A074872 Inverse BinomialMean transform of the Fibonacci sequence A000045 (with the initial 0 omitted).

1, 1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125, 30517578125, 152587890625

Offset: 1

Author

John W. Layman, Sep 12 2002

Keywords

• nonn
• easy

Comments

See A075271 for the definition of the BinomialMean transform.

The inverse binomial transform of 2^n*c(n+1), where c(n) is the solution to c(n) = c(n-1) + k*c(n-2), a(0)=0, a(1)=1 is 1, 1, 4k+1, 4k+1, (4k+1)^2, ... - Paul Barry, Feb 12 2004

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..2000
• Index entries for linear recurrences with constant coefficients, signature (0,5).

Crossrefs

Cf. A000045, A075271, A056451, A063727, A016116, A108411.

Programs

Magma

[5^Floor((n-1)/2): n in [1..40]]; // Vincenzo Librandi, Aug 16 2011

Mathematica

a[1] := 1; a[2] := 1; a[n_] := 5a[n - 2]; Table[a[n], {n, 30}] (* Alonso del Arte, Mar 04 2011 *)

PARI

a(n)=5^((n-1)\2) \\ Charles R Greathouse IV, Oct 03 2016

Formula

a(n) = 5^floor((n-1)/2).

a(1)=1, a(2)=1 and, for n > 2, a(n) = 5*a(n-2).

From Paul Barry, Feb 12 2004: (Start)

G.f.: x*(1+x)/(1-5*x^2);

a(n) = (1/(2*sqrt(5))*((1+sqrt(5))*(sqrt(5))^n - (1-sqrt(5))*(-sqrt(5))^n)).

Inverse binomial transform of A063727 (2^n*Fibonacci(n+1)). (End)

a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011

E.g.f.: (cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x) - 1)/5. - Stefano Spezia, May 24 2024