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A113534
Ascending descending base exponent transform of the flipped tribonacci substitution (A092782).
1
1, 3, 6, 7, 20, 10, 39, 12, 26, 19, 20, 43, 21, 78, 24, 53, 30, 57, 43, 88, 61, 59, 56, 43, 90, 42, 155, 46, 109, 53, 122, 75, 105, 114, 73, 122, 62, 197, 63, 172, 71, 136, 96, 183, 140, 122, 139, 86, 179, 81, 304, 83, 185, 98, 153, 162, 160, 261, 121, 192, 107, 236, 126
OFFSET
1,2
COMMENTS
The flipped tribonacci substitution (A092782) b(n) is the fixed point of the morphism 1 -> 12, 2 -> 13, 3 -> 1, starting from b(1) = 1. The transformed sequence a(n) satisfies n <= a(n) <= 27 n but the bound can be determined to be much tighter.
LINKS
V. F. Sirvent, Semigroups and the self-similar structure of the flipped tribonacci substitution, Applied Math. Letters, 12 (1999), 25-29. [Contains many further references.]
FORMULA
a(n) = Sum_{k=1..n} A092782(k)^(A092782(n-k+1)). - G. C. Greubel, May 17 2017
EXAMPLE
a(1) = A092782(1)^A092782(1) = 1^1 = 1.
a(2) = A092782(1)^A092782(2) + A092782(2)^A092782(1) = 1^2 + 2^1 = 3.
a(3) = 1^1 + 2^2 + 1^1 = 6.
a(4) = 1^3 + 2^1 + 1^2 + 3^1 = 7.
a(5) = 1^1 + 2^3 + 1^1 + 3^2 + 1^1 = 20.
a(6) = 1^2 + 2^1 + 1^3 + 3^1 + 1^2 + 2^1 = 10.
a(7) = 1^1 + 2^2 + 1^1 + 3^3 + 1^1 + 2^2 + 1^1 = 39.
a(8) = 1^1 + 2^1 + 1^2 + 3^1 + 1^3 + 2^1 + 1^2 + 1^1 = 12.
a(9) = 1^2 + 2^1 + 1^1 + 3^2 + 1^1 + 2^3 + 1^1 + 1^2 + 2^1 = 26.
a(10) = 1^1 + 2^2 + 1^1 + 3^1 + 1^2 + 2^1 + 1^3 + 1^1 + 2^2 + 1^1 = 19.
a(11) = 1^3 + 2^1 + 1^2 + 3^1 + 1^1 + 2^2 + 1^1 + 1^3 + 2^1 + 1^2 + 3^1 = 20.
a(12) = 1^1 + 2^3 + 1^1 + 3^2 + 1^1 + 2^1 + 1^2 + 1^1 + 2^3 + 1^1 + 3^2 + 1^1 = 43.
MATHEMATICA
A092782[n_] := Nest[Function[l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {1, 3}, 3 -> {1}})]}], {1}, n][[1]]; Table[Sum[(A092782[k][[k]])^((A092782[n - k + 1][[n - k + 1]])), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Jan 13 2006
EXTENSIONS
a(3) corrected by Giovanni Resta, Jun 13 2016
a(13) onward from G. C. Greubel, May 18 2017
STATUS
approved