A048761 -id:A048761 - cp's OEIS frontend

A232395 (ceiling(sqrt(n^3 + n^2 + n + 1)))^2 - (n^3 + n^2 + n + 1).

0, 0, 1, 9, 15, 13, 30, 0, 40, 21, 45, 57, 51, 21, 70, 105, 120, 109, 66, 156, 43, 77, 81, 49, 216, 108, 217, 9, 36, 21, 293, 192, 31, 189, 309, 385, 411, 381, 289, 129, 408, 112, 281, 396, 451, 440, 357, 196, 624, 309, 613, 120, 276, 360, 366, 288, 120, 725

Offset: 0

Vladimir Shevelev, Nov 23 2013

nonn

a(n)=0, iff 1 + n + n^2 + n^3 is a perfect square. For example, a(7)=0 and we have 1 + 7 + 7^2 + 7^3 = 20^2.

a(n) = Difference between smallest square >= (n^3 + n^2 + n + 1) and (n^3 + n^2 + n + 1) - Antti Karttunen, Nov 27 2013

Ivan Panchenko, Table of n, a(n) for n = 0..1000

Cf. A053698, A068527.

a(n) = ceil(sqrt(n^3+n^2+n+1))^2 - (n^3+n^2+n+1); \\ Michel Marcus, Nov 23 2013

Contribution from Antti Karttunen, Nov 27 2013: (Start)
a(n) = A000290(⌈sqrt(A053698(n))⌉) - A053698(n). Where ⌈x⌉ stands for ceiling(x). This further reduces as:
a(n) = A000290(A135034(A053698(n))) - A053698(n).
a(n) = A048761(A053698(n)) - A053698(n).
a(n) = A068527(A053698(n)).
(End)

More terms from Peter J. C. Moses

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).

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

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

A256098 Denominators for the numerators A256097.

Original entry on oeis.org

1, 2, 1, 1, 4, 2, 4, 1, 1, 6, 3, 2, 3, 6, 1, 1, 8, 4, 8, 2, 8, 4, 8, 1, 1, 10, 5, 10, 5, 2, 5, 10, 5, 10, 1, 1, 12, 6, 4, 3, 12, 2, 12, 3, 4, 6, 12, 1, 1, 14, 7, 14, 7, 14, 7, 2, 7, 14, 7, 14, 7, 14, 1, 1, 16, 8, 16, 4, 16, 8, 16, 2, 16, 8, 16, 4, 16, 8, 16, 1

Offset: 1

Wolfdieter Lang, Mar 24 2015

The corresponding numerators are given in A256097.

See A256097 for comments, references and links.

			See A256097.

Stefano Spezia, Table of n, a(n) for n = 1..10000

Cf. A000196, A048760, A048761, A256097.

A000196[n_]:=Floor[Sqrt[n]]; a[n_]:=Denominator[(A000196[n]+n/A000196[n])/2]; Array[a,80](* Stefano Spezia, Feb 15 2025 *)

a(n) = denominator(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.

a(61)-a(80) from Stefano Spezia, Feb 15 2025

A077503 Smallest n-digit square beginning with n.

Original entry on oeis.org

1, 25, 324, 4096, 50176, 600625, 7001316, 80013025, 900000000, 1000014129, 11000024161, 120000580921, 1300001310976, 14000004588964, 150000007007601, 1600000000000000, 17000000233634401, 180000000244436761

Offset: 1

Amarnath Murthy, Nov 08 2002

Cf. A000290, A048761, A055642, A087095.

a(n) = ceiling(sqrt(n*10^d))^2 where d = n - digit(n), digit(n) = number of digits in n. - Ray Chandler, Aug 11 2003

More terms from Ray Chandler, Aug 11 2003

A099135 a(n) = smallest number m, not occurring earlier, such that a(k)+m is a square for some k

Original entry on oeis.org

1, 3, 6, 8, 10, 13, 12, 4, 5, 11, 14, 2, 7, 9, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72

Offset: 1

Reinhard Zumkeller, Sep 29 2004

Permutation of the natural numbers with inverse A099136;

a(n) <> n iff 1 < n <= 14.

Cf. A048761, A068527.

a(n)=if(n>14,n,[1, 3, 6, 8, 10, 13, 12, 4, 5, 11, 14, 2, 7, 9][n]) \\ Charles R Greathouse IV, Sep 02 2011

A140597 Squares nearest to and > terms in A098562.

Original entry on oeis.org

16, 20736, 76176, 239121, 2211169, 3583449, 29203216, 40005625, 45454564, 55606849, 77299264, 108097609, 115004176, 155650576, 226231681, 302934025, 324684361, 519703209, 551357361, 618367689, 797045824, 944025625, 1039740025

Offset: 1

Enoch Haga, May 17 2008

			The first term of A098562 is 13, the prime sum of 2^2 + 3^2, where 4+9=13. The square just exceeding 13 is 16, the first term of this sequence.

Carlos Rivera, Puzzle 128. Sum of consecutive squared primes a square, The Prime Puzzles & Problems Connection.
Carlos Rivera, Puzzle 443. Sum of cubes of consecutive primes, The Prime Puzzles and Problems Connection.

Cf. A048761, A140596, A066525, A098561, A098562, A140250, A140251.

a(n) = A048761(A098562(n)). - Jason Yuen, Sep 30 2024

A166068 a(n) = a(n-1)+ [least square > a(n-1)].

Original entry on oeis.org

1, 5, 14, 30, 66, 147, 316, 640, 1316, 2685, 5389, 10865, 21890, 43794, 87894, 176103, 352503, 705339, 1410939, 2822283, 5644683, 11290059, 22586380, 45177389, 90362673, 180726709, 361467845, 722962014, 1445926558, 2891903234

Offset: 1

Ctibor O. Zizka, Oct 06 2009

nonn

This sequence is the base sequence of the map: a(n) = a(n-1)+ [least square > a(n-1)] if a(n) is not divisible by Y, else a(n)=a(n-1)/Y, where Y is a positive integer.
Experimental results shows this map converges to a periodic orbit for all Y.
What is the number and length of periodic orbits for different Y?
What is the trajectory of some input under the map? If Y=2, the map converges to two periodic orbits, {1-5-14-7-16-8-4-2} and {11-27-63-127-271-560-280-140-70-35-71-152-76-38-19-44-22} whose length is L1=8, L2=17.
Two examples of trajectories for initial value 9 resp. 13 under the map for Y=2 are 9-25-61-125-269-558-279-568-284-142-{76-38-19-44-22-11-27-63-127-271-560-280-140-70-35-71-152} and 13-29-65-146-73-154-77-158-79-160-80-40-20-10-{5-14-7-16-8-4-2-1}.

Robert Israel, Table of n, a(n) for n = 1..2656
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.

Cf. A006370, A048761

A[1]:= 1:
for n from 1 to 100 do
  A[n+1]:= A[n] + (floor(sqrt(A[n]))+1)^2
od:
seq(A[n],n=1..100); # Robert Israel, Oct 06 2014

lista(n) = {na = 0; for (i=1, n, na += ceil(sqrt(na+1))^2; print1(na, ", "););} \\ Michel Marcus, Jun 02 2013

Typo in data corrected by D. S. McNeil, Aug 17 2010

A259225 Smallest oblong number greater than or equal to n.

Original entry on oeis.org

0, 2, 2, 6, 6, 6, 6, 12, 12, 12, 12, 12, 12, 20, 20, 20, 20, 20, 20, 20, 20, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 56, 72, 72, 72, 72, 72, 72, 72, 72, 72, 72

Offset: 0

Michel Marcus, Jun 21 2015

nonn

Casey Douglas, The Next Square or Pronic, June 2012. [Wayback Machine copy]

Cf. A002378 (oblong numbers), A048761 (similar, with square instead).

f[n_] := n*(n + 1); g[n_] := (Sqrt[4*n + 1] - 1)/2; a[n_] := f[Ceiling[g[n]]]; Array[a, 100, 0] (* Amiram Eldar, Aug 16 2022 *)

PARI
```
a(n) = my(k = 0); while(k*(k+1)
				
```

Sum_{n>=1} 1/a(n)^2 = 4 - Pi^2/3. - Amiram Eldar, Aug 16 2022

cp's OEIS Frontend

A232395 (ceiling(sqrt(n^3 + n^2 + n + 1)))^2 - (n^3 + n^2 + n + 1).

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

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

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Haskell

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

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Mathematica

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A256098 Denominators for the numerators A256097.

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Mathematica

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A077503 Smallest n-digit square beginning with n.

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A099135 a(n) = smallest number m, not occurring earlier, such that a(k)+m is a square for some k

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PARI

A140597 Squares nearest to and > terms in A098562.

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