cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A019554 Smallest number whose square is divisible by n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 12, 5, 26, 9, 14, 29, 30, 31, 8, 33, 34, 35, 6, 37, 38, 39, 20, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 18, 55, 28, 57, 58, 59, 30, 61, 62, 21, 8, 65, 66, 67, 34, 69, 70, 71, 12, 73, 74, 15, 38, 77
Offset: 1

Views

Author

R. Muller

Keywords

Comments

A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), and b*c = A019554(n) = "outer square root" of n.
Instead of the terms "inner square root" and "outer square root", we may use the terms "lower square root" and "upper square root", respectively. Upper k-th roots have been studied by Broughan (2002, 2003, 2006). - Petros Hadjicostas, Sep 15 2019
The number of times each number k appears in this sequence is A034444(k). The first time k appears is at position A102631(k). - N. J. A. Sloane, Jul 28 2021

Crossrefs

Cf. A000188 (inner square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Programs

  • Haskell
    a019554 n = product $ zipWith (^)
                (a027748_row n) (map ((`div` 2) . (+ 1)) $ a124010_row n)
    -- Reinhard Zumkeller, Apr 13 2013
    (Python 3.8+)
    from math import prod
    from sympy import factorint
    def A019554(n): return n//prod(p**(q//2) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021
  • Maple
    with(numtheory):A019554 := proc(n) local i: RETURN(op(mul(i,i=map(x->x[1]^ceil(x[2]/2),ifactors(n)[2])))); end;
  • Mathematica
    Flatten[Table[Select[Range[n],Divisible[#^2,n]&,1],{n,100}]] (* Harvey P. Dale, Oct 17 2011 *)
    f[p_, e_] := p^Ceiling[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)
  • PARI
    a(n)=n/core(n,1)[2] \\ Charles R Greathouse IV, Feb 24 2011
    

Formula

Replace any square factors in n by their square roots.
Multiplicative with a(p^e) = p^ceiling(e/2).
Dirichlet series:
Sum_{n>=1} a(n)/n^s = zeta(2*s-1)*zeta(s-1)/zeta(2*s-2), (Re(s) > 2);
Sum_{n>=1} (1/a(n))/n^s = zeta(2*s+1)*zeta(s+1)/zeta(2*s+2), (Re(s) > 0).
a(n) = n/A000188(n).
a(n) = denominator of n/n^(3/2). - Arkadiusz Wesolowski, Dec 04 2011
a(n) = Product_{k=1..A001221(n)} A027748(n,k)^ceiling(A124010(n,k)/2). - Reinhard Zumkeller, Apr 13 2013
Sum_{k=1..n} a(k) ~ 3*zeta(3)*n^2 / Pi^2. - Vaclav Kotesovec, Sep 18 2020
Sum_{k=1..n} 1/a(k) ~ 3*log(n)^2/(2*Pi^2) + (9*gamma/Pi^2 - 36*zeta'(2)/Pi^4)*log(n) + 6*gamma^2/Pi^2 - 108*gamma*zeta'(2)/Pi^4 + 432*zeta'(2)^2/Pi^6 - 36*zeta''(2)/Pi^4 - 15*sg1/Pi^2, where gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). - Vaclav Kotesovec, Jul 27 2021
a(n) = sqrt(n*A007913(n)). - Jianing Song, May 08 2022
a(n) = sqrt(A053143(n)). - Amiram Eldar, Sep 02 2023
From Mia Boudreau, Jul 17 2025: (Start)
a(n^2) = n.
a(A005117(n)) = A005117(n).
a(A133466(n)) = A133466(n)/2.
a(A195085(n)) = A195085(n)/3. (End)

A057918 Number of pairs of numbers (r,s) each less than n such that (r,s,n) is in geometric progression.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 3, 0, 2, 0, 1, 0, 0, 0, 1, 4, 0, 2, 1, 0, 0, 0, 3, 0, 0, 0, 5, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 3, 6, 4, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 2, 7, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 4, 1, 0, 0, 0, 3, 8, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 3, 0, 6, 2, 9, 0, 0, 0, 1, 0
Offset: 1

Views

Author

Henry Bottomley, Nov 22 2000

Keywords

Comments

Also, the number of integers k in {1,2,...,n-1} such that k*n is square. - John W. Layman, Sep 08 2011

Examples

			a(72)=5 since (2,12,72), (8,24,72), (18,36,72), (32,48,72), (50,60,72) are the possible three term geometric progressions.
		

Crossrefs

Cf. A132345 (partial sums).

Programs

  • Haskell
    a057918 n = sum $ map ((0 ^) . (`mod` n) . (^ 2)) [1..n-1]
    -- Reinhard Zumkeller, Mar 27 2012

Formula

a(n) = A000188(n) - 1.
a(A005117(n)) = 0; a(A013929(n)) > 0; A008966(n) = A000007(a(n)); a(A133466(n)) = 1; a(A195085(n)) = 2. - Reinhard Zumkeller, Mar 27 2012

A133466 Positive integers k for which there is exactly one integer i in {1,2,3,...,k-1} such that i*k is a square.

Original entry on oeis.org

4, 8, 12, 20, 24, 28, 40, 44, 52, 56, 60, 68, 76, 84, 88, 92, 104, 116, 120, 124, 132, 136, 140, 148, 152, 156, 164, 168, 172, 184, 188, 204, 212, 220, 228, 232, 236, 244, 248, 260, 264, 268, 276, 280, 284, 292, 296, 308, 312, 316, 328, 332, 340, 344, 348, 356
Offset: 1

Views

Author

John W. Layman, Nov 28 2007

Keywords

Comments

It appears that all terms of this sequence are exactly four times those of the squarefree integers (A005117).
The observed behavior is true for all n. All positive integers n are written uniquely as k*m^2 where k is squarefree, k >=1, m >= 1. The square multiples of n are j^2*k*n, j >= 1. We seek n with exactly 1 multiple that is square and less than n^2. If m = 1, there are no such multiples as we have k = n, so the least square multiple is n^2. If m >= 2, k*n is square and less than n^2. However, 4*k*n also qualifies as square and less than n^2 if m > 2. So the qualifying values of n are those with m=2. - Peter Munn, Nov 28 2019
The asymptotic density of this sequence is 3/(2*Pi^2). - Amiram Eldar, Mar 08 2021

Examples

			4 is in the sequence because among the products 1*4,2*4,3*4 = 4,8,12 there is exactly one square.
		

Crossrefs

Programs

Formula

A057918(a(n)) = 1. - Reinhard Zumkeller, Mar 27 2012
From Peter Munn, Nov 28 2019: (Start)
a(n) = 4 * A005117(n).
{a(n)} = {A225546(A007283(n)) : n >= 0}, where {a(n)} denotes the set of integers in the sequence.
(End)
Sum_{n>=1} 1/a(n)^s = zeta(s)/(4^s*zeta(2*s)), s>1. - Amiram Eldar, Sep 26 2023

A335437 Numbers k with a partition into two distinct parts (s,t) such that k | s*t.

Original entry on oeis.org

9, 16, 18, 25, 27, 32, 36, 45, 48, 49, 50, 54, 63, 64, 72, 75, 80, 81, 90, 96, 98, 99, 100, 108, 112, 117, 121, 125, 126, 128, 135, 144, 147, 150, 153, 160, 162, 169, 171, 175, 176, 180, 189, 192, 196, 198, 200, 207, 208, 216, 224, 225, 234, 240, 242, 243, 245, 250, 252, 256
Offset: 1

Views

Author

Wesley Ivan Hurt, Jun 10 2020

Keywords

Comments

All values of this sequence are nonsquarefree (A013929).
From Peter Munn, Nov 23 2020: (Start)
Numbers whose square part is greater than 4. [Proof follows from s and t having to be multiples of A019554(k), the smallest number whose square is divisible by k.]
Compare with A116451, numbers whose odd part is greater than 3. The self-inverse function A225546(.) maps the members of either one of these sets 1:1 onto the other set.
Compare with A028983, numbers whose squarefree part is greater than 2.
(End)
The asymptotic density of this sequence is 1 - 15/(2*Pi^2). - Amiram Eldar, Mar 08 2021
From Bernard Schott, Jan 09 2022: (Start)
Numbers of the form u*m^2, for u >= 1 and m >= 3 (union of first 2 comments).
A geometric application: in trapezoid ABCD, with AB // CD, the diagonals intersect at E. If the area of triangle ABE is u and the area of triangle CDE is v, with u>v, then the area of trapezoid ABCD is w = u + v + 2*sqrt(u*v); in this case, u, v, w are integer solutions iff (u,v,w) = (k*s^2,k*t^2,k*(s+t)^2), with s>t and k positives; hence, w is a term of this sequence (see IMTS link). (End)

Examples

			16 is in the sequence since it has a partition into two distinct parts (12,4), such that 16 | 12*4 = 48.
		

References

  • S. Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 of International Mathematical Talent Search, round 7, page 285.

Crossrefs

Complement of A133466 within A013929.
A038838, A046101, A062312\{1}, A195085 are subsequences.
Related to A116451 via A225546.

Programs

  • Mathematica
    Table[If[Sum[(1 - Ceiling[(i*(n - i))/n] + Floor[(i*(n - i))/n]), {i, Floor[(n - 1)/2]}] > 0, n, {}], {n, 300}] // Flatten
    f[p_, e_] := p^(2*Floor[e/2]); sqpart[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[256], sqpart[#] > 4 &] (* Amiram Eldar, Mar 08 2021 *)

A328285 Smallest positive number k >= 2 for which there exist exactly n >= 1 integers m in M = {1, 2, 3, ..., k-1} such that k*m is a perfect power.

Original entry on oeis.org

12, 4, 8, 48, 16, 32, 49, 640, 108, 162, 64, 121, 243, 144, 196, 225, 867, 289, 324, 361, 256, 400, 484, 529, 512, 1250, 676, 625, 576, 1682, 784, 900, 961, 1458, 729, 1156, 1225, 2312, 1369, 1024, 1521, 2048, 1681, 1600, 1849, 1936, 6348, 2025, 2209, 4232
Offset: 1

Views

Author

Marius A. Burtea, Nov 29 2019

Keywords

Examples

			For n = 1 and k = 12 the set M = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} and we obtain only 12 * 3 = 36 = 6^2, so a(1) = 12.
For n = 2 and k = 4 the set M = {1, 2, 3} and we obtain 4 * 1 = 4 = 2^2 and 4 * 2 = 8 = 2^3 so a(2) = 4.
For n = 3 and k = 8 the set M = {1, 2, 3, 4, 5, 6, 7}. The powers 8 * 1 = 2^3, 8 * 2 = 16 = 2^4 and 8 * 4 = 32 = 2^5 are obtained, so a(3) = 8.
		

Crossrefs

Programs

  • Magma
    a:=[]; for n in [1..40] do k:=1; while #[m:m in [1..k-1]| IsPower(m*k)] ne n do k:=k+1; end while; Append(~a,k); end for; a;
    
  • Mathematica
    ppQ[n_] := 1 < GCD @@ FactorInteger[n][[All, 2]]; cnt[k_] := cnt[k] = Length[ Select[ Range[k-1], ppQ[k #] &]]; a[n_] := Block[{k = n + 1}, While[ cnt[k] != n, k++]; k]; Array[a, 40] (* Giovanni Resta, Dec 05 2019 *)
  • PARI
    a(n) = {my(k=2); while (sum(m=1, k-1, ispower(m*k) != 0) != n, k++); k;} \\ Michel Marcus, Dec 05 2019
Showing 1-5 of 5 results.