A141125 -id:A141125 - cp's OEIS frontend

A056450 a(n) = (3*2^n - (-2)^n)/2.

1, 4, 4, 16, 16, 64, 64, 256, 256, 1024, 1024, 4096, 4096, 16384, 16384, 65536, 65536, 262144, 262144, 1048576, 1048576, 4194304, 4194304, 16777216, 16777216, 67108864, 67108864, 268435456, 268435456, 1073741824, 1073741824, 4294967296

Offset: 0

Marks R. Nester

Number of palindromes of length n using a maximum of four different symbols.
Number of achiral rows of n colors using up to four colors. - Robert A. Russell, Nov 09 2018
Interleaving of A000302 and 4*A000302.
Unsigned version of A141125.
Binomial transform is A164907. Second binomial transform is A164908. Third binomial transform is A057651. Fourth binomial transform is A016129.

			At length n=1 there are a(1)=4 palindromes, A, B, C, D.
At length n=2, there are a(2)=4 palindromes, AA, BB, CC, DD.
At length n=3, there are a(3)=16 palindromes, AAA, BBB, CCC, DDD, ABA, BAB, ... , CDC, DCD.

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]

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Sean A. Irvine, Walks on Graphs.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,4).

Column k=4 of A321391.
Cf. A000302 (powers of 4), A056450, A141125, A164907, A164908, A057651, A016129.
Cf. A016116.
Essentially the same as A213173.
Cf. A000302 (oriented), A032121 (unoriented), A032087(n>1) (chiral).

Magma
```
[ (3*2^n-(-2)^n)/2: n in [0..31] ];
```

[4^Floor((n+1)/2): n in [0..40]]; // Vincenzo Librandi, Aug 16 2011

Table[4^Ceiling[n/2], {n,0,40}] (* or *)
CoefficientList[Series[(1 + 4 x)/((1 + 2 x) (1 - 2 x)), {x, 0, 31}], x] (* or *)
LinearRecurrence[{0, 4}, {1, 4}, 40] (* Robert A. Russell, Nov 07 2018 *)

a(n)=4^((n+1)\2) \\ Charles R Greathouse IV, Apr 08 2012

a(n)=(3*2^n-(-2)^n)/2 \\ Charles R Greathouse IV, Oct 03 2016

a(n) = 4^floor((n+1)/2).
a(n) = 4*a(n-2) for n > 1; a(0) = 1, a(1) = 4.
G.f.: (1+4*x) / (1-4*x^2). - R. J. Mathar, Jan 19 2011 [Adapted to offset 0 by Robert A. Russell, Nov 07 2018]
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = 4*abs(A164111(n-1)). - R. J. Mathar, Jan 19 2011
a(n) = C(4,0)*A000007(n) + C(4,1)*A057427(n) + C(4,2)*A056453(n) + C(4,3)*A056454(n) + C(4,4)*A056455(n). - Robert A. Russell, Nov 08 2018

a(0)=1 prepended by Robert A. Russell, Nov 07 2018

Edited by N. J. A. Sloane, Sep 29 2019

A100095 An inverse Chebyshev transform of the Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 5, 7, 25, 41, 125, 225, 625, 1195, 3125, 6227, 15625, 32059, 78125, 163727, 390625, 831505, 1953125, 4206145, 9765625, 21215481, 48828125, 106782837, 244140625, 536618341, 1220703125, 2693492305, 6103515625, 13507578125

Offset: 0

Paul Barry, Nov 03 2004

Image of x/(1-x-x^2) under the transform g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. This is the inverse of the Chebyshev transform which takes A(x) to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).
Hankel transform is (-1)^n*(2^n-0^n)/2. Hankel transform of a(n+1) is A141125. - Paul Barry, Jun 05 2008
Basically A000351 interleaved with A144635. - Peter Luschny, May 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000351, A144635, A000045, A100096, A100097.

CoefficientList[Series[(x^2*Sqrt[1-4*x^2]+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

a(n):=sum(4^(j)*binomial((n-1)/2,j),j,0,(n-1)/2); /* Vladimir Kruchinin, May 31 2014 */

G.f.: (x^2*sqrt(1-4*x^2)+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*Fibonacci(n-2*k).
Conjecture: (-n+2)*a(n) +(-n+3)*a(n-1) +(9*n-22)*a(n-2) +(9*n-31)*a(n-3) +20*(-n+3)*a(n-4) +20*(-n+4)*a(n-5)=0. - R. J. Mathar, Nov 24 2012
Recurrence: (n-2)*a(n) = (9*n-22)*a(n-2) - 20*(n-3)*a(n-4). - Vaclav Kotesovec, Feb 12 2014
a(n) ~ 5^((n-1)/2). - Vaclav Kotesovec, Feb 12 2014
a(n) = sum(j=0..(n-1)/2, 4^(j)*binomial((n-1)/2,j)). - Vladimir Kruchinin, May 31 2014
a(2*n) = 5^n/sqrt(5) - 2^n * (2*n-1)!! * hypergeom([1, n+1/2], [n+1], 4/5)/(5*n!), a(2*n+1) = 5^n. - Vladimir Reshetnikov, Oct 13 2016

cp's OEIS Frontend

A056450 a(n) = (3*2^n - (-2)^n)/2.

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