A000267 - OEIS

%I #97 Nov 24 2024 01:49:31

%S 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,

%T 11,11,11,11,11,12,12,12,12,12,12,13,13,13,13,13,13,13,14,14,14,14,14,

%U 14,14,15,15,15,15,15,15,15,15,16,16,16,16,16,16,16,16,17,17,17,17,17

%N Integer part of square root of 4n+1.

%C 1^1, 2^1, 3^2, 4^2, 5^3, 6^3, 7^4, 8^4, 9^5, 10^5, ...

%C 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

%C Triangle A094727 read by diagonals. - _Philippe Deléham_, Mar 21 2014

%C Partial sums of A240025; a(n) = number of quarter squares <= n. - _Reinhard Zumkeller_, Jul 05 2014

%C Every number k is present consecutively (floor((2*k+3)/4)) times. - _Bernard Schott_, Jun 08 2019

%D Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 20.

%D Bruce C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, 1994, see p. 77, Entry 23.

%H T. D. Noe, <a href="/A000267/b000267.txt">Table of n, a(n) for n = 0..10000</a>

%H Gal Cohensius, Urban Larsson, Reshef Meir, and David Wahlstedt, <a href="https://arxiv.org/abs/1805.09368">Cumulative subtraction games</a>, arXiv:1805.09368 [math.CO], 2018-2020.

%H S. Ramanujan, <a href="http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/question/q723.htm">Question 723</a>, J. Ind. Math. Soc., Vol. 7 (1915), p. 240, Vol. 10 (1918), pp. 357-358.

%F floor(a(n)/2) = A000196(n).

%F a(n) = 1 + a(n - floor(n^(1/2))), if n>0. - _Michael Somos_, Jul 22 2002

%F a(n) = floor( 1 / ( sqrt(n + 1) - sqrt(n) ) ). - Robert A. Stump (bob_ess107(AT)yahoo.com), Apr 07 2003

%F a(n) = |{floor(n/k): k in Z+}|. - _David W. Wilson_, May 26 2005

%F a(n) = ceiling(2*sqrt(n+1) - 1). - _Mircea Merca_, Feb 03 2012

%F a(n) = A000196(A016813(n)). - _Reinhard Zumkeller_, Dec 13 2012

%F a(n) = A070939(A227368(n+1)), conjectured. - _Antti Karttunen_, Dec 28 2013

%F a(n) = floor( sqrt(n) + sqrt(n+2) ). [_Bruno Berselli_, Jan 08 2015]

%F a(n) = floor( sqrt(4*n + k) ) where k = 1, 2, or 3. - _Michael Somos_, Mar 11 2015

%F G.f.: (Sum_{k>0} x^floor(k^2 / 4)) / (1 - x). - _Michael Somos_, Mar 11 2015

%F a(n) = 1 + A055086(n). - _Michael Somos_, Sep 02 2017

%F a(n) = floor(sqrt(n+1)+1/2) + floor(sqrt(n)). - _Ridouane Oudra_, Jun 07 2019

%F Sum_{k>=0} (-1)^k/a(k) = Pi/8 + log(2)/4. - _Amiram Eldar_, Jan 26 2024

%e From _Philippe Deléham_, Mar 21 2014: (Start)

%e Triangle A094727 begins:

%e   1;

%e   2,  3;

%e   3,  4,  5;

%e   4,  5,  6,  7;

%e   5,  6,  7,  8,  9;

%e   6,  7,  8,  9, 10, 11; ...

%e Read by diagonals:

%e    1;

%e    2;

%e    3,  3;

%e    4,  4;

%e    5,  5,  5;

%e    6,  6,  6;

%e    7,  7,  7,  7;

%e    8,  8,  8,  8;

%e    9,  9,  9,  9,  9;

%e   10, 10, 10, 10, 10; (End)

%e 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 + ...

%p A000267:=seq(floor(sqrt(4*n+1)), n=0..100); // _Bernard Schott_, Jun 08 2019

%t Table[Floor[Sqrt[4*n + 1]], {n, 0, 100}] (* _T. D. Noe_, Jun 19 2012 *)

%o (PARI) {a(n) = if( n<0, 0, sqrtint(4*n + 1))};

%o (Haskell)

%o a000267 = a000196 . a016813  -- _Reinhard Zumkeller_, Dec 13 2012

%o (Magma) [Floor(Sqrt(4*n+1)): n in [0..100]]; // _Vincenzo Librandi_, Jun 08 2019

%o (Python)

%o from math import isqrt

%o def A000267(n): return isqrt((n<<2)|1) # _Chai Wah Wu_, Nov 23 2024

%Y Cf. A055086, A080037, A227368.

%Y Cf. A000196, A002620, A016813, A070939, A094727, A240025.

%K nonn,easy,nice,tabf

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Michael Somos_, Jun 13 2000