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Numbers k such that floor((3/2)^(k+1))/floor((3/2)^k) = 3/2.
10

%I #26 Dec 09 2024 22:03:59

%S 2,9,11,13,24,29,31,36,37,40,41,43,49,50,51,67,68,70,72,73,77,79,80,

%T 86,88,91,92,95,101,102,103,115,121,126,127,132,134,136,142,145,146,

%U 151,154,156,162,165,167,171,172,176,178,179,181,191,193,194,195,198,199

%N Numbers k such that floor((3/2)^(k+1))/floor((3/2)^k) = 3/2.

%C Also k such that A002380(k+1) = 3*A002380(k). - _Benoit Cloitre_, Apr 21 2003

%C It appears that lim_{n->oo} a(n)/n = 3. - _Benoit Cloitre_, Jan 29 2006

%H Harry J. Smith, <a href="/A065554/b065554.txt">Table of n, a(n) for n = 1..1000</a>

%t a[1] = 2; a[n_ ] := a[n] = Block[ {k = a[n - 1] + 1}, While[ Floor[(3/2)^(k + 1)] / Floor[(3/2)^k] != 3/2, k++ ]; Return[k]]; Table[ a[n], {n, 1, 70} ]

%o (PARI) isok(k) = { my(f=3/2); floor(f^(k+1))/floor(f^k) == f } \\ _Harry J. Smith_, Oct 22 2009

%Y Cf. A002379, A002380.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Nov 28 2001

%E More terms from _Robert G. Wilson v_, Nov 30 2001