A056487 a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd.

1, 3, 5, 15, 25, 75, 125, 375, 625, 1875, 3125, 9375, 15625, 46875, 78125, 234375, 390625, 1171875, 1953125, 5859375, 9765625, 29296875, 48828125, 146484375, 244140625, 732421875, 1220703125, 3662109375, 6103515625, 18310546875, 30517578125, 91552734375

Offset: 0

Author

Marks R. Nester

Keywords

• nonn
• easy

Comments

Apparently identical to A111386! Is this a theorem? - Klaus Brockhaus, Jul 21 2009

For n > 1, number of necklaces with n-1 beads and 5 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

References

• M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Links

• Index entries for linear recurrences with constant coefficients, signature (0,5).

Crossrefs

Column 5 of A284855.

Cf. A029744, A038754, A056391, A056451, A087205, A087206, A111386.

Programs

Magma

[n le 2 select 2*n-1 else 5*Self(n-2): n in [1..28]]; // Bruno Berselli, Mar 24 2011

Maple

A056487:=n->3^((1-(-1)^n)/2)*5^((2*n+(-1)^n-1)/4): seq(A056487(n), n=0..40); # Wesley Ivan Hurt, Nov 24 2016

Mathematica

Table[3^((1 - (-1)^n)/2)*5^((2*n + (-1)^n - 1)/4), {n, 0, 30}] (* Wesley Ivan Hurt, Nov 24 2016 *)
CoefficientList[Series[(1 + 3 x)/(1 - 5 x^2), {x, 0, 31}], x] (* Michael De Vlieger, Nov 24 2016 *)
LinearRecurrence[{0, 5}, {1, 3}, 35] (* Vincenzo Librandi, Nov 25 2016 *)
k=5; Table[(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2]) / 2, {n, -1, 30}] (* Robert A. Russell, Sep 21 2018 *)

PARI

a(n)=if(n%2,3,1)*5^(n\2) \\ Charles R Greathouse IV, Oct 07 2015

Python

def A056487(n): return 5**(n>>1)*(3 if n&1 else 1) # Chai Wah Wu, Oct 27 2022

Formula

a(n+2) = 5*a(n), a(0)=1, a(2)=3.

Binomial transform of A087205. Binomial transform is A087206. - Paul Barry, Aug 25 2003

G.f.: (1+3*x)/(1-5*x^2); a(n) = 5^(n/2)(1/2 + 3*sqrt(5)/10 + (1/2 - 3*sqrt(5)/10)(-1)^n). - Paul Barry, Mar 19 2004

2nd inverse binomial transform of Fibonacci(3n+2). - Paul Barry, Apr 16 2004

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

a(n) = 3^((1 - (-1)^n)/2) * 5^((2*n + (-1)^n-1)/4). - Bruno Berselli, Mar 24 2011

a(n+1) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 2, where k=5 is the number of possible colors. - Robert A. Russell, Sep 22 2018

E.g.f.: cosh(sqrt(5)*x) + 3*sinh(sqrt(5)*x)/sqrt(5). - Stefano Spezia, Jun 06 2023

Extensions

Changed one 'even' to 'odd' in the definition. - R. J. Mathar, Oct 06 2010