A048760 -id:A048760 - cp's OEIS frontend

A061886 Largest square less than or equal to sum of previous terms.

1, 1, 1, 1, 4, 4, 9, 16, 36, 64, 121, 256, 484, 961, 1936, 3844, 7569, 15129, 30276, 60516, 121104, 242064, 483025, 966289, 1932100, 3865156, 7728400, 15452761, 30902481, 61811044, 123609924, 247212729, 494439696, 988850916, 1977669841

Offset: 0

Henry Bottomley, May 12 2001

nonn

			a(6) = 9 since 1+1+1+1+4+4 = 12 and 9 is the largest square less than or equal to this.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A061883.

a061886 n = a061886_list !! n
a061886_list = 1 : zipWith (-) (tail a060984_list) a060984_list
-- Reinhard Zumkeller, Dec 24 2013

For n > 0: a(n) = A060984(n+1)-A060984(n) = A048760(A060984(n)).

Formula corrected by Reinhard Zumkeller, Dec 24 2013

A061887 n + largest square less than or equal to n; numbers in the range [2k^2,2k^2+2k] for some k.

Original entry on oeis.org

0, 2, 3, 4, 8, 9, 10, 11, 12, 18, 19, 20, 21, 22, 23, 24, 32, 33, 34, 35, 36, 37, 38, 39, 40, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 128, 129

Offset: 0

Henry Bottomley, May 12 2001

nonn

			a(15)=15+9=24; a(16)=16+16=32; a(17)=17+16=33.

Cf. A060984, A061885.

Table[n+Floor[Sqrt[n]]^2,{n,0,70}] (* Harvey P. Dale, Aug 23 2012 *)

a(n) = n+[sqrt(n)]^2 = n+A048760(n) = 2n-A053186(n).

A064672 a(0) = 0, a(1) = 1; for a(n), n >= 2, write n = x^2 + y with y >= 0 as small as possible, then a(n) = a(x) + a(y).

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 6, 7, 2, 3, 4, 5, 4, 5, 6, 7, 6, 3, 4, 5, 6, 5, 6, 7, 8, 7, 6, 7, 4, 5, 6, 7, 6, 7, 8, 9, 8, 7, 8, 9, 10, 5, 6, 7, 8, 7, 8, 9, 10, 9, 8, 9, 10, 11, 10, 11, 4, 5, 6, 7, 6, 7, 8, 9, 8, 7, 8, 9, 10, 9, 10, 11, 6, 3, 4, 5, 6, 5, 6, 7, 8, 7, 6, 7, 8, 9, 8, 9, 10, 5, 6

Offset: 0

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 09 2001

Because of the definition of a(n), a(n^2) = a(n) and more generally a(n^(2m)) = a(n), so the sequence recursively contains itself.

a(A064689(n)) = n and a(m) < n for m < A064689(n).

			a(7) = 5 because 7 = 2^2 + 3, a(2) = 2 and a(3) = 3, giving 5

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A064689.

Cf. A048760.

a064672 n = a064672_list !! n
a064672_list = 0 : 1 : f (drop 2 a000196_list) 1 1 (tail a064672_list)
   where f (r:rs) r' u (v:vs)
           | r == r' = (u + v) : f rs r u vs
           | r /= r' = u' : f rs r u' (tail a064672_list)
           where u' = a064672 $ fromInteger r
-- Reinhard Zumkeller, Apr 27 2012

a[0]=0; a[1]=1; a[n_] := a[n] = a[ Floor[ Sqrt[n] ] ] + a[ n - Floor[ Sqrt[n] ]^2 ]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, May 23 2012, after Reinhard Zumkeller *)

For n > 1: a(n) = a(A000196(n)) + a(A053186(n)), a(0) = 0, a(1) = 1. [Reinhard Zumkeller, Apr 27 2012]

A065732 Largest square <= 2^n.

Original entry on oeis.org

1, 4, 4, 16, 25, 64, 121, 256, 484, 1024, 2025, 4096, 8100, 16384, 32761, 65536, 131044, 262144, 524176, 1048576, 2096704, 4194304, 8386816, 16777216, 33547264, 67108864, 134212225, 268435456, 536848900, 1073741824, 2147395600, 4294967296, 8589767761, 17179869184

Offset: 1

Labos Elemer, Nov 15 2001

Harry J. Smith, Table of n, a(n) for n = 1..200

Cf. A048760, A000079, A065730-A065741.

Table[Floor[Sqrt[2^w]//N]^2, {w, 1, 50}]

a(n) = { sqrtint(2^n)^2 } \\ Harry J. Smith, Oct 28 2009

a(n) = A048760(A000079(n)) = A048760(2^n).

A065736 Largest square <= 10^n.

Original entry on oeis.org

9, 100, 961, 10000, 99856, 1000000, 9998244, 100000000, 999950884, 10000000000, 99999515529, 1000000000000, 9999995824729, 100000000000000, 999999961946176, 10000000000000000, 99999999989350756, 1000000000000000000, 9999999998935075600, 100000000000000000000, 999999999956753113201

Offset: 1

Labos Elemer, Nov 15 2001

Harry J. Smith, Table of n, a(n) for n = 1..100

Cf. A048760, A011557, A065730-A065741.

Table[Floor[Sqrt[10^n]]^2,{n,20}] (* Harvey P. Dale, Dec 04 2014 *)

a(n) = { sqrtint(10^n)^2 } \\ Harry J. Smith, Oct 28 2009

a(n) = A048760(A011557(n)) = A048760(10^n).

A072689 Difference between (least square >= n) and (largest square <= n).

Original entry on oeis.org

0, 3, 3, 0, 5, 5, 5, 5, 0, 7, 7, 7, 7, 7, 7, 0, 9, 9, 9, 9, 9, 9, 9, 9, 0, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 0, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 0, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 0, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17

Offset: 1

Reinhard Zumkeller, Jul 02 2002

nonn

a(n) = 0 iff n is a square.

a(n) = 1+2*A000196(n) if n is not a square. - Robert Israel, Sep 22 2020

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000196, A000290, A048760, A048761, A072690.

f:= proc(n) local t; t:= floor(sqrt(n));
  if n = t^2 then 0 else 1 + 2*t fi
end proc:
map(f, [$1..100]); # Robert Israel, Sep 22 2020

ds[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],0,1+2Floor[s]]]; Array[ds,80] (* Harvey P. Dale, Dec 05 2013 *)

a(n) = A057427(n - A048760(n)) * (A000196(A048760(n))*2 + 1).

a(n) = A048761(n) - A048760(n).

A227453 Numbers k such that the distance to the largest square less than k is a multiple of 4.

Original entry on oeis.org

8, 13, 20, 24, 29, 33, 40, 44, 48, 53, 57, 61, 68, 72, 76, 80, 85, 89, 93, 97, 104, 108, 112, 116, 120, 125, 129, 133, 137, 141, 148, 152, 156, 160, 164, 168, 173, 177, 181, 185, 189, 193, 200, 204, 208, 212, 216, 220, 224, 229, 233, 237, 241, 245, 249, 253, 260, 264, 268, 272, 276, 280

Offset: 1

Ralf Stephan, Sep 22 2013

nonn

A071797(a(n)) = 4*m, A053186(a(n)+1) = 4*m, m > 0.

Apparently a bisection of A079896. While it may not be difficult to prove that the sequence is a subsequence of A079896, the apparent fact that a(n) = A079896(2n-1) is by no means obvious.

			8 - 2^2 = 1*4 and 24 - 4^2 = 2*4 so 8 and 24 are in the sequence.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A048760, A053186, A071797, A079896.

lsm4Q[n_]:=Module[{s=Floor[Sqrt[n]]^2},sHarvey P. Dale, Jun 20 2014 *)

PARI
```
is(n)=(n-sqrtint(n-1)^2)%4==0
```

A232501 Numbers k such that distances from k to three nearest squares are three triangular numbers.

Original entry on oeis.org

1, 10, 15, 19, 26, 197, 253, 325, 631, 1090, 1522, 2395, 3601, 4434, 4625, 6571, 9026, 11026, 11116, 14631, 15454, 19045, 22501, 35722, 38431, 41210, 53036, 61505, 65521, 66239, 69697, 69949, 70291, 85384, 99226, 110890, 152101, 152803, 160021, 168101, 181801, 189631

Offset: 1

Alex Ratushnyak, Feb 23 2014

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A000217, A000290, A234335.
Cf. A232608 (terms that are triangular numbers).
Cf. A000196, A010054, A048760, A048761.

import Data.List (sort)
a232501 n = a232501_list !! (n-1)
a232501_list = filter f [1..] where
   f x = all ((== 1) . a010054) $ init $ sort $
         map (abs . (x -) . (^ 2) . (+ (a000196 x))) [-1..2]
-- Reinhard Zumkeller, Mar 16 2014

A233401 Numbers k such that k^3 - b2 is a triangular number (A000217), where b2 is the largest square less than k^3.

Original entry on oeis.org

1, 4, 8, 21, 37, 40, 56, 112, 113, 204, 280, 445, 481, 560, 688, 709, 1933, 1945, 3601, 3805, 3861, 4156, 4333, 4365, 7096, 8408, 8516, 11064, 12688, 13609, 13945, 16501, 17080, 18901, 21464, 23125, 27244, 27364, 28141, 45228, 45549, 58321, 60061, 66245, 70585, 78688

Offset: 1

Alex Ratushnyak, Dec 09 2013

nonn

The cubes k^3 begin: 1, 64, 512, 9261, 50653, 64000, 175616, 1404928, ...
The squares b2 begin: 0, 49, 484, 9216, 50625, 63504, 175561, 1404225, ...
Their square roots are 0, 7, 22, 96, 225, 252, 419, 1185, 1201, 2913, 4685, 9387, ...

Cf. A000217, A000290, A000578, A048760, A233400.

f(k) = if (issquare(k), sqrtint(k-1)^2, sqrtint(k)^2); \\ adapted from A048760
isok(k) = my(b2 = sqrtint(k^3-1)^2); (k^3-b2) && ispolygonal(k^3-b2, 3); \\ Michel Marcus, Jan 26 2019

from math import isqrt
def isTriangular(a):
  a+=a
  sr = isqrt(a)
  return (a==sr*(sr+1))
for n in range(1,79999):
  n3 = n*n*n
  b = isqrt(n3)
  if b*b==n3: b-=1
  if isTriangular(n3-b*b):  print(n, end=', ')

A256097 Numerators of a rational guess r(n) for the input for Newton's algorithm to find sqrt(n).

Original entry on oeis.org

1, 3, 2, 2, 9, 5, 11, 3, 3, 19, 10, 7, 11, 23, 4, 4, 33, 17, 35, 9, 37, 19, 39, 5, 5, 51, 26, 53, 27, 11, 28, 57, 29, 59, 6, 6, 73, 37, 25, 19, 77, 13, 79, 20, 27, 41, 83, 7, 7, 99, 50, 101, 51, 103, 52, 15, 53, 107, 54, 109, 55, 111, 8, 8, 129, 65, 131, 33, 133, 67

Offset: 1

Wolfdieter Lang, Mar 24 2015

The corresponding denominators are given in A256098.
This educated guess for the rational input R(n) = x(n;k=0) for the so-called Babylonian (also called Heron's) iteration to find sqrt(n) (Newton's method for sqrt(n)), x(n; k+1) = (x(n; k) + n/x(n; k))/2, k >= 0, was used in Vedic Mathematics (see the H.-W. Alhen et al. reference, pp. 145-146, and the MacTutor link on Sulbasutras). In the Wikipedia link on Shulba Sutras another suggestion is given how the approximation 1 + 1/3 + 1/(3*4) - 1/(3*4*34) for sqrt(2) was obtained in Sulbasutras. The explanation given in the H.-W. Alten et al. reference seems to me more convincing.
This R(n) is obtained by n = s(n)^2 + r(n) with s(n)^2 = A048760(n) (largest square not exceeding n) and the remainder r(n). Then the approximation of the square root is used sqrt(n) = sqrt(s(n)^2 + r(n)) approximately s(n)*(1 + r(n)/(2*s(n)^2)) = s(n) + r(n)/(2*s(n)). Note that A048760(n) = A000196(n)^2, that is, s(n) = floor(sqrt(n)).

			n = 2: s(n) = floor(sqrt(2)) = sqrt(A048760(2)) = 1, r(n) = 2 - 1^2 = 1. R(2) = s(2) + r(2)/(2*s(2)) = 1 + 1/(2*1) = 3/2. That is a(2) = 3 and A256098(2) = 2.
n = 17: s(n) = floor(sqrt(17)) = sqrt(A048760(17)) = 4 , r(n) = 17 - 4^2 = 1. R(17) = s(17) + r(17)/(2*s(17)) = 4 + 1/(2*4) = 33/8. That is, a(n) = 33 and A256098(17) = 8.
The rationals R(n) for n = 1..60 are: [1, 3/2, 2, 2, 9/4, 5/2, 11/4, 3, 3, 19/6, 10/3, 7/2, 11/3, 23/6, 4, 4, 33/8, 17/4, 35/8, 9/2, 37/8, 19/4, 39/8, 5, 5, 51/10, 26/5, 53/10, 27/5, 11/2, 28/5, 57/10, 29/5, 59/10, 6, 6, 73/12, 37/6, 25/4, 19/3, 77/12, 13/2, 79/12, 20/3, 27/4, 41/6, 83/12, 7, 7, 99/14, 50/7, 101/14, 51/7, 103/14, 52/7, 15/2, 53/7, 107/14, 54/7, 109/14,...]
For n=2 the Newton (Babylonian also called Heron) iteration produces. with x(2; k=0) = R(2) = 3/2: x(2; 1) = (3/2 + 4/3)/2 = 17/12 = 1 + 5/2  = 1 + 1/3 + 1/(3*4).
  x(2; 2) = (17/12 + 24/17)/2 = 577/408 = 17/12 + (577/408 - 17*34/408)  = 17/12 - 1/408 = 1 + 1/3 + 1/(3*4) - 1/(3*4*34) = 1.4142156... versus sqrt(2) = 1.4142135... (see A002193).

H.-W. Alten et al., 4000 Jahre Algebra, 2. Auflage, Springer, 2014, p. 145-146.

Stefano Spezia, Table of n, a(n) for n = 1..10000
The MacTutor History of Mathematics archive, Sulbasutras.
Wikipedia, Shulba Sutras.

Cf. A000196, A002193, A048760, A048761, A256098.

A000196[n_]:=Floor[Sqrt[n]]; a[n_]:=Numerator[(A000196[n]+n/A000196[n])/2]; Array[a,70](* Stefano Spezia, Feb 15 2025 *)

a(n) = numerator(R(n)) with the rational (in lowest terms) R(n) = f(n) + (n - f(n)^2)/(2*f(n)) = (f(n) + n/f(n))/2 with f(n) := floor(sqrt(n)) = A000196(n), for n >= 1. See the comment above for this formula.

a(61)-a(70) from Stefano Spezia, Feb 15 2025

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A061886 Largest square less than or equal to sum of previous terms.

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A061887 n + largest square less than or equal to n; numbers in the range [2k^2,2k^2+2k] for some k.

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A064672 a(0) = 0, a(1) = 1; for a(n), n >= 2, write n = x^2 + y with y >= 0 as small as possible, then a(n) = a(x) + a(y).

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A065732 Largest square <= 2^n.

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A065736 Largest square <= 10^n.

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A072689 Difference between (least square >= n) and (largest square <= n).

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A227453 Numbers k such that the distance to the largest square less than k is a multiple of 4.

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A232501 Numbers k such that distances from k to three nearest squares are three triangular numbers.

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