A113535 Ascending descending base exponent transform of the tribonacci substitution (A100619).
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
Examples
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.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..500
- 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.]
Crossrefs
Programs
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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 *)
Formula
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
Comments