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A113535
Ascending descending base exponent transform of the tribonacci substitution (A100619).
2
1, 3, 8, 19, 32, 9, 11, 16, 26, 19, 29, 24, 47, 70, 28, 31, 58, 89, 35, 50, 65, 108, 65, 51, 52, 90, 101, 82, 101, 88, 122, 63, 81, 92, 153, 110, 89, 125, 110, 92, 101, 155, 90, 127, 196, 142, 87, 138, 207, 112, 112, 135, 217, 150, 124, 115, 204, 245, 139, 158, 189, 268, 121, 155, 154
OFFSET
1,2
COMMENTS
Sirvent comments that in spite of the similarity of this map to the one in A092782, the two sequences have very different properties. They have different complexities, different Rauzy fractals, etc.
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} A100619(k)^(A100619(n-k-1)). - G. C. Greubel, May 18 2017
EXAMPLE
a(1) = A100619(1)^A100619(1) = 1^1 = 1.
a(2) = A100619(1)^A100619(2) + A100619(2)^A100619(1) = 1^2 + 2^1 = 3.
a(3) = 1^3 + 2^2 + 3^1 = 8.
a(4) = 1^1 + 2^3 + 3^2 + 1^1 = 19.
a(5) = 1^1 + 2^1 + 3^3 + 1^2 + 1^1 = 32.
a(6) = 1^1 + 2^1 + 3^1 + 1^3 + 1^2 + 1^1 = 9.
a(7) = 1^2 + 2^1 + 3^1 + 1^1 + 1^3 + 1^2 + 2^1 = 11.
a(8) = 1^1 + 2^2 + 3^1 + 1^1 + 1^1 + 1^3 + 2^2 + 1^1 = 16.
a(9) = 1^1 + 2^1 + 3^2 + 1^1 + 1^1 + 1^1 + 2^3 + 1^2 + 2^1 = 26.
a(10) = 1^1 + 2^2 + 3^1 + 1^2 + 1^1 + 1^1 + 2^1 + 1^3 + 2^2 + 1^1 = 19.
a(11) = 1^2 + 2^1 + 3^2 + 1^1 + 1^2 + 1^1 + 2^1 + 1^1 + 2^3 + 1^2 + 2^1 = 29.
a(12) = 1^3 + 2^2 + 3^1 + 1^2 + 1^1 + 1^2 + 2^1 + 1^1 + 2^1+ 1^3 + 2^2 + 3^1 = 24.
MATHEMATICA
A100619:= Nest[Function[l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {3, 1}, 3 -> {1}})]}], {1}, 8][[1]]; Table[Sum[(A100619[[k]])^(A100619[[n-k+1]]), {k, 1, n}], {n, 1, 100}] (* G. C. Greubel, May 18 2017 *)
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Jan 13 2006
EXTENSIONS
Terms a(13) to a(50) from G. C. Greubel, May 18 2017
Terms a(51) onward added by G. C. Greubel, Jan 03 2019
STATUS
approved