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A081853
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Consider recurrence b(0) = (2n+1)/2, b(n) = b(n-1)*ceiling(b(n-1)); sequence gives first integer reached.
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3
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3, 60, 14, 268065, 33, 2093, 60, 1204154941925628, 95, 13398, 138, 701600900, 189, 47415, 248, 1489788110004539889867929328515560588293, 315, 123728, 390, 34427225343, 473, 268065, 564, 19873182780430314444725, 663, 512298, 770, 467193780498, 885
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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LINKS
| J. C. Lagarias and N. J. A. Sloane, Approximate squaring (pdf, ps), Experimental Math., 13 (2004), 113-128.
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FORMULA
| 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.
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CROSSREFS
| A001511 gives number of steps to reach an integer. Cf. A081849, A073524, A074078.
Cf. A081854.
Sequence in context: A201184 A046024 A184951 * A183286 A195560 A002738
Adjacent sequences: A081850 A081851 A081852 * A081854 A081855 A081856
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KEYWORD
| nonn
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Apr 13 2003
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