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A363070 Take the terms 0..n of the infinite Fibonacci word A003849, regard them as a number in Fibonacci base.

%I A363070 #65 Jun 08 2023 10:34:23
%S A363070 0,1,2,3,6,10,17,28,45,74,120,194,315,510,826,1337,2163,3501,5665,
%T A363070 9167,14833,24000,38834,62835,101669,164505,266175,430681,696857,
%U A363070 1127538,1824396,2951935,4776331,7728267,12504599,20232867,32737467,52970334,85707802,138678137,224385940,363064078
%N A363070 Take the terms 0..n of the infinite Fibonacci word A003849, regard them as a number in Fibonacci base.
%H A363070 Gandhar Joshi, <a href="/A363070/b363070.txt">Table of n, a(n) for n = 0..1000</a>
%H A363070 Agnibha Banerjee, <a href="/A363070/a363070.txt">Python Program</a>, May 16 2023.
%F A363070 a(n) = Sum_{i=0..n} A003849(i)*Fibonacci(n-i+2).
%F A363070 a(n) = a(n-1) + a(n-2) + A003849(n) + A003849(n-1).
%F A363070 a(n) = a(n-1) + a(n-2) + A005713(n-1). - _Kevin Ryde_, May 20 2023
%e A363070 0 ->                            0 -> a(0) = 0,
%e A363070 0,1 ->                         01 -> a(1) = 1,
%e A363070 0,1,0 ->                      010 -> a(2) = 2,
%e A363070 0,1,0,0 ->                   0100 -> a(3) = 3,
%e A363070 0,1,0,0,1 ->                01001 -> a(4) = 6,
%e A363070 0,1,0,0,1,0 ->             010010 -> a(5) = 10,
%e A363070 0,1,0,0,1,0,1 ->          0100101 -> a(6) = 17,
%e A363070 0,1,0,0,1,0,1,0 ->       01001010 -> a(7) = 28,
%e A363070 0,1,0,0,1,0,1,0,0 ->    010010100 -> a(8) = 45,
%e A363070 0,1,0,0,1,0,1,0,0,1 -> 0100101001 -> a(9) = 74.
%o A363070 (Python) # see linked program
%o A363070 (Python)
%o A363070 def aupto(n): # produces n terms, indices 0..n-1
%o A363070     F1, F, a = [0], [0, 1], [0, 1]
%o A363070     while len(F) < n:
%o A363070         F1, F = F, F+F1
%o A363070     [a.append(a[-2]+a[-1]+F[i]+F[i-1]) for i in range(2, n)]
%o A363070     return a
%o A363070 print(aupto(42)) # _Michael S. Branicky_, May 17 2023
%Y A363070 Cf. A003849 (Fibonacci word), A005713, A189920 (Zeckendorf digits).
%Y A363070 Cf. A182028.
%K A363070 base,nonn
%O A363070 0,3
%A A363070 _Gandhar Joshi_, May 16 2023