a(n) = a[n/4] + a[n/8] + a[n/16] + ..., where a(0) = 0, a(1) = 1, and [ ] = floor.

2

`%I #14 May 30 2024 22:25:58
`

`%S 0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,
`

`%T 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,5,5,5,
`

`%U 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5
`

`%N a(n) = a[n/4] + a[n/8] + a[n/16] + ..., where a(0) = 0, a(1) = 1, and [ ] = floor.
`

`%C Every term is a Fibonacci number (A000045), and every nonnegative Fibonacci number occurs.
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`%F The initial 16 terms (0s and 1s) are followed by 16 twos, then 32 threes, then 64 fives,... . Specifically, for m>=3, there are 2^(m+1) F(m)'s.
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`%e a(20) = a(5) + a(2) + a(1) = 1 + 0 + 1 = 2.
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`%t a[0] = 0; a[1] = 1;
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`%t a[n_] := a[n] = Sum[a[Floor[n/2^k]], {k, 2, n}]
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`%t Table[a[n], {n, 0, 570}]
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`%Y Cf. A000045, A005187, A072376 (sum starts with a[n/2]), A373096, A373097.
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`%K nonn
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`%O 0,17
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`%A _Clark Kimberling_, May 28 2024
`

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