A217575 -id:A217575 - cp's OEIS frontend

A063657 Numbers with property that truncated square root is unequal to rounded square root.

3, 7, 8, 13, 14, 15, 21, 22, 23, 24, 31, 32, 33, 34, 35, 43, 44, 45, 46, 47, 48, 57, 58, 59, 60, 61, 62, 63, 73, 74, 75, 76, 77, 78, 79, 80, 91, 92, 93, 94, 95, 96, 97, 98, 99, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 133, 134, 135, 136, 137, 138, 139, 140

Offset: 1

Floor van Lamoen, Jul 24 2001

Also: skip 1, take 0, skip 2, take 1, skip 3, take 2, ...
Integers for which the periodic part of the continued fraction for the square root of n begins with a 1. - Robert G. Wilson v, Nov 01 2001
a(n) belongs to the sequence if and only if a(n) > floor(sqrt(a(n))) * ceiling(sqrt(a(n))), i.e. a(n) in (k*(k+1),k^2), k >= 0. - Daniel Forgues, Apr 17 2011
Any integer between (k - 1/2)^2 and k^2 exclusive, for k > 1, is in this sequence. If we take this sequence and remove each term that is one more than the previous term, we obtain the central polygonal numbers (A002061). If instead we remove each term that is one less than the next term, we obtain numbers that are one less than squares (A005563). - Alonso del Arte, Dec 28 2013

			7 is in the sequence because its square root is 2.64575..., which truncates to 2 but rounds to 3.
8 is in the sequence because its square root is 2.828427..., which also truncates to 2 but rounds to 3.
9 is not in the sequence because its square root is 3 exactly, which truncates and rounds the same.
Here is the example per Lamoen's skip n, take n - 1 process: starting at 0, we skip one integer (0) but take zero integers for our sequence. Then we skip two integers (1 and 2) and take one integer (3) for our sequence. Then we skip three integers (4, 5, 6) and take two integers for our sequence (7 and 8, so the sequence now stands as 3, 7, 8). Then we skip four integers (9, 10, 11, 12) and so on and so forth.
From _Seiichi Manyama_, Sep 19 2017: (Start)
See R. B. Nelsen's paper.
   k|            A063656(n)         |            a(n)
   -------------------------------------------------------------------
   0|                             0
   1|                        1 +  2 =  3
   2|                   4 +  5 +  6 =  7 +  8
   3|              9 + 10 + 11 + 12 = 13 + 14 + 15
   4|        16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24
    | ...
(End)
The triangle begins as:
   3;
   7,  8;
  13, 14, 15;
  21, 22, 23, 24;
  31, 32, 33, 34, 35;
  43, 44, 45, 46, 47, 48;
  57, 58, 59, 60, 61, 62, 63;
  73, 74, 75, 76, 77, 78, 79, 80;
  91, 92, 93, 94, 95, 96, 97, 98, 99;
  ... - _Stefano Spezia_, Oct 20 2024

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)
Roger B. Nelsen, Proof without Words: Consecutive sums of consecutive integers, Mathematics Magazine, Vol. 63, No. 1 (1990), p. 25.

Cf. A002620, A004201-A004202, A063656, A189151.

Cf. A005563 (main diagonal), A059270 (row sums), A217575.

a063657 n = a063657_list !! n
a063657_list = f 0 [0..] where
   f k (_:xs) = us ++ f (k + 1) (drop (k + 1) vs) where
                        (us, vs) = splitAt k xs
-- Reinhard Zumkeller, Jun 20 2015

A063657:=n->`if`(floor(floor(sqrt(n+1)) * (1+floor(sqrt(n+1)))/(n+1))=1, NULL, n+1); seq(A063657(n), n=1..200); # Wesley Ivan Hurt, Dec 28 2013

Select[ Range[200], Floor[ Sqrt[ # ]] != Floor[ Sqrt[ # ] + 1/2] & ] (* or *) Select[ Range[200], First[ Last[ ContinuedFraction[ Sqrt[ # ]]]] == 1 & ]

{ n=0; for (m=0, 10^9, if (sqrt(m)%1 > .5, write("b063657.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 27 2009

a(n) = A217575(n) + 1. - Reinhard Zumkeller, Jun 20 2015
From Stefano Spezia, Oct 20 2024: (Start)
As a triangle:
T(n,k) = n^2 + n + k with 1 <= k <= n.
G.f.: x*y*(3 + x^2*(1 - 4*y) - x*(2 + y) + x^3*y*(1 + 2*y))/((1 - x)^3*(1 - x*y)^3). (End)

A063656 Numbers k such that the truncated square root of k is equal to the rounded square root of k.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 9, 10, 11, 12, 16, 17, 18, 19, 20, 25, 26, 27, 28, 29, 30, 36, 37, 38, 39, 40, 41, 42, 49, 50, 51, 52, 53, 54, 55, 56, 64, 65, 66, 67, 68, 69, 70, 71, 72, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 121

Offset: 0

Floor van Lamoen, Jul 24 2001

Also: take 1, skip 0, take 2, skip 1, take 3, skip 2, ...
The union of sets of numbers in closed intervals [k^2,k^2+k], k >= 0, intervals 0 to 1, 1 to 2, 4 to 6, 9 to 12 etc. - J. M. Bergot, Jun 27 2013
Conjecture: the following definition produces a(n) for n >= 1: a(1) = 1; for n > 1, smallest number > a(n-1) satisfying the condition that a(n) is a square if and only if n is a triangular number. - J. Lowell, May 13 2014
Thus a(2) = 2, because 2 is not a triangular number and not a square; a(3) != 3, because 3 is not a square but is a triangular number; a(3) = 4 is OK because 4 is a square and 3 is a triangular number; etc. [Examples supplied by N. J. A. Sloane, May 13 2014]

			The triangle begins as:
   0;
   1,  2;
   4,  5,  6;
   9, 10, 11, 12;
  16, 17, 18, 19, 20;
  25, 26, 27, 28, 29, 30;
  36, 37, 38, 39, 40, 41, 42;
  49, 50, 51, 52, 53, 54, 55, 56;
  ... - _Stefano Spezia_, Oct 19 2024

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Harry J. Smith)

Cf. A063657, A004201-A004202.
Cf. A128217, A217575.
Essentially partial sums of A051340.

a063656 n = a063656_list !! n
a063656_list = f 1 [0..] where
   f k xs = us ++ f (k + 1) (drop (k - 1) vs) where
                    (us, vs) = splitAt k xs
-- Reinhard Zumkeller, Jun 20 2015

Select[Range[121],Floor[Sqrt[#]]==Round[Sqrt[#]] &] (* Stefano Spezia, Oct 19 2024 *)

{ n=-1; for (m=0, 10^9, if (sqrt(m)%1 < .5, write("b063656.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 27 2009

As a triangle from Stefano Spezia, Oct 19 2024: (Start)
T(n,k) = n^2 + k with 0 <= k <= n.
G.f.: x*(1 + x + 2*y - 4*x*y + 3*x^3*y^2 - x^2*y*(2 + y))/((1 - x)^3*(1 - x*y)^3). (End)

A217570 Numbers n such that floor(sqrt(n)) = floor(n/(floor(sqrt(n))-1))-1.

Original entry on oeis.org

9, 16, 17, 25, 26, 27, 36, 37, 38, 39, 49, 50, 51, 52, 53, 64, 65, 66, 67, 68, 69, 81, 82, 83, 84, 85, 86, 87, 100, 101, 102, 103, 104, 105, 106, 107, 121, 122, 123, 124, 125, 126, 127, 128, 129, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 169, 170, 171, 172, 173

Offset: 1

Takumi Sato, Oct 07 2012

nonn

The sequence consists of numbers n^2+k, 0<=k<=n-3, n=3,4,5,... - M. F. Hasler, Oct 09 2012

One of four sequences given by classifying natural numbers according to the value of floor(sqrt(n)). See the paper in Link lines and A005563, A217571, A217575. - Takumi Sato, Oct 09 2012

			As a triangle (see the first comment) this begins:
9;
16, 17;
25, 26, 27;
36, 37, 38, 39;
49, 50, 51, 52, 53;
64, 65, 66, 67, 68, 69;
81, 82, 83, 84, 85, 86, 87;
100, 101, 102, 103, 104, 105, 106, 107; etc.
[_Bruno Berselli_, Oct 12 2012]

Robert Israel, Table of n, a(n) for n = 1..10000
Takumi Sato, Classification of Natural Numbers

Cf. A005563, A217571, A217575.

seq($n^2 .. n^2 + n - 3, n=3..20); # Robert Israel, Dec 23 2024

is_A217570(n)={ n>3 & n\(n=sqrtint(n)-1)==n+2}  \\ - M. F. Hasler, Oct 09 2012

A217571 a(n) = (2n(n+5) + (2n+1)(-1)^n - 1)/8.

Original entry on oeis.org

1, 4, 5, 10, 11, 18, 19, 28, 29, 40, 41, 54, 55, 70, 71, 88, 89, 108, 109, 130, 131, 154, 155, 180, 181, 208, 209, 238, 239, 270, 271, 304, 305, 340, 341, 378, 379, 418, 419, 460, 461, 504, 505, 550, 551, 598, 599, 648, 649, 700, 701, 754, 755, 810, 811, 868

Offset: 1

Takumi Sato, Oct 07 2012

One of four sequences given by classifying natural numbers according to the value of floor(sqrt(n)). See Sato link and sequences A005563, A217570, A217575.

Numbers n such that floor(sqrt(n)) = floor(n/floor(sqrt(n))) = floor(n/(floor(sqrt(n)) + 2)) + 1.

			From _Stefano Spezia_, Dec 14 2019: (Start)
Illustration of the initial terms:
o      o        o        o           o
     o o o    o o o    o o o       o o o
                o        o           o
                     o o o o o   o o o o o
                                     o
(1)   (4)      (5)     (10)        (11)
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Takumi Sato, Classification of Natural Numbers [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A005563, A028387, A028552, A217570, A217575.

List([1..60], n-> (2*n^2 +10*n -1 +(-1)^n*(2*n+1))/8 ); # G. C. Greubel, Dec 19 2019

[n: n in [1..900] | Floor(n/Isqrt(n)) eq Floor(n/(Isqrt(n)+2))+1]; // Bruno Berselli, Oct 10 2012

I:=[1, 4, 5, 10, 11]; [n le 5 select I[n] else Self(n-1) + 2*Self(n-2) - 2*Self(n-3) - Self(n-4) + Self(n-5): n in [1..60]]; // Vincenzo Librandi, Dec 15 2012

seq( (2*n^2 +10*n -1 +(-1)^n*(2*n+1))/8, n=1..60); # G. C. Greubel, Dec 19 2019

CoefficientList[Series[(1 + 3*x - x^2 - x^3)/((1 + x)^2*(1 - x)^3), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 15 2012 *)
a[1]=1;a[n_]:=If[EvenQ[n],a[n-1]+1+n,a[n-1]+1]; Array[a,56] (* Stefano Spezia, Dec 18 2019 *)

makelist((2*n*(n+5)+(2*n+1)*(-1)^n-1)/8, n, 1, 56); /* Martin Ettl, Oct 15 2012 */

vector(60, n, (2*n^2 +10*n -1 +(-1)^n*(2*n+1))/8 ) \\ G. C. Greubel, Dec 19 2019

[(2*n^2 +10*n -1 +(-1)^n*(2*n+1))/8 for n in (1..60)] # G. C. Greubel, Dec 19 2019

G.f.: x*(1+3*x-x^2-x^3)/((1+x)^2*(1-x)^3). - Bruno Berselli, Oct 11 2012
From Stefano Spezia, Dec 14 2019: (Start)
E.g.f.: (x*(5+x)*cosh(x) - (1-7*x-x^2)*sinh(x))/4.
a(n) = a(n-1) + 1 for n odd.
a(n) = a(n-1) + n + 1 for n even.
a(2*n) = A028552(n).
a(2*n+1) = A028387(n).
(End)

Definition by Bruno Berselli, Oct 11 2012

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A063657 Numbers with property that truncated square root is unequal to rounded square root.

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A063656 Numbers k such that the truncated square root of k is equal to the rounded square root of k.

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A217570 Numbers n such that floor(sqrt(n)) = floor(n/(floor(sqrt(n))-1))-1.

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A217571 a(n) = (2*n*(n+5) + (2*n+1)*(-1)^n - 1)/8.

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A217571 a(n) = (2n(n+5) + (2n+1)(-1)^n - 1)/8.