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A000267
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Integer part of square root of 4n+1.
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25
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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
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OFFSET
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0,2
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COMMENTS
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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
Every number k is present consecutively (floor((2*k+3)/4)) times. - Bernard Schott, Jun 08 2019
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REFERENCES
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Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 20.
Bruce C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, 1994, see p. 77, Entry 23.
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LINKS
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S. Ramanujan, Question 723, J. Ind. Math. Soc., Vol. 7 (1915), p. 240, Vol. 10 (1918), pp. 357-358.
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FORMULA
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a(n) = 1 + a(n - floor(n^(1/2))), if n>0. - Michael Somos, Jul 22 2002
a(n) = floor( 1 / ( sqrt(n + 1) - sqrt(n) ) ). - Robert A. Stump (bob_ess107(AT)yahoo.com), Apr 07 2003
a(n) = ceiling(2*sqrt(n+1) - 1). - Mircea Merca, Feb 03 2012
a(n) = floor( sqrt(4*n + k) ) where k = 1, 2, or 3. - Michael Somos, Mar 11 2015
G.f.: (Sum_{k>0} x^floor(k^2 / 4)) / (1 - x). - Michael Somos, Mar 11 2015
a(n) = floor(sqrt(n+1)+1/2) + floor(sqrt(n)). - Ridouane Oudra, Jun 07 2019
Sum_{k>=0} (-1)^k/a(k) = Pi/8 + log(2)/4. - Amiram Eldar, Jan 26 2024
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EXAMPLE
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1;
2, 3;
3, 4, 5;
4, 5, 6, 7;
5, 6, 7, 8, 9;
6, 7, 8, 9, 10, 11; ...
Read by diagonals:
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; (End)
G.f. = 1 + 2*x + 3*x^2 + 3*x^3 + 4*x^4 + 4*x^5 + 5*x^6 + 5*x^7 + 5*x^8 + 6*x^9 + ...
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MAPLE
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MATHEMATICA
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Table[Floor[Sqrt[4*n + 1]], {n, 0, 100}] (* T. D. Noe, Jun 19 2012 *)
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PROG
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(PARI) {a(n) = if( n<0, 0, sqrtint(4*n + 1))};
(Haskell)
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CROSSREFS
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KEYWORD
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nonn,easy,nice,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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