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A077957
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Powers of 2 alternating with zeros.
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60
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1, 0, 2, 0, 4, 0, 8, 0, 16, 0, 32, 0, 64, 0, 128, 0, 256, 0, 512, 0, 1024, 0, 2048, 0, 4096, 0, 8192, 0, 16384, 0, 32768, 0, 65536, 0, 131072, 0, 262144, 0, 524288, 0, 1048576, 0, 2097152, 0, 4194304, 0, 8388608, 0, 16777216, 0, 33554432, 0, 67108864, 0, 134217728, 0, 268435456
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OFFSET
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0,3
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COMMENTS
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Normally sequences like this are not included, since with the alternating 0's deleted it is already in the database.
0,1,0,2,0,4,0,8,0,16,... is the inverse binomial transform of A000129 (Pell numbers). - Philippe Deléham, Oct 28 2008
Number of maximal self-avoiding walks from the NW to SW corners of a 3 X n grid.
Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012
This sequence occurs in the length L(n) = sqrt(2)^n of Lévy's C-curve at the n-th iteration step. Therefore, L(n) is the Q(sqrt(2)) integer a(n) + a(n-1)*sqrt(2), with a(-1) = 0. For a variant of this C-curve see A251732 and A251733. - Wolfdieter Lang, Dec 08 2014
a(n) counts walks (closed) on the graph G(1-vertex,2-loop,2-loop). Equivalently the middle entry (2,2) of A^n where the adjacency matrix of digraph is A=(0,1,0;1,0,1;0,1,0). - David Neil McGrath, Dec 19 2014
a(n-2) is the number of compositions of n into even parts. For example, there are 4 compositions of 6 into even parts: (6), (222), (42), and (24). - David Neil McGrath, Dec 19 2014
a(n) counts degree n fixed points of GF(2)[x]'s automorphisms. Proof: given a field k, k[x]'s automorphisms are determined by k's automorphisms and invertible affine maps x -> ax + b. GF(2) is rigid and has only one unit so its only nontrivial automorphism is x -> x + 1. For n = 0 we have 1 fixed point, the constant polynomial 1. (Taking the convention that 0 is not a degree 0 polynomial.) For n = 1 we have 0 fixed points as x -> x + 1 -> x are the only degree 1 polynomials. Note that if f(x) is a fixed point, then f(x) + 1 is also a fixed point. Given f(x) a degree n fixed point, we can assume WLOG x | f(x). Applying the automorphism, we then have x + 1 | f(x). Now note that f(x) / (x^2 + x) must be a fixed point, so any fixed point of degree n must either be of the form g(x) * (x^2 + x) or g(x) * (x^2 + x) + 1 for a unique degree n - 2 fixed point g(x). Therefore we have the recurrence relation a(n) = 2 * a(n - 2) as desired. - Keith J. Bauer, Mar 19 2024
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LINKS
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FORMULA
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G.f.: 1/(1-2*x^2).
E.g.f.: cosh(x*sqrt(2)).
a(n) = (1 - n mod 2) * 2^floor(n/2).
a(n) = sqrt(2)^n*(1+(-1)^n)/2. - Paul Barry, May 13 2003
a(n) = 2*a(n-2) with a(0)=1, a(1)=0. - Jim Singh, Jul 12 2018
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MAPLE
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MATHEMATICA
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a077957[n_] := Riffle[Table[2^i, {i, 0, n - 1}], Table[0, {n}]]; a077957[29] (* Michael De Vlieger, Dec 22 2014 *)
CoefficientList[Series[1/(1 - 2*x^2), {x, 0, 50}], x] (* G. C. Greubel, Apr 12 2017 *)
Riffle[2^Range[0, 30], 0, {2, -1, 2}] (* Harvey P. Dale, Jan 06 2022 *)
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PROG
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(PARI) a(n)=if(n<0||n%2, 0, 2^(n/2))
(Haskell)
(Sage)
x, y = -1, 1
while True:
yield -x
x, y = x + y, x - y
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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