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%I #6 Oct 04 2012 10:28:43
%S 3,60,14,268065,33,2093,60,1204154941925628,95,13398,138,701600900,
%T 189,47415,248,1489788110004539889867929328515560588293,315,123728,
%U 390,34427225343,473,268065,564,19873182780430314444725,663,512298,770,467193780498,885
%N Consider recurrence b(0) = (2n+1)/2, b(n) = b(n-1)*ceiling(b(n-1)); sequence gives first integer reached.
%H J. C. Lagarias and N. J. A. Sloane, Approximate squaring (<a href="http://neilsloane.com/doc/apsq.pdf">pdf</a>, <a href="http://neilsloane.com/doc/apsq.ps">ps</a>), Experimental Math., 13 (2004), 113-128.
%F Define F(x) = x(x+1)/2. Write 2n+1 = 2^i*m + 2^(i-1) + 1, then a(n) = (1/2)F^(i-1)(2n+1). E.g. n=4, 2n+1 = 9 = 2^4*0 + 2^3 + 1, so i=4, m=0 and F(F(F(9))) = F(F(45)) = F(1035) = 536130, a(4) = 536130/2 = 268065.
%Y A001511 gives number of steps to reach an integer. Cf. A081849, A073524, A074078.
%Y Cf. A081854.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, Apr 13 2003
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