A087710 -id:A087710 - cp's OEIS frontend

A087666 Consider recurrence b(0) = n/3, b(k+1) = b(k)*floor(b(k)); a(n) is the least k such that b(k) is an integer, or -1 if no integer is ever reached.

0, 3, 4, 0, 1, 1, 0, 3, 2, 0, 3, 7, 0, 1, 1, 0, 2, 3, 0, 2, 2, 0, 1, 1, 0, 5, 5, 0, 5, 6, 0, 1, 1, 0, 9, 2, 0, 8, 3, 0, 1, 1, 0, 2, 5, 0, 2, 2, 0, 1, 1, 0, 3, 3, 0, 6, 3, 0, 1, 1, 0, 4, 2, 0, 6, 4, 0, 1, 1, 0, 2, 4, 0, 2, 2, 0, 1, 1, 0, 6, 4, 0, 3, 6, 0, 1, 1, 0, 3, 2, 0, 3, 4, 0, 1, 1, 0, 2, 3, 0, 2, 2, 0, 1, 1, 0, 4, 7, 0, 6, 6, 0, 1, 1, 0, 5, 2, 0, 4, 3, 0, 1, 1, 0, 2

Offset: 6

N. J. A. Sloane, Sep 27 2003

nonn

It is conjectured that an integer is always reached if the initial value n/3 is >= 2.

Benoit Cloitre, Graph of (sum(k=6,n,a(k))-2n)*n^(-1/2), pdf
J. C. Lagarias and N. J. A. Sloane, Approximate squaring (pdf, ps), Experimental Math., 13 (2004), 113-128.

Cf. A083863 (integer reached), A086336 and A087663 (records), A057016, A087710, A088706 (inverse).

# Gives right answer as long as answer is < M.
# This is better than the Mathematica or PARI programs.
M := 50; f := proc(n) local c,k,tn,tf; global M; k := n/3; c := 0; while whattype(k) <> 'integer' do tn := floor(k); tf := k-tn; tn := tn mod 3^50; k := tn*(tn+tf); c := c+1; od: c; end; # N. J. A. Sloane

f[n_] := If[ Mod[3n, 3] == 0, 0, Length[ NestWhileList[ #1*Floor[ #1] &, n, !IntegerQ[ #2] &, 2]] - 1]; Table[f[n/3], {n, 6, 120}] (* Robert G. Wilson v *)

a(n)=if(n<0,0,c=n/3; x=0; while(frac(c)>0,c=c*floor(c); x++); x) \\ Benoit Cloitre, Sep 29 2003

def A087666(n):
    c, x = 0, n
    a, b = divmod(x,3)
    while b != 0:
        x *= a
        c += 1
        a, b = divmod(x,3)
    return c # Chai Wah Wu, Mar 01 2021

a(n)=0 iff n == 0 (mod 3), a(n)==1 iff n == 1 or 2 (mod 3^2), a(n)=2 iff n == 14,22,25,26 (mod 3^3). In general a(n)=m iff n == x (mod 3^m) where x pertains to a set of 2^m distinct elements included in {1,2,...,(3^m)-1}. Conjecture: a(6) + a(7) + a(8) + ... + a(n) = 2n + O(sqrt(n)). - Benoit Cloitre, Sep 24 2012

More terms from Benoit Cloitre, Sep 29 2003

cp's OEIS Frontend

A088706 Duplicate of A087710.

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A087666 Consider recurrence b(0) = n/3, b(k+1) = b(k)*floor(b(k)); a(n) is the least k such that b(k) is an integer, or -1 if no integer is ever reached.

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