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A016116
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2^floor(n/2).
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119
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1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024, 2048, 2048, 4096, 4096, 8192, 8192, 16384, 16384, 32768, 32768, 65536, 65536, 131072, 131072, 262144, 262144, 524288, 524288, 1048576, 1048576, 2097152
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OFFSET
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0,3
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COMMENTS
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Powers of 2 doubled up. The usual OEIS policy is to omit the duplicates in such cases (when this would become A000079). This is an exception.
Number of symmetric partitions of n: e.g. 5 = 2+1+2 = 1+3+1 = 1+1+1+1+1 so a(5) = 4; 6 = 3+3 = 2+2+2 = 1+4+1 = 2+1+1+2 = 1+2+2+1 = 1+1+2+1+1 = 1+1+1+1+1+1 so a(6) = 8. - Henry Bottomley, Dec 10 2001
This sequence is the number of digits of each term of A061519. [From Dmitry Kamenetsky, Jan 17 2009]
Starting with offset 1 = binomial transform of [1, 1, -1, 3, -7, 17, -41,...]; where A001333 = (1, 1, 3, 7, 17, 41,...). [From Gary W. Adamson, Mar 25 2009]
a(n+1) is the number of symmetric subsets of [n]={1,2,...,n}. A subset S of [n] is symmetric if k is an element of S implies (n-k+1) is an element of S. - Dennis P. Walsh, Oct 27 2009
INVERT and inverse INVERT transforms give A006138, A039834(n-1).
The Kn21 sums, see A180662, of triangle A065941 equal the terms of this sequence. [Johannes W. Meijer, Aug 15 2011]
First differences of A027383. - Jason Kimberley, Nov 01 2011
Run lengths in A079944. - Jeremy Gardiner, Nov 21 2011
Number of binary palindromes (A006995) between 2^(n-1) and 2^n (for n>1). - Hieronymus Fischer, Feb 17 2012
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
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REFERENCES
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Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
E. Deutsch, Problem 1633, Math. Mag., 74 #5 (2001), p. 403.
D. Merlini, F. Uncini and M. C. Verri, A unified approach to the study of general and palindromic compositions, Integers 4 (2004), A23, 26 pp.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..5000
S. Heubach and T. Mansour, Counting rises, levels and drops in compositions
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1067
Dennis Walsh, Notes on symmetric subsets of {1, 2, ..., n} [From Dennis P. Walsh, Oct 27 2009]
Index to divisibility sequences
Index entries for sequences related to linear recurrences with constant coefficients, signature (0,2).
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FORMULA
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a(n) = a(n-1)*a(n-2)/a(n-3) = 2*a(n-2) = 2^A004526(n).
G.f.: (1+x)/(1-2*x^2)
(1/2+sqrt(1/8))*sqrt(2)^n+(1/2-sqrt(1/8))*(-sqrt(2))^n. - Ralf Stephan, Mar 11 2003
E.g.f.: cosh(sqrt(2)*x)+sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, Jul 16 2003
The signed sequence (-1)^n2^[n/2] has a(n)=(sqrt(2))^n(1/2-sqrt(2)/4)+(-sqrt(2))^n(1/2+sqrt(2)/4). It is the inverse binomial transform of A000129(n-1). - Paul Barry, Apr 21 2004
Diagonal sums of A046854. a(n)=sum{k=0..n, binomial(floor(n/2), k)}. - Paul Barry, Jul 07 2004
a(n) = a(n-2)+2^floor((n-2)/2) - Paul Barry, Jul 14 2004
a(n) = sum{k=0..floor(n/2), binomial(floor(n/2), floor(k/2)) } - Paul Barry, Jul 15 2004
E.g.f.: cosh(asinh(1)+sqrt(2)*x)/sqrt(2). - Michael Somos Feb 28 2005
a(n) = Sum_{k, 0<=k<=n}A103633(n,k). - Philippe DELEHAM, Dec 03 2006
a(n) = 2^(n/2)*((1+(-1)^n)/2+(1-(-1)^n)/(2*sqrt(2))). [From Paul Barry, Nov 12 2009]
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EXAMPLE
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For n=5 the a(5)=4 symmetric subsets of [4] are {1,4}, {2,3}, {1,2,3,4} and the empty set. [Dennis P. Walsh, Oct 27 2009]
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MAPLE
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A016116:= proc(n): 2^floor(n/2) end: seq(A016116(n), n=0..42); [From Dennis P. Walsh, Oct 27 2009]
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MATHEMATICA
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Table[ 2^Floor[n/2], {n, 0, 42}] (from Robert G. Wilson v Jun 05 2004)
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PROG
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(PARI) a(n)=if(n<0, 0, 2^(n\2))
(MAGMA) [2^Floor(n/2): n in [0..50]]; // Vincenzo Librandi, Aug 16 2011
(Maxima) makelist(2^floor(n/2), n, 0, 50); [Martin Ettl, Oct 17 2012]
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CROSSREFS
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Cf. A006995, A057148, A079944, A112030, A112033.
a(n) = A094718(3, n).
Cf. A001333.
See A052955 for partial sums (without the initial term).
Sequence in context: A117575 A131572 A152166 * A060546 A163403 A222955
Adjacent sequences: A016113 A016114 A016115 * A016117 A016118 A016119
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane.
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STATUS
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approved
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