A109965 -id:A109965 - cp's OEIS frontend

A002984 a(0) = 1; for n > 0, a(n) = a(n-1) + floor(sqrt(a(n-1))).

1, 2, 3, 4, 6, 8, 10, 13, 16, 20, 24, 28, 33, 38, 44, 50, 57, 64, 72, 80, 88, 97, 106, 116, 126, 137, 148, 160, 172, 185, 198, 212, 226, 241, 256, 272, 288, 304, 321, 338, 356, 374, 393, 412, 432, 452, 473, 494, 516, 538, 561, 584, 608, 632, 657, 682, 708, 734

Offset: 0

N. J. A. Sloane

For n > 3 we have a(n) < n^2/4; for n > 44 we have a(n) > n^2/5. - Stefan Steinerberger, Apr 17 2006
This sequence contains infinitely many squares. - Philippe Deléham, Apr 03 2009
The squares in this sequence are precisely the powers of 4. - Franklin T. Adams-Watters, Jan 06 2014

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A000302 (subsequence of squares).

Essentially the same as A109965.

a002984 n = a002984_list !! n
a002984_list = iterate (\x -> x + a000196 x) 1
-- Reinhard Zumkeller, Dec 28 2011

[n le 0 select 1 else Self(n)+Floor(Sqrt(Self(n))): n in [0..60]]; // Bruno Berselli, Feb 15 2013

NestList[ # + Floor[ Sqrt[ # ] ] &, 1, 50 ]

a(n+1) = a(n) + A000196(a(n)). - Reinhard Zumkeller, Dec 28 2011

Conjecture: a(n) ~ n^2/4. - José María Grau Ribas, Feb 13 2024

More terms from Larry Reeves (larryr(AT)acm.org), Dec 14 2000

A109964 a(n) = floor(sqrt(Sum_{i

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 32, 33, 33, 34, 34, 35, 35

Offset: 0

Henry Bottomley, Jul 06 2005

nonn

			a(5) = floor(sqrt(1+1+1+1+2)) = floor(sqrt(8)) = 2.
From _Paul Weisenhorn_, Jun 22, Jun 26 2010: (Start)
n=21; j=3; k=1; a(21)=2^3+1=9;
j=3; k=4; a(27)=a(28)=12.
(End)

Related problem was offered at XXIX Moscow Mathematical Olympiad (1966).

sumr:=0: a(0):=1: for n from 1 to 1000 do sumr:=sumr+a(n-1): a(n):=floor(sqrt(sumr)): end do: # Paul Weisenhorn, Jun 22 2010
a(0..1)=1; for n from 2 to 100 do j:=floor(log[2](n))-1: k:=iquo(n-2^(j+1)-j,2): a(n):=2^j+k: end do: # Paul Weisenhorn, Jun 26 2010

lst={1};Nest[AppendTo[lst,Floor[Sqrt[Total[lst]]]]&,1,85] (* Harvey P. Dale, May 24 2012 *)

a(n) = floor(sqrt(A109965(n))) = A109965(n+1)-A109965(n). Roughly (n-log_2(n))/2. 1 appears four times, other powers of 2 appear three times, other numbers appear twice.
From Paul Weisenhorn, Jun 22, Jun 26 2010: (Start)
For n>1, a(n)=2^j+k where j=floor(log_2(n))-1 and k=(n-2^(j+1)-j) mod 2.
a(2^(j+1)+j+2*k) = a(2^(j+1)+j+2*k+1) = 2^j+k; a(2^(j+1)+j-1) = 2^j for all j=0..infinity, k=0..(2^j-1).
(End)

Formulas corrected by Paul Weisenhorn, Aug 11 2010

Formula a(0..3)=1; a(n)=iquo(n+1-floor(log[2](n-2)),2); n=4..infinity; deleted and second Maple program changed Paul Weisenhorn, Aug 22 2010

cp's OEIS Frontend

A002984 a(0) = 1; for n > 0, a(n) = a(n-1) + floor(sqrt(a(n-1))).

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