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A005578
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a(2*n) = 2*a(2*n-1), a(2*n+1) = 2*a(2*n)-1.
(Formerly M0788)
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60
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1, 1, 2, 3, 6, 11, 22, 43, 86, 171, 342, 683, 1366, 2731, 5462, 10923, 21846, 43691, 87382, 174763, 349526, 699051, 1398102, 2796203, 5592406, 11184811, 22369622, 44739243, 89478486, 178956971, 357913942, 715827883, 1431655766
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OFFSET
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0,3
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COMMENTS
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Might be called the "Arima sequence" after Yoriyuki Arima who in 1769 constructed this sequence as the number of moves of the outer ring in the optimal solution for the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1) = u(k) + v(k), v(k+1) = u(k) + w(k), w(k+1) = v(k) + w(k); let M(k) = Max(u(k), v(k), w(k)); then a(n) = M(n). - Benoit Cloitre, Mar 25 2002
Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..n/2} A071922(n-k, n-2*k). Based on the observation that F_{n+1} = Sum_{k} binomial (n-k, k). - Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 30 2002
Numbers n at which the length of the symmetric signed digit expansion of n with q=2 (i.e., the length of the representation of n in the (-1,0,1)_2 number system) increases. - Ralf Stephan, Jun 30 2003
Row sums of Riordan array (1/(1-x), x/(1-2*x^2)). - Paul Barry, Apr 24 2005
2^(n+1) = 2*a(n) + 2*A001045(n) + A000975(n-1); e.g., 2^6 = 64 = 2*a(5) + 2*A001045(5) + 2*A000975(4) = 2*11 + 2*11 + 2*10. Let a(n), A001045(n) and A000975(n-1) = the legs of a triangle (a, b, c). Then a(n-1), A001045(n-1) and A000975(n-2) = (S-c), (S-b), (S-a), where S = the triangle semiperimeter. Example: a(5), A001045(5) and A000975(4) = triangle (a, b, c) = (11, 11, 10). Then a(4), A001045(4), A000975(3) = (S-c), (S-b), (S-a) = (6, 5, 5). - Gary W. Adamson, Dec 24 2007
a(n) is the number of length-n binary representations of a nonnegative integer that is divisible by 3. The initial digits are allowed to be 0's. a(4) = 6 because we have 0000, 0011, 0110, 1001, 1100, 1111. - Geoffrey Critzer, Jan 13 2014
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 0, 1; 0, 1, 1; 1, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 1, 0, 1; 0, 1, 1]. - R. J. Mathar, Feb 04 2014
With 0 prefixed, this sequence is an autosequence of the first kind because the sequence of first differences A001045 is. Its companion is A052950. - Paul Curtz, Dec 18 2018, edited by M. F. Hasler, Dec 21 2018
Apparently, the sequence gives the distinct values taken by A129761, the first differences of fibbinary numbers. - Rémy Sigrist, Oct 26 2019
The sequence with offset 1 can be generated in three steps starting with A158780. First, put in alternate signs (1, -1, 1, -2, 2, -4, ...) and take the inverse; getting (1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, ...). Take the invert transform of the latter, resulting in the sequence. It follows from the inverti transform being 1, 1, 0, 1, 1, 2, 3, ... that (for example), a(9) = 171 = (1, 1, 0, 1, 1, 2, 3, 5, 8) dot (86, 43, 0, 11, 6, 6, 6, 5, 8) = (86 + 43 + 0 + 11 + 6 + 6 + 6 + 5 + 8). A similar procedure is shown in the Aug 08 2019 comment of A006356. - Gary W. Adamson, Feb 04 2022
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REFERENCES
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R. K. Guy, Graphs and the strong law of small numbers. Graph theory, combinatorics and applications. Vol. 2 (Kalamazoo, MI, 1988), 597-614, Wiley-Intersci. Publ., Wiley, New York, 1991.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Sujith Uthsara Kalansuriya Arachchi, Hung Viet Chu, Jiasen Liu, Qitong Luan, Rukshan Marasinghe, and Steven J. Miller, On a Pair of Diophantine Equations, arXiv:2309.04488 [math.NT], 2023.
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FORMULA
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a(n) = ceiling(2^n/3).
a(n) = 1 + floor((2^n)/3) (proof by mathematical induction).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
G.f.: (1 - x - x^2)/((1-x^2)*(1-2*x)). [Guy, 1988];
E.g.f.: (exp(2*x) - exp(-x))/3 + cosh(x) = (2*exp(2*x) + 3*exp(x) + exp(-x))/6. (End)
The 30 listed terms are given by a(0)=1, a(1)=1 and, for n > 1, by a(n) = a(n-1) + a(n-2) + Sum_{i=0..n-4} Fibonacci(i)*a(n-4-i). - John W. Layman, Jan 07 2000
a(n) = Sum_{k=0..n} (-1)^k*Sum_{j=0..n-k} (if((j-k) mod 2)=0, binomial(n-k, j), 0). - Paul Barry, Jan 25 2005
Let M = the 6 X 6 adjacency matrix of a benzene ring, (reference): [0,1,0,0,0,1; 1,0,1,0,0,0; 0,1,0,1,0,0; 0,0,1,0,1,0; 0,0,0,1,0,1; 1,0,0,0,1,0]. Then a(n) = leftmost nonzero term of M^n * [1,0,0,0,0,0]. E.g.: a(6) = 22 since M^6 * [1,0,0,0,0,0] = [22,0,21,0,21,0]. - Gary W. Adamson, Jun 14 2006
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007
a(n) = 1 + 2^(n-1) - a(n-1) = a(n-1) + 2*a(n-2) - 1 = a(n-2) + 2^(n-2). - Paul Curtz, Jan 31 2009
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MAPLE
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with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL3), b=ZL3], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n), n=2..34); # Zerinvary Lajos, Mar 08 2008
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MATHEMATICA
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LinearRecurrence[{2, 1, -2}, {1, 1, 2}, 40] (* G. C. Greubel, Aug 26 2019 *)
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PROG
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(GAP) List([0..40], n->(2^(n+1)+3+(-1)^n)/6); # Muniru A Asiru, Dec 22 2018
(Sage) [(2^(n+1)+3+(-1)^n)/6 for n in (0..40)] # G. C. Greubel, Aug 26 2019
(Python)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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