A247282 Run Length Transform of A001317.
1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 15, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 15, 17, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 15, 3, 3, 3, 9, 3, 3, 9, 15, 5, 5, 5, 15, 15, 15, 17, 51, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 15, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 15, 17, 3, 3, 3, 9, 3, 3, 9, 15, 3, 3, 3, 9, 9, 9, 15, 45
Offset: 0
Keywords
Examples
115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus a(115) = A001317(2-1) * A001317(3-1) = 3*5 = 15. From _Omar E. Pol_, Feb 15 2015: (Start) Written as an irregular triangle in which row lengths are the terms of A011782: 1; 1; 1,3; 1,1,3,5; 1,1,1,3,3,3,5,15; 1,1,1,3,1,1,3,5,3,3,3,9,5,5,15,17; 1,1,1,3,1,1,3,5,1,1,1,3,3,3,5,15,3,3,3,9,3,3,9,15,5,5,5,15,15,15,17,51; ... Right border gives 1 together with A001317. (End)
Links
- Antti Karttunen, Table of n, a(n) for n = 0..8192
Crossrefs
Programs
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Mathematica
a1317[n_] := FromDigits[ Table[ Mod[Binomial[n-1, k], 2], {k, 0, n-1}], 2]; Table[ Times @@ (a1317[Length[#]]&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 100}] (* Jean-François Alcover, Jul 11 2017 *)
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Python
# uses RLT function from A278159 def A247282(n): return RLT(n,lambda m: int(''.join(str(int(not(~(m-1)&k))) for k in range(m)),2)) # Chai Wah Wu, Feb 04 2022
Comments