Valtteri Raiko - cp's OEIS frontend

A256173 Numbers k such that ceiling(sqrt(k))^2 - k is a square.

0, 1, 3, 4, 5, 8, 9, 12, 15, 16, 21, 24, 25, 27, 32, 35, 36, 40, 45, 48, 49, 55, 60, 63, 64, 65, 72, 77, 80, 81, 84, 91, 96, 99, 100, 105, 112, 117, 120, 121, 128, 135, 140, 143, 144, 153, 160, 165, 168, 169, 171, 180, 187, 192, 195, 196, 200, 209, 216, 221, 224, 225, 231, 240, 247, 252, 255, 256, 264, 273, 280, 285, 288, 289, 299

Offset: 1

Valtteri Raiko, Mar 17 2015

nonn

Numbers k such that A068527(k) is a square. k is in the sequence if and only if k - ceiling(sqrt(k))^2 + ceiling(sqrt(ceiling(sqrt(k))^2 - k))^2 = 0.
A000290 is a subsequence since for a square k, ceiling(sqrt(k))^2 - k = 0, a square too.
Also, numbers k such that A249298(k) is 1.
Also, numbers k such that A249142(k) is 0.
The only prime numbers in the sequence are 3 and 5.
No number from A016825 appears in the sequence.
If p and q are terms of A065091 and if q satisfies the inequality p - 2*sqrt(2p) + 2 < q < p + 2*sqrt(2p) + 2, then p*q is in the sequence. Thus infinitely many numbers from A046315 appear in the sequence.

			Ceiling(sqrt(27))^2 - 27 = 9 = 3^2, so 27 is in the sequence.

Cf. A000290, A068527, A145236, A249142, A249298, A046315.

[n: n in [0..200] | IsSquare(Ceiling(Sqrt(n))^2-n)]; // Vincenzo Librandi, Mar 18 2015

Flatten[Position[Table[n - Ceiling[Sqrt[n]]^2 + Ceiling[Sqrt[-n + Ceiling[Sqrt[n]]^2]]^2, {n, 0, 300}], 0]] - 1
Select[Range[0,300],IntegerQ[Sqrt[Ceiling[Sqrt[#]]^2-#]]&] (* Harvey P. Dale, Sep 06 2023 *)

isok(n) = issquare(ceil(sqrt(n))^2-n); \\ Michel Marcus, Mar 18 2015

A249298 Smallest positive integer k, such that s-kn is a square where s is the smallest square >= kn.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 3, 1, 1, 4, 5, 1, 5, 6, 1, 1, 9, 2, 9, 2, 1, 12, 13, 1, 1, 14, 1, 3, 17, 2, 19, 1, 3, 20, 1, 1, 23, 24, 3, 1, 25, 2, 27, 6, 1, 30, 31, 1, 1, 2, 3, 7, 35, 4, 1, 2, 3, 40, 41, 1, 41, 42, 1, 1, 1, 4, 47, 10, 5, 2, 51, 1, 51, 52, 3, 12, 1, 6, 57, 1, 1, 60, 61, 1, 3, 62, 7, 3, 65, 2

Offset: 1

Valtteri Raiko, Oct 24 2014

nonn

For any n>=3, there exists at least one positive integer k, 1 <= k <= n-1 such that the difference between the smallest square >= k*n and k*n is a square. To prove this, consider the multiplier k = n-2. Then (n-2)*n = (n-1)^2-1, thus the difference from the next square is 1, which is a square. If n = 1, k = 1 and if n = 2, k = 2.

Smallest positive integer k such that ceiling(sqrt(k*n))^2-k*n is a square.

			a(10) = 4, for ceiling(sqrt(10))^2-10 = 6, ceiling(sqrt(2*10))^2-2*10 = 5, ceiling(sqrt(3*10))^2-3*10 = 6 and ceiling(sqrt(4*10))^2-4*10 = 9 = 3^2.

Cf. A000290, A145236 (equals a(A000040)), A068527 (difference for k=1).

Cf. A145016, A145022, A145023, A145047, A145048, A145049, A145050, A145215.

dif[n_] := Ceiling[Sqrt[n]]^2 - n;a[k_] := Module[{n = 1}, While[dif[dif[n*k]] != 0, n++]; Return[n]];Table[a[k], {k, 1, 90}]

a(n) = {k=1; while(!issquare(ceil(sqrt(k*n))^2-k*n), k++); k;} \\ Michel Marcus, Oct 24 2014

A249160 Smallest number of iterations k such that A068527^(k)(n)=A068527^(k+1)(n).

Original entry on oeis.org

1, 0, 2, 1, 2, 3, 1, 2, 1, 4, 3, 2, 3, 1, 2, 1, 3, 2, 4, 3, 2, 3, 1, 2, 1, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 2, 4, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 2, 3, 2, 4, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 3, 4, 2, 3, 2, 4, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1

Offset: 1

Valtteri Raiko, Oct 22 2014

Given a number n, denote its distance from next perfect square >= n as R(n), sequence A068527. The function R(n) has two fixed points, 0 and 2, and for all n>=3, R(n)=0, there exists a k>=0 such that R^(k)(n)=R^(k+1)(n)=0 or 2. This sequence gives the number of iterations needed to reach the fixed point starting at n.

This sequence is unbounded, but grows very slowly, reaching records of 1, 2, 3, 4, 6 etc at n=1, 3, 6, 10, 26, 170, 7226, etc.

			R(10) = 6, R(6) = 3, R(3) = 1, R(1) = 0, R(0) = 0. Thus a(10) = 4.

Cf. A068527.

A249160 := proc(n)
    local k,prev,this;
    prev := n ;
    for k from 1 do
        this := A068527(prev) ;
        if this = prev then
            return k-1;
        end if;
        prev := this ;
    end do:
end proc:
seq(A249160(n),n=1..80) ; # R. J. Mathar, Nov 17 2014

r[n_]:=Ceiling[Sqrt[n]]^2-n;Table[Length[FixedPointList[r,n]]-2,{n,1,100}]

r(n)=if(issquare(n),0,(sqrtint(n)+1)^2-n);
le(n)=b=0;while(n!=0&&n!=2,b=b+1;n=r(n));return(b);
range(n) = c=List(); for(a = 1, n, listput(c,a)); return(c);
apply(le, range(100))

A249142 Let k be the difference between the smallest square >= n and n. Sequence gives difference between the smallest square >= k and k.

Original entry on oeis.org

0, 2, 0, 0, 0, 1, 2, 0, 0, 3, 4, 0, 1, 2, 0, 0, 1, 2, 3, 4, 0, 1, 2, 0, 0, 6, 0, 1, 2, 3, 4, 0, 1, 2, 0, 0, 4, 5, 6, 0, 1, 2, 3, 4, 0, 1, 2, 0, 0, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 0, 1, 2, 0, 0, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 0, 1, 2, 0, 0, 7, 8, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 0, 1, 2, 0, 0

Offset: 1

Valtteri Raiko, Oct 22 2014

Equals A068527 applied to itself.

			For n = 13 the next biggest square is 16, thus k = 16 - 13 = 3 and for 3 the next biggest square is 4, thus a(14) = 3 - 2 = 1.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A068527.

[n - Ceiling(Sqrt(n))^2 + Ceiling(Sqrt(-n+Ceiling(Sqrt(n))^2))^2: n in [1..100]]; // Vincenzo Librandi, Oct 23 2014

A068527:= n -> ceil(sqrt(n))^2 - n:
map(A068527@@2, [$1..100]); # Robert Israel, Nov 02 2017

Table[n - Ceiling[Sqrt[n]]^2 + Ceiling[Sqrt[-n + Ceiling[Sqrt[n]]^2]]^2, {n, 1, 100}]

A068527(n)=if(issquare(n), 0, (sqrtint(n)+1)^2-n)
a(n)=A068527(A068527(n)) \\ Charles R Greathouse IV, Oct 22 2014

a(n) = A068527(A068527(n)).
a(n) = n - ceiling(sqrt(n))^2 + ceiling(sqrt(-n+ceiling(sqrt(n))^2))^2.
a(n) < (64n)^(1/4). - Charles R Greathouse IV, Oct 22 2014

Edited, old crossrefs entry moved to Comments, and first two formula lines interchanged by Wolfdieter Lang, Nov 10 2014

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A256173 Numbers k such that ceiling(sqrt(k))^2 - k is a square.

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A249298 Smallest positive integer k, such that s-k*n is a square where s is the smallest square >= k*n.

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A249160 Smallest number of iterations k such that A068527^(k)(n)=A068527^(k+1)(n).

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A249142 Let k be the difference between the smallest square >= n and n. Sequence gives difference between the smallest square >= k and k.

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A249298 Smallest positive integer k, such that s-kn is a square where s is the smallest square >= kn.