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A345290 a(n) is obtained by replacing 2^k in binary expansion of n with Fibonacci(-k-2). 3

%I #13 Jun 17 2021 09:11:56

%S 0,-1,2,1,-3,-4,-1,-2,5,4,7,6,2,1,4,3,-8,-9,-6,-7,-11,-12,-9,-10,-3,

%T -4,-1,-2,-6,-7,-4,-5,13,12,15,14,10,9,12,11,18,17,20,19,15,14,17,16,

%U 5,4,7,6,2,1,4,3,10,9,12,11,7,6,9,8,-21,-22,-19,-20,-24

%N a(n) is obtained by replacing 2^k in binary expansion of n with Fibonacci(-k-2).

%C This sequence is a variant of A022290; here we consider Fibonacci numbers with negative indices (A039834), there Fibonacci numbers with positive indices (A000045).

%C After the initial 0, the sequence alternates runs of positive terms and runs of negative terms, the k-th run having 2^(k-1) terms.

%H Rémy Sigrist, <a href="/A345290/b345290.txt">Table of n, a(n) for n = 0..8191</a>

%F a(n) = A022290(A063695(n)) - A022290(A063694(n)).

%F a(n) = A022290(n) iff n belongs to A062880.

%F a(n) = -A022290(n) iff n belongs to A000695.

%F a(n) = 0 iff n = 0.

%F a(n) = 1 iff n belongs to A072197.

%F a(n) = 2 iff n belongs to A080675.

%F a(n) = -1 iff n belongs to A020989.

%F a(n) = -2 iff n belongs to A136412.

%e For n = 3:

%e - 3 = 2^1 + 2^0,

%e - so a(3) = A039834(2+1) + A039834(2+0) = 2 - 1 = 1.

%o (PARI) a(n) = { my (v=0, e); while (n, n-=2^e=valuation(n, 2); v+=fibonacci(-2-e)); v }

%Y Cf. A000045, A000695, A020989, A022290, A039834, A062880, A063694, A063695, A072197, A080675, A136412, A345291, A345292.

%K sign,base

%O 0,3

%A _Rémy Sigrist_, Jun 13 2021

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Last modified March 28 07:33 EDT 2024. Contains 371235 sequences. (Running on oeis4.)