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Search: id:A000267
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| A000267 |
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Integer part of square root of 4n+1. |
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+0
9
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| 1, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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1^1, 2^1, 3^2, 4^2, 5^3, 6^3, 7^4, 8^4, 9^5, 10^5, ...
Start with n, repeatedly subtract the square root of the previous term; a(n) gives number of steps to reach 0. - Robert G. Wilson v, Jul 22, 2002.
a(n) = 1+a(n-[n^(1/2)]), n>0. - Michael Somos, Jul 22, 2002
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REFERENCES
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T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 20.
B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 77, Entry 23.
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LINKS
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S. Ramanujan, Question 723, J. Ind. Math. Soc.
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FORMULA
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a(n) = int( 1 / ( sqrt(n + 1) - sqrt(n) ) ) - Robert A. Stump (bob_ess107(AT)yahoo.com), Apr 07 2003
a(n) = |{floor(n/k): k in Z+}| - David W. Wilson (davidwwilson(AT)comcast.net), May 26 2005
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PROGRAM
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(PARI) a(n)=if(n<0, 0, sqrtint(4*n+1))
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CROSSREFS
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[a(n)/2]=A000196(n). Cf. A080037.
Adjacent sequences: A000264 A000265 A000266 this_sequence A000268 A000269 A000270
Sequence in context: A086592 A132663 A023964 this_sequence A060020 A143502 A070984
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KEYWORD
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nonn,easy,nice
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AUTHOR
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njas
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EXTENSIONS
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More terms from Michael Somos, Jun 13, 2000
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