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Search: id:A073121
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| A073121 |
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Recurrence occurring in the analysis of a regular expression algorithm. |
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+0 2
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| 1, 4, 10, 16, 28, 40, 52, 64, 88, 112, 136, 160, 184, 208, 232, 256, 304, 352, 400, 448, 496, 544, 592, 640, 688, 736, 784, 832, 880, 928, 976, 1024, 1120, 1216, 1312, 1408, 1504, 1600, 1696, 1792, 1888, 1984, 2080, 2176, 2272, 2368, 2464, 2560, 2656, 2752
(list; graph; listen)
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OFFSET
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1,2
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REFERENCES
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J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
K. Ellul, J. Shallit and M.-w. Wang, Regular expressions: new results and open problems, in Descriptional Complexity of Formal Systems (DCFS), Proceedings of workshop, London, Ontario, Canada, 21-24 August 2002, pp. 17-34.
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LINKS
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J.-P. Allouche and J. Shallit, The Ring of k-regular Sequences, II
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions
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FORMULA
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a(n) == 1 (mod 3), a(n+1)-a(n) = 3* A053644(n). If k>=1: a(2^k)=4^k, a(3*2^k)=(10/9)*4^k. More generally a(m*2^k) = a(m)*4^k. Hence for any n, n^2 <= a(n) <= C*n^2 where C is a constant 1.125 < C < 1.14 and it seems that C = lim k -> infinity a(A001045(k))/A001045(k)^2 where A001045(k) ={2^n - (-1)^n}/3 is the Jacobsthal sequence. In other words, in the range 2^k<=n<=2^(k+1) the maximum of a(n)/n^2 is reached for the only possible n in the Jacobsthal sequence. - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 26 2002
a(n) = 2*(a(floor(n/2))+a(ceil(n/2))) for n >= 2; alternatively a(n) = 2^c(n+2b) where n = 2^c + b, 0 <= b < 2^c
G.f.: 3/(1-x)^2 * ((2x+1)/3 + sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 18 2003
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EXAMPLE
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a(1)=1, a(2) = 2(a(1)+a(1)) = 4, a(3) = 2(a(2)+a(1)) = 10
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CROSSREFS
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Sequence in context: A089340 A132227 A112984 this_sequence A167346 A027430 A027425
Adjacent sequences: A073118 A073119 A073120 this_sequence A073122 A073123 A073124
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KEYWORD
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nonn
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AUTHOR
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Jeffrey Shallit (shallit(AT)graceland.uwaterloo.ca), Aug 25 2002
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