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.

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A015049 Let m = A013929(n); then a(n) = smallest k such that m divides k^2.

Original entry on oeis.org

2, 4, 3, 6, 4, 6, 10, 12, 5, 9, 14, 8, 6, 20, 22, 15, 12, 7, 10, 26, 18, 28, 30, 21, 8, 34, 12, 15, 38, 20, 9, 42, 44, 30, 46, 24, 14, 33, 10, 52, 18, 28, 58, 39, 60, 11, 62, 25, 42, 16, 66, 45, 68, 70, 12, 21, 74, 30, 76, 51, 78, 40, 18, 82, 84, 13, 57, 86
Offset: 1

Views

Author

R. Muller

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^Ceiling[e/2]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; s /@ Select[Range[200], !SquareFreeQ[#] &] (* Amiram Eldar, Feb 10 2021 *)
  • PARI
    lista(kmax) = {my(f); for(k = 2, kmax, f = factor(k); if(!issquarefree(f), print1(k/core(f, 1)[2], ", ")));} \\ Amiram Eldar, Jan 06 2024

Formula

a(n) = A019554(A013929(n)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(2)*(zeta(3)-1)/(zeta(2)-1)^2 = 0.799082... . - Amiram Eldar, Jan 06 2024

Extensions

Description corrected by Diego Torres (torresvillarroel(AT)hotmail.com), Jun 23 2002
Offset corrected by Amiram Eldar, Feb 10 2021

A235918 Largest m such that 1, 2, ..., m divide n^2.

Original entry on oeis.org

1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 4, 1, 2, 1
Offset: 1

Views

Author

Michel Marcus, Jan 17 2014

Keywords

Comments

Note that a(n) is equal to A071222(n-1) = A053669(n)-1 for the first 209 values of n. The first difference occurs at n=210, where a(210)=7, while A071222(209)=10. A235921 lists all n where a(n) differs from A071222(n-1). (Note also that a(n) is equal to A071222(n+29) for n=1..179.) - [Comment revised by Antti Karttunen, Jan 26 2014 because of the changed definition of A235921 and newly inserted a(0)=1 term of A071222.]
See A055874 for a similar comment concerning the difference between A055874 and A232098.
Average value is 1.9124064... = sum_{n>=1} 1/A019554(A003418(n)). - Charles R Greathouse IV, Jan 24 2014

Crossrefs

One less than A236454.

Programs

  • Mathematica
    a[n_] := Module[{m = 1}, While[Divisible[n^2, m++]]; m - 2]; Array[a, 100] (* Jean-François Alcover, Mar 07 2016 *)
  • PARI
    a(n) = my(m = 1); while ((n^2 % m) == 0, m++); m - 1; \\ Michel Marcus, Jan 17 2014

Formula

a(n) = A055874(n^2).
a(n) = A236454(n)-1.

A236454 Smallest number not dividing n^2.

Original entry on oeis.org

2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2
Offset: 1

Views

Author

Antti Karttunen, Jan 26 2014

Keywords

Comments

Differs from A053669, "smallest prime not dividing n", for the first time at n=210, where a(210)=8, while A053669(210)=11. A235921 lists all n for which a(n) differs from A053669(n).
Differs from A214720 at n=2, 210, 630, 1050, 1470, 1890, 2310,.... - R. J. Mathar, Mar 30 2014

Crossrefs

One more than A235918.

Programs

  • Maple
    A236454 := proc(n)
        for m from 2 do
            if modp(n^2,m) <> 0 then
                return m;
            end if;
        end do:
    end proc:# R. J. Mathar, Mar 30 2014
  • Mathematica
    Join[{2,3},Table[Complement[Range[n],Divisors[n^2]][[1]],{n,3,90}]] (* Harvey P. Dale, Mar 18 2018 *)
  • Scheme
    (define (A236454 n) (A007978 (A000290 n)))

Formula

a(n) = A007978(A000290(n)) = A007978(n^2).
a(n) = A235918(n)+1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=0} 1/A019554(A003418(k)) = 2.91240643793540602415... . - Amiram Eldar, Jan 14 2024

A254732 a(n) is the least k > n such that n divides k^2.

Original entry on oeis.org

2, 4, 6, 6, 10, 12, 14, 12, 12, 20, 22, 18, 26, 28, 30, 20, 34, 24, 38, 30, 42, 44, 46, 36, 30, 52, 36, 42, 58, 60, 62, 40, 66, 68, 70, 42, 74, 76, 78, 60, 82, 84, 86, 66, 60, 92, 94, 60, 56, 60, 102, 78, 106, 72, 110, 84, 114, 116, 118, 90, 122, 124, 84, 72
Offset: 1

Views

Author

Peter Kagey, Feb 06 2015

Keywords

Comments

A073353(n) <= a(n) <= 2*n. Any prime that divides n must also divide a(n), and because n divides (2*n)^2.
Are all terms even? -Harvey P. Dale, Aug 07 2025

Examples

			a(12) = 18 because 12 divides 18^2, but 12 does not divide 13^2, 14^2, 15^2, 16^2, or 17^2.
		

Crossrefs

Cf. A254733 (similar, with k^3), A254734 (similar, with k^4), A073353 (similar, with limit m->infinity of k^m).
Cf. A253905.

Programs

  • Haskell
    a254732 n = head [k | k <- [n + 1 ..], mod (k ^ 2) n == 0]
    -- Reinhard Zumkeller, Feb 07 2015
    
  • Mathematica
    lk[n_]:=Module[{k=n+1},While[!Divisible[k^2,n],k++];k]; Array[lk,70] (* Harvey P. Dale, Nov 05 2017 *)
    Table[Module[{k=n+1},While[PowerMod[k,2,n]!=0,k++];k],{n,70}] (* Harvey P. Dale, Aug 07 2025 *)
  • PARI
    a(n)=for(k=n+1,2*n,if(k^2%n==0,return(k)))
    vector(100,n,a(n)) \\ Derek Orr, Feb 06 2015
    
  • PARI
    a(n)=my(t=factorback(factor(n)[,1])); forstep(k=n+t,2*n,t,if(k^2%n==0, return(k))) \\ Charles R Greathouse IV, Feb 07 2015
    
  • Python
    def A254732(n):
        k = n + 1
        while pow(k,2,n):
            k += 1
        return k # Chai Wah Wu, Feb 15 2015
  • Ruby
    def a(n)
      (n+1..2*n).find { |k| k**2 % n == 0 }
    end
    

Formula

a(n) = sqrt(n*A072905(n)).
a(n) = A019554(n)*(A000188(n)+1).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + zeta(3)/zeta(2) = 1 + A253905 = 1.73076296940143849872... . - Amiram Eldar, Feb 17 2024

A365491 The number of divisors of the smallest number whose 4th power is divisible by n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 3, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 3, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Sep 05 2023

Keywords

Comments

First differs from A365210 at n = 25 and from A034444 at n = 32.
The number of divisors of the smallest 4th divisible by n, A053167(n), is A365492(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Ceiling[e/4] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
    With[{c=Range[100]^4},Table[DivisorSigma[0,Surd[SelectFirst[c,Mod[#,n]==0&],4]],{n,90}]] (* Harvey P. Dale, Jul 09 2024 *)
  • PARI
    a(n) = vecprod(apply(x -> (x-1)\4 + 2, factor(n)[, 2]));

Formula

a(n) = A000005(A053166(n)).
Multiplicative with a(p^e) = ceiling(e/4) + 1.
a(n) <= A000005(n) with equality if and only if n is squarefree (A005117).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 1/p^s - 1/p^(4*s)).
From Vaclav Kotesovec, Sep 06 2023: (Start)
Dirichlet g.f.: zeta(s)^2 * zeta(4*s) * Product_{p prime} (1 - 1/p^(2*s) - 1/p^(4*s) + 1/p^(5*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) - 1/p^(4*s) + 1/p^(5*s)).
Sum_{k=1..n} a(k) ~ zeta(4) * f(1) * n * (log(n) + 2*gamma - 1 + 4*zeta'(4)/zeta(4) + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/p^2 - 1/p^4 + 1/p^5) = 0.57615273538566705952061107826411727540624711680289618854325028459572487...,
f'(1) = f(1) * Sum_{p prime} (-5 + 4*p + 2*p^3) * log(p) / (1 - p - p^3 + p^5) = f(1) * 1.3011434396559802378314782600747661399223385669839998680418996210...
and gamma is the Euler-Mascheroni constant A001620. (End)
a(n) = A322483(A019554(n)) (the number of exponentially odd divisors of the smallest number whose square is divisible by n). - Amiram Eldar, Sep 08 2023

A065887 Smallest number whose square is divisible by n!.

Original entry on oeis.org

1, 1, 2, 6, 12, 60, 60, 420, 1680, 5040, 5040, 55440, 332640, 4324320, 8648640, 43243200, 172972800, 2940537600, 8821612800, 167610643200, 335221286400, 7039647014400, 14079294028800, 323823762662400, 647647525324800, 3238237626624000, 6476475253248000
Offset: 0

Views

Author

Henry Bottomley, Nov 27 2001

Keywords

Examples

			a(10) = 5040 since 10! = 3628800 and the smallest square divisible by this is 25401600 = 3628800*7 = 5040^2.
		

Crossrefs

Programs

  • Maple
    a:= n-> mul(i[1]^ceil(i[2]/2), i=ifactors(n!)[2]):
    seq(a(n), n=0..26);  # Alois P. Heinz, Jan 24 2022
  • Mathematica
    f[p_, e_] := p^Ceiling[e/2]; a[0] = a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 30, 0] (* Amiram Eldar, Feb 11 2024 *)

Formula

a(n) = A019554(A000142(n)) = sqrt(A065886(n)) = A000142(n)/A055772(n).

Extensions

Missing a(0) inserted, formula corrected, and a(25)-a(26) added by Kevin P. Thompson, Jan 24 2022

A066069 a(n) is the smallest positive integer m such that n divides (n + m)^m.

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, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38
Offset: 1

Views

Author

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Dec 02 2001

Keywords

Comments

Essentially the same as A019530. - R. J. Mathar, Sep 30 2008

Examples

			a(8) = 4 because (8+1)^1 mod 8 = 1, (8+2)^2 mod 8 = 4, (8+3)^3 mod 8 = 3 and (8+4)^4 mod 8 = 0.
		

Crossrefs

First 23 terms are the same as A019554.

Programs

  • PARI
    { for (n=1, 1000, m=1; while ((n+m)^m % n, m++); write("b066069.txt", n, " ", m) ) } \\ Harry J. Smith, Nov 09 2009

A083482 Square root of smallest square of the type n(n+1)*k.

Original entry on oeis.org

2, 6, 6, 10, 30, 42, 28, 12, 30, 110, 66, 78, 182, 210, 60, 68, 102, 114, 190, 210, 462, 506, 276, 60, 130, 234, 126, 406, 870, 930, 248, 264, 1122, 1190, 210, 222, 1406, 1482, 780, 820, 1722, 1806, 946, 330, 690, 2162, 564, 84, 70, 510, 1326, 1378, 954, 990
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 03 2003

Keywords

Comments

Squares pertaining to A083481.
a(n) == (p*q*r... ) where p,q,r are prime factors of n(n+1).

Crossrefs

Cf. A083481.

Programs

  • Mathematica
    Table[Times @@ ((a = Transpose[FactorInteger[n (n + 1)]])[[1]]^Quotient[a[[2]] + 1, 2]), {n, 54}] (* Ivan Neretin, May 20 2015 *)
  • PARI
    a(n)=sqrt(n*(n+1)*core(n*(n+1)))
    
  • Python
    from math import prod
    from sympy import factorint
    def A083482(n): return n*(n+1)//prod(p**(q>>1) for p, q in factorint(n*(n+1)).items()) # Chai Wah Wu, Mar 20 2023

Formula

a(n) = sqrt(A002378(n)*A083481(n)) = sqrt(A002378(n)*A007913(A002378(n))). a(n) = A019554(A002378(n)). - David Wasserman, Nov 16 2004

Extensions

More terms from Benoit Cloitre, May 04 2003
More terms from David Wasserman, Nov 16 2004

A327170 Number of divisors d of n such that A327171(d) (= phi(d)*core(d)) is equal to n.

Original entry on oeis.org

1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Sep 28 2019

Keywords

Comments

From any solution (*) to A327171(d) = d*phi(d) = n, we obtain a solution for core(d')*phi(d') = n by forming a "pumped up" version d' of d, by replacing each exponent e_i in the prime factorization of d = p_1^e_1 * p_2^e_2 * ... * p_k^e_k, with exponent 2*e_i - 1 so that d' = p_1^(2*e_1 - 1) * p_2^(2*e_2 - 1)* ... * p_k^(2*e_k - 1) = A102631(d) = d*A003557(d), and this d' is also a divisor of n, as n = d' * A173557(d). Generally, any product m = p_1^(2*e_1 - x) * p_2^(2*e_2 - y)* ... * p_k^(2*e_k - z), where each x, y, ..., z is either 0 or 1 gives a solution for core(m)*phi(m) = n, thus every nonzero term in this sequence is a power of 2, even though not all such m's might be divisors of n.
(* by necessity unique, see Franz Vrabec's Dec 12 2012 comment in A002618).
On the other hand, if we have any solution d for core(d)*phi(d) = n, we can find the unique such divisor e of d that e*phi(e) = n by setting e = A019554(d).
Thus, it follows that the nonzero terms in this sequence occur exactly at positions given by A082473.
Records (1, 2, 4, 8, 16, ...) occur at n = 1, 12, 504, 223200, 50097600, ...

Examples

			For n = 504 = 2^3 * 3^2 * 7, it has 24 divisors, out of which four divisors: 42 (= 2^1 * 3^1 * 7^1), 84 (= 2^2 * 3^1 * 7^1), 126 (= 2^1 * 3^2 * 7^1), 252 (= 2^2 * 3^2 * 7^1) are such that A007913(d)*A000010(d) = 504, thus a(504) = 4.
		

Crossrefs

Programs

  • Mathematica
    With[{s = Array[EulerPhi[#] (Sqrt@ # /. (c_: 1) a_^(b_: 0) :> (c a^b)^2) &, 120]}, Table[DivisorSum[n, 1 &, s[[#]] == n &], {n, Length@ s}]] (* Michael De Vlieger, Sep 29 2019, after Bill Gosper at A007913 *)
  • PARI
    A327170(n) = sumdiv(n,d,eulerphi(d)*core(d) == n);

Formula

a(n) = Sum_{d|n} [A000010(d)*A007913(d) == n], where [ ] is the Iverson bracket.

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 *)
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