A120385 - OEIS

A120385

If a(n-1) = 1 then largest value so far + 1, otherwise floor(a(n-1)/2); or table T(n,k) with T(n,0) = n, T(n,k+1) = floor(T(n,k)/2).

1, 2, 1, 3, 1, 4, 2, 1, 5, 2, 1, 6, 3, 1, 7, 3, 1, 8, 4, 2, 1, 9, 4, 2, 1, 10, 5, 2, 1, 11, 5, 2, 1, 12, 6, 3, 1, 13, 6, 3, 1, 14, 7, 3, 1, 15, 7, 3, 1, 16, 8, 4, 2, 1, 17, 8, 4, 2, 1, 18, 9, 4, 2, 1, 19, 9, 4, 2, 1, 20, 10, 5, 2, 1, 21, 10, 5, 2, 1, 22, 11, 5, 2, 1, 23, 11, 5, 2, 1, 24, 12, 6, 3, 1, 25

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OFFSET

1,2

COMMENTS

Although not strictly a fractal sequence as defined in the Kimberling link, this sequence has many fractal properties. If the first instance of each value is removed, the result is the original sequence with each row repeated twice. Removing all odd-indexed instances of each value does give the original sequence.

LINKS

Alois P. Heinz, Rows n = 1..2048, flattened

C. Kimberling, Fractal sequences

FORMULA

T(n,k) = floor(n/2^(k-1)).

From Peter Bala, Feb 02 2013: (Start)

The n-th row polynomial R(n,t) = Sum_{k>=0} t^k*floor(n/2^k) and satisfies the recurrence equation R(n,t) = t*R(floor(n/2),t) + n, with R(1,t) = 1.

O.g.f. Sum_{n>=1} R(n,t)*x^n = 1/(1-x)*Sum_{n>=0} t^n*x^(2^n)/(1 - x^(2^n)).

Product_{n>=1} ( 1 + x^((t^n - 2^n)/(t-2)) ) = 1 + Sum_{n>=1} x^R(n,t) = 1 + x + x^(2 + t) + x^(3 + t) + x^(4 + 2*t + t^2) + .... For related sequences see A050292 (t = -1), A001477(t = 0), A005187 (t = 1) and A080277 (t = 2).

(End)

EXAMPLE

The table starts:

2, 1;

3, 1;

4, 2, 1;

5, 2, 1;

6, 3, 1;

7, 3, 1;

8, 4, 2, 1;

MAPLE

T:= proc(n) T(n):= `if`(n=1, 1, [n, T(iquo(n, 2))][]) end: