A081849 Consider recurrence b(0) = (2n+1)/2, b(n) = b(0)*ceiling(b(n-1)); sequence gives first integer reached.

3, 20, 14, 468, 33, 299, 60, 47328, 95, 1218, 138, 25475, 189, 3161, 248, 20830128, 315, 6512, 390, 181138, 473, 11655, 564, 9015167, 663, 18974, 770, 671745, 885, 28853, 1008, 38906570560, 1139, 41676, 1278, 1799888, 1425, 57827, 1580, 110341278, 1743, 77690

Offset: 1

Author

N. J. A. Sloane, Apr 13 2003

Keywords

• nonn
• easy

Comments

k = A001511(n) is the number of steps to reach an integer b(k).

Crossrefs

Cf. A001511, A081853, A073524, A074078.

Programs

Maple

Digits := 100: c := ceil; A081849 := proc(a) local i,t0,t; t0 := a; t := 0; for i from 1 to 100 do if whattype(t0) <> integer then t0 := a*c(t0); t := t+1; else RETURN(t0); fi; od; RETURN('FAIL'); end;

Mathematica

a[n_]:=Module[{b=b0=(2n+1)/2},While[!IntegerQ[b],b=b0*Ceiling[b]]; b]; Array[a,42] (* Stefano Spezia, Jun 26 2024 *)

PARI

a(n) = if(n==1,3, my(t=2*n+1, k=1+valuation(n,2)); n*t^(k+1) >>k \ (t-2)); \\ Kevin Ryde, Jun 30 2024

Python

from math import ceil
from fractions import Fraction
def a(n):
  b0 = b = Fraction((2*n+1), 2)
  while b.denominator != 1: b = b0*ceil(b)
  return b.numerator
print([a(n) for n in range(1, 43)]) # Michael S. Branicky, Mar 20 2021

Formula

a(n) = s*(n*s^k - 1/2) / (s-1) where s = b(0) = (2*n+1)/2 and k = A001511(n). - Kevin Ryde, Jun 30 2024