathOverflow 444094.
R. P. Stanley, Invariants of finite groups and their applications to combinatorics, Bull. Amer. Math. Soc. 1 (1979), 475-511.

Cf. A000079, A000292, A007582, A093123.

a[n_, m_] := ((n + 2)/2)^m*Sum[1/Sin[k*(Pi/(n + 2))]^(2*m), {k, 1, n + 1}]; Table[a[n - m, m] // FullSimplify, {n, 0, 10}, {m, 0, n}] // Flatten

a(n,m)={t=Pi/(n+2);u=1+n/2;round(sum(k=1,n+1,(u/sin(k*t)^2)^m))} \\ M. F. Hasler, Dec 03 2014

First formulas for rows:
  a(0,m) = 1.
  a(1,m) = 2^(m + 1).
  a(2,m) = 2^m + 2^(2*m + 1).
  a(3,m) = 2*((5 - sqrt(5))^m + (5 + sqrt(5))^m).
  a(4,m) = 2^(2*m + 1) + 3^m + 2^(2*m + 1)*3^m.
First formulas for columns:
  a(n,0) = n + 1.
  a(n,1) = (n + 1)*(n + 2)*(n + 3)/6.
  a(n,2) = coefficient of x^n in the expansion of (1 - x^4)/(1 - x)^8.
Let b(N,m) be (N/2)^m times the coefficient of x^(2*m) in 1-N*x*cot(N*arcsin(x))/ sqrt(1-x^2). Then for m>0, a(n,m) = b(n+2,m). - Ira M. Gessel, Apr 04 2023

cp's OEIS Frontend

A093129 Binomial transform of Fibonacci(2n-1) (A001519).

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A247239 Array a(n,m) = ((n+2)/2)^mSum_{k=1..n+1} 1/sin(kPi/(n+2))^(2m), n>=0, k>=0, read by ascending antidiagonals.

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cp's OEIS Frontend

A093129 Binomial transform of Fibonacci(2n-1) (A001519).

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A247239 Array a(n,m) = ((n+2)/2)^m*Sum_{k=1..n+1} 1/sin(k*Pi/(n+2))^(2m), n>=0, k>=0, read by ascending antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A247239 Array a(n,m) = ((n+2)/2)^mSum_{k=1..n+1} 1/sin(kPi/(n+2))^(2m), n>=0, k>=0, read by ascending antidiagonals.