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A333994
Arithmetical complexity of the regular paperfolding sequence (A014577).
2
1, 2, 4, 8, 16, 24, 32, 44, 52, 64, 76, 86, 96, 106, 116, 124, 132, 140, 148, 156, 164, 172, 180, 188, 196, 204, 212, 220, 228, 236, 244, 252, 260, 268, 276, 284, 292, 300, 308, 316, 324, 332, 340, 348, 356, 364, 372, 380, 388, 396, 404, 412, 420, 428, 436
OFFSET
0,2
COMMENTS
Avgustinovich, Fon-Der-Flaas, and Frid define arithmetical complexity of a sequence t as the number of distinct subwords of length n formed by taking terms in arithmetic progression, so t(s), t(s+d), t(s+2*d), ..., t(s+(n-1)*d), each term a step d>=1 apart. For d=1, these are the ordinary subwords (factors) so that arithmetical complexity >= factor complexity, which here is a(n) >= A337120(n).
LINKS
S. V. Avgustinovich, D. G. Fon-Der-Flaas, and A. E. Frid, Arithmetical Complexity of Infinite Words, in Proceedings of the International Colloquium on Words, Languages & Combinatorics III (ICWLC), Kyoto, March 2000, World Scientific Publishing, 2003, pages 51-62. Also third author's copy. See section 4 final example 2, a(n) = f^A(n).
FORMULA
a(1..13) = 2,4,8,16,24, 32,44,52,64,76, 86,96,106, and a(n) = 8*n + 4 for n >= 14. [Avgustinovich, Fon-Der-Flaas, and Frid]
From Colin Barker, Sep 05 2020: (Start)
G.f.: (1 + x^2 + 2*x^3 + 4*x^4 + 4*x^7 - 4*x^8 + 4*x^9 - 2*x^11 - 2*x^15) / (1 - x)^2.
a(n) = 2*a(n-1) - a(n-2) for n >= 16. (End)
EXAMPLE
For n=4, all subwords of length 4 occur in arithmetic progressions so a(4)=16. These are the 12 ordinary subwords of the paperfolding sequence (A337120(4) = 12) and the 4 further 0000, 0101, 1010, 1111 which are arithmetic progressions in the odd terms. (Odd terms alternate 0,1.)
MATHEMATICA
LinearRecurrence[{2, -1}, {1, 2, 4, 8, 16, 24, 32, 44, 52, 64, 76, 86, 96, 106, 116, 124}, 100] (* Paolo Xausa, Feb 29 2024 *)
CROSSREFS
Cf. A014577, A337120 (factor complexity), A214613 (Abelian complexity).
Sequence in context: A376617 A318654 A375979 * A376065 A305656 A005943
KEYWORD
nonn,easy
AUTHOR
Kevin Ryde, Sep 04 2020
STATUS
approved