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%I #17 Aug 17 2024 03:22:07
%S 1,3,4,7,8,13,14,20,26,50,55,58,90,140,270,314,603
%N Numbers k such that the number of terms in the Zeckendorf representation of 2^k equals the binary weight of Fibonacci(k).
%C Numbers k such that A007895(A000079(k)) = A000120(A000045(k)), or equivalently A020908(k) = A011373(k).
%C The corresponding values of A020908(k) = A011373(k) are 1, 1, 2, 3, 3, 5, 6, 8, 9, 18, 22, 24, 33, 53, 106, 122, 232, ... .
%C a(18) > 63000, if it exists.
%C a(18) > 333333, if it exists. - _Lucas A. Brown_, Aug 13 2024
%H Lucas A. Brown, <a href="https://github.com/lucasaugustus/oeis/blob/main/A374962.py">Python program</a>.
%e n | k = a(n) | 2^k | A014417(2^k) | F(k) | A007088(F(k)) | Number of 1's
%e --+----------+-----+--------------+------+---------------+--------------
%e 1 | 1 | 2 | 10 | 1 | 1 | 1
%e 2 | 3 | 8 | 10000 | 2 | 10 | 1
%e 3 | 4 | 16 | 100100 | 3 | 11 | 2
%e 4 | 7 | 128 | 1010001000 | 13 | 1101 | 3
%e 5 | 8 | 256 | 100001000010 | 21 | 10101 | 3
%t z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* _Alonso del Arte_ at A007895 *)
%t Select[Range[700], z[2^#] == DigitCount[Fibonacci[#], 2, 1] &]
%o (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 is(k) = A007895(2^k) == hammingweight(fibonacci(k));
%Y Cf. A000045, A000079, A000120, A007088, A007895, A011373, A014417, A020908.
%K nonn,base,more
%O 1,2
%A _Amiram Eldar_, Jul 25 2024
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