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
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Programs
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Maple
# 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
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Mathematica
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 *) -
PARI
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
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Python
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
Formula
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
Extensions
More terms from Benoit Cloitre, Sep 29 2003
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