A061420 a(n) = a(ceiling((n-1)*2/3)) + 1 with a(0) = 0.
0, 1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
Offset: 0
Keywords
Examples
a(10) = a(ceiling(9*2/3)) + 1 = a(6) + 1 = 4 + 1 = 5.
Links
- Clark Kimberling, Table of n, a(n) for n = 0..2000
- K. A. C. Bickerstaff, M. Schulte and E. E. Swartzlander, Reduced area multipliers, Proceedings of International Conference on Application Specific Array Processors (ASAP '93), Venice, Italy, 1993, pp. 478-489. See Table 1 p. 480.
- William J. Gilbert, Radix Representations of Quadratic Fields, Journal of Mathematical Analysis and Applications 83 (1981) pp 264-274. Gilbert (page 273) cites Wang and Washburn (below) in connection with the length of the base 3/2 expansion of the even positive integers.
- A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235-240, 1991.
- E. T. H. Wang, Phillip C. Washburn, Problem E2604, American Mathematical Monthly 84 (1977) pp. 821-822.
- Index entries for sequences related to the Josephus Problem
Programs
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Magma
[IsZero(n) select 0 else Self(Floor(2*n/3)+1)+1: n in [0..90]]; // Bruno Berselli, Oct 31 2012
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Maple
a:= n-> `if`(n=0, 0, a(ceil((n-1)*2/3))+1): seq(a(n), n=0..100); # Alois P. Heinz, Oct 29 2012
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Mathematica
(* 1st program, using the alternative definition *) a[0] = 0; a[n_] := a[Floor[2 n/3]] + 1; Table[a[n], {n, 0, 120}] (* 2nd program, using Cloitre's recurrence *) f[x_] := Floor[2 x/3]; g[0, x_] := f[x]; g[k_, x_] := f[g[k - 1, x]]; u[n_] := Flatten[Table[g[k, n], {k, 0, 12}]] v[n_] := First[Position[u[n], 0]]; Flatten[Table[v[n], {n, 1, 120}]] (* 3rd program, using the constant c *) f[n_] := -Floor[-Log[3/2, (n + 1)/1.62227050288476731595695098289932]] Table[f[n], {n, 1, 120}] (* Clark Kimberling, Oct 23 2012 *) -
PARI
a(n) = if(n<0, 0, s=n; c=0; while(floor(s)>0, s=floor(2/3*s); c++); c) \\ Benoit Cloitre, May 26 2007
Formula
a(n) = a(n-1) + 1 if n is in A061419; a(n) = a(n-1) otherwise.
From Clark Kimberling, Oct 19 2012: (Start)
a(n) = a(floor(2*n/3)) + 1, where a(0) = 0 (alternative definition).
Washburn's solution of Problem E2604 (see References) shows that (for n>0), a(n) = -floor(-L((n+1)/c)), where L is the logarithm with base 3/2 and
c = lim_{n->infinity} (2/3)^n*s(n) where s(n) = floor(3*s(n-1)/2) + 1 and s(0)=0. The editors state that "It may be interesting to know whether c is irrational or even transcendental"; c = 1.62227050288476731595695098289932... .
Odlyzko and Wilf also discuss the defining recurrence, and they, after Washburn, give a formula for the sequence using c, as in the third Mathematica program below.
(End)
Comments