A068119 Number of steps to reach an integer starting with n + 1/4 and iterating the map x -> x*ceiling(x).

3, 3, 1, 3, 2, 2, 1, 7, 4, 4, 1, 2, 2, 4, 1, 6, 3, 5, 1, 5, 2, 2, 1, 4, 6, 3, 1, 2, 2, 3, 1, 7, 3, 4, 1, 3, 2, 2, 1, 7, 4, 7, 1, 2, 2, 5, 1, 3, 3, 10, 1, 4, 2, 2, 1, 3, 5, 11, 1, 2, 2, 3, 1, 5, 3, 3, 1, 3, 2, 2, 1, 4, 4, 6, 1, 2, 2, 4, 1, 4, 3, 6, 1, 6, 2, 2, 1, 6, 7, 3, 1, 2, 2, 3, 1, 4, 3, 5, 1, 3, 2, 2, 1, 4

Offset: 1

Author

Benoit Cloitre, Aug 30 2002

Keywords

• nonn

Comments

If the initial value is n + 1/2 we get A001511.

S(n)=sum(k=1, n, a(k)) seems to be asymptotic to 3*n. S(n)=3n for in A074069.

The sign of 3n-S(n) seems to change often: 3n-S(n) gives A074077. Is 3n-S(n) bounded? - Benoit Cloitre, Sep 05 2002

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• J. C. Lagarias and N. J. A. Sloane, Approximate squaring (pdf, ps), Experimental Math., 13 (2004), 113-128.

Crossrefs

Cf. A073524, A074069, A074077

Programs

Haskell

import Data.Ratio ((%), denominator)
a068119 n = fst $ until ((== 1) . denominator . snd)
                        (\(i, x) -> (i + 1, f x)) (0, fromInteger n + 1%4)
   where f x = x * fromIntegral (ceiling x)
-- Reinhard Zumkeller, May 26 2013

Mathematica

ce[n_] := Length[NestWhileList[#*Ceiling[#] &, n + 1/4, ! IntegerQ[#] &]] - 1; ce /@ Range[104] (* Jayanta Basu, Jul 29 2013 *)

PARI

a(n)=if(n<0,0,s=n+1/4; c=0; while(frac(s)>0,s=s*ceil(s); c++); c)

Formula

a(n) = 1 if n == 3 (mod 4); a(n) = 2 if n == 5, 6, 12, 13 (mod 16); a(n) = 3 if n == 1, 2, 4, 17, 26, 30, 33, 36, 48, 49, 56, 62 (mod 64);...

Extensions

Corrected by Diego Torres (torresvillarroel(AT)hotmail.com), Aug 31 2002