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A374960 Numbers k such that 2^k and 2^(k+1) have the same number of terms in their Zeckendorf representation (A007895).

%I A374960 #11 Jul 26 2024 08:57:43
%S A374960 0,5,6,7,11,18,20,25,39,52,61,96,104,157,176,199,206,210,279,326,333,
%T A374960 339,369,380,397,411,426,473,542,576,743,860,898,921,961,970,993,1024,
%U A374960 1043,1049,1100,1121,1176,1184,1193,1199,1206,1230,1253,1376,1380,1387,1435
%N A374960 Numbers k such that 2^k and 2^(k+1) have the same number of terms in their Zeckendorf representation (A007895).
%C A374960 Numbers k such that A020908(k) = A020908(k+1).
%C A374960 The corresponding values of A020908(k) are 1, 3, 3, 3, 6, 7, 8, 9, 18, 20, 28, 44, 37, ... .
%H A374960 Amiram Eldar, <a href="/A374960/b374960.txt">Table of n, a(n) for n = 1..550</a> (terms below 10^5)
%e A374960 0 is a term since the Zeckendorf representation of 2^0 = 1 is A014417(1) = 1, and the Zeckendorf representation of 2^1 = 2 is A014417(2) = 10, so A020908(0) = A020908(1) = 1.
%e A374960 5 is a term since the Zeckendorf representation of 2^5 = 32 is A014417(32) = 1010100, and the Zeckendorf representation of 2^6 = 64 is A014417(64) = 100010001, so A020908(5) = A020908(6) = 3.
%t A374960 z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* _Alonso del Arte_ at A007895 *)
%t A374960 s[n_] := s[n] = z[2^n]; Select[Range[0, 1500], s[#] == s[# + 1] &]
%o A374960 (PARI) A007895(n)=if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ _Charles R Greathouse IV_ at A007895
%o A374960 lista(kmax) = {my(z1 = A007895(1), z2); for(k = 1, kmax, z2 = A007895(2^k); if(z1 == z2, print1(k-1 , ", ")); z1 = z2);}
%Y A374960 Cf. A007895, A014417, A020908, A353986.
%Y A374960 A374961 is a subsequence.
%K A374960 nonn,base
%O A374960 1,2
%A A374960 _Amiram Eldar_, Jul 25 2024