A147612 If n is a Jacobsthal number then 1 else 0.
1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0
Keywords
Links
Programs
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Mathematica
With[{s = LinearRecurrence[{1, 2}, {0, 1}, 24]}, Array[If[FreeQ[s, #], 0, 1] &, 105, 0]] (* Michael De Vlieger, Oct 09 2017 *)
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PARI
A147612aux(n,i) = if(!(n%2),n,A147612aux((n+i)/2,-i)); A147612(n) = 0^(A147612aux(n,1)*A147612aux(n,-1)); \\ Antti Karttunen, Oct 09 2017
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Python
def A147612(n): return int((m:=3*n+1).bit_length()>(m-3).bit_length()) if n else 1 # Chai Wah Wu, Apr 18 2025
Formula
a(n) = 0^(j(n,1)*j(n,-1)) with j(n,i) = if n mod 2 = 0 then n else j((n+i)/2,-i).
a(n) = A105348(n), for n <> 1. - R. J. Mathar, Nov 19 2008
Comments