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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A019555 Smallest number whose cube is divisible by n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 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, 12, 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, 77, 78
Offset: 1

Views

Author

R. Muller

Keywords

Comments

This can be thought as an "upper 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "lower 3rd root" of positive integers is given by A053150. - Petros Hadjicostas, Sep 15 2019

Crossrefs

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Programs

  • Maple
    f:= n -> mul(t[1]^ceil(t[2]/3), t = ifactors(n)[2]):
    map(f, [$1..100]); # Robert Israel, Sep 22 2015
  • Mathematica
    cubes=Range[85]^3; Table[Position[Divisible[cubes,i],True,1,1][[1,1]],{i,85}] (* Harvey P. Dale, Jan 12 2011 *)
    f[p_, e_] := p^Ceiling[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]  (* Amiram Eldar, Jan 06 2024 *)
  • PARI
    a(n)=my(r=1);while(r^3%n!=0,r++);r \\ Anders Hellström, Sep 22 2015
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 + p*X + p*X^2)/(1 - p*X^3))[n], ", ")) \\ Vaclav Kotesovec, Aug 30 2021
    
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^ceil(f[i,2]/3));} \\ Amiram Eldar, Jan 06 2024
    
  • Python
    from math import prod
    from sympy import factorint
    def A019555(n): return prod(p**((q%3 != 0)+(q//3)) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021
  • Sage
    [prod([t[0]^(ceil(t[1]/3)) for t in factor(n)]) for n in range(1,79)] # Danny Rorabaugh, Sep 22 2015
    

Formula

Replace any cubic factors in n by their cube roots.
a(n) = n/A000189(n).
Multiplicative with a(p^e) = p^ceiling(e/3). - R. J. Mathar, May 29 2011
From Vaclav Kotesovec, Aug 30 2021: (Start)
Dirichlet g.f.: zeta(3*s-1) * Product_{p prime} (1 + p^(1 - s) + p^(1 - 2*s)).
Dirichlet g.f.: zeta(3*s-1) * zeta(s-1) * Product_{p prime} (1 - p^(2 - 3*s) + p^(1 - 2*s) - p^(2 - 2*s)).
Sum_{k=1..n} a(k) ~ c * zeta(5) * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.684286924186862318141968725791218083472312736723163777284618226290055... (End)

Extensions

Corrected and extended by David W. Wilson

A053143 Smallest square divisible by n.

Original entry on oeis.org

1, 4, 9, 4, 25, 36, 49, 16, 9, 100, 121, 36, 169, 196, 225, 16, 289, 36, 361, 100, 441, 484, 529, 144, 25, 676, 81, 196, 841, 900, 961, 64, 1089, 1156, 1225, 36, 1369, 1444, 1521, 400, 1681, 1764, 1849, 484, 225, 2116, 2209, 144, 49, 100, 2601, 676, 2809
Offset: 1

Views

Author

Henry Bottomley, Feb 28 2000

Keywords

Crossrefs

Programs

  • Mathematica
    Table[i = 1; While[Mod[i^2, n] > 0, i++]; i^2, {n, 100}] (* T. D. Noe, Oct 30 2011 *)
    f[p_, e_] := p^(e + Mod[e, 2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
  • PARI
    a(n) = n*core(n); /* Joerg Arndt, Aug 02 2012 */

Formula

a(n) = n*A007913(n) = A019554(n)^2 = n^2/A008833(n) = (n/A000188(n))^2.
Multiplicative with p^e -> p^(e + e mod 2), p prime. - Reinhard Zumkeller, Feb 09 2003
Dirichlet g.f.: zeta(2s-2)*zeta(s-2)/zeta(2s-4). - R. J. Mathar, Oct 31 2011
Sum_{k=1..n} a(k) ~ Pi^2 * n^3 / 45. - Vaclav Kotesovec, Feb 08 2019
Sum_{n>=1} 1/a(n) = 5/2. - Amiram Eldar, Jul 29 2022

A053166 Smallest positive integer for which n divides a(n)^4.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 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

Henry Bottomley, Feb 29 2000

Keywords

Comments

According to Broughan (2002, 2003, 2006), a(n) is the "upper 4th root of n". The "lower 4th root of n" is sequence A053164. - Petros Hadjicostas, Sep 15 2019

Crossrefs

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A015052 (outer 5th root), A015053 (outer 6th root).

Programs

  • Mathematica
    f[p_, e_] := p^Ceiling[e/4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)
  • PARI
    a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/4)); factorback(f); \\ Michel Marcus, Jun 09 2014

Formula

a(n) = n/A000190(n) = A019554(n)/(A008835(A019554(n)^2))^(1/4).
If n is 5th-power-free (i.e., not 32, 64, 128, 243, ...) then a(n) = A007947(n).
Multiplicative with a(p^e) = p^(ceiling(e/4)). - Christian G. Bower, May 16 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(7)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3528057925... . - Amiram Eldar, Oct 27 2022

A268390 Products of an even number of distinct primes and the square of a number in the sequence (including 1).

Original entry on oeis.org

1, 6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 36, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95, 100, 106, 111, 115, 118, 119, 122, 123, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 166, 177, 178, 183, 185, 187, 194, 196, 201, 202, 203, 205, 206, 209, 210
Offset: 1

Views

Author

Antti Karttunen, Feb 05 2016

Keywords

Comments

Old name: 'Positions of zeros in A268387: numbers n such that when the exponents e_1 .. e_k in their prime factorization n = p_1^e_1 * ... * p_k^e_k are bitwise-xored together, the result is zero.
From Peter Munn, Sep 14 2019 and Dec 01 2019: (Start)
When trailing zeros are removed from the terms written in base p, for any prime p, every positive integer not divisible by p appears exactly once. This is the lexicographically earliest sequence with this property.
The closure of A238748 with respect to the commutative binary operation A059897(.,.). As integers are self-inverse under A059897(.,.), the sequence thereby forms a subgroup, denoted H, of the positive integers under A059897(.,.). H is a subgroup of A000379.
(The symbol ^ can take on a meaning in relation to a group operation. However, in this comment ^ denotes the power operator for standard integer multiplication.) For any prime p, the subgroup {p^k : k >= 0} and H are each a (left and right) transversal of the other. For k >= 0 and primes p_1 and p_2, the cosets (p_1^k)H and (p_2^k)H are the same.
(End)
From Peter Munn, Dec 01 2021: (Start)
If we take the square root of the square terms we reproduce the sequence itself. The set of all products of a square term and a squarefree term is the sequence as a set.
The terms are the elements of the ideal generated by {6} in the ring defined in A329329. Similarly, the ideal generated by {8} gives A262675. 6 and 8 are images of each other under A225546(.), which is an automorphism of the ring. So this sequence and A262675, as sets, are images of each other under A225546(.). The elements of the ideal generated by {6,8} form the notable set A000379.
(End)

Examples

			1 has an empty factorization, and as XOR of an empty set is zero, 1 is included.
6 = 2^1 * 3^1 and as XOR(1,1) = 0, 6 is included.
30 = 2^1 * 3^1 * 5^1 is NOT included, as XOR(1,1,1) = 1.
360 = 2^3 * 3^2 * 5^1 is included, as the bitwise-XOR of exponents 3, 2 and 1 ("11", "10" and "01" in binary) results zero.
10, 15, 36 and 216 are in A238748. 360 = A059897(10, 36) = A059897(15, 216) and 540 = A059897(15, 36) = A059897(10, 216). So 360 and 540 are in the closure of A238748 under A059897(.,.), so in this sequence although absent from A238748. - _Peter Munn_, Oct 30 2019
		

Crossrefs

Positions of 0's in A268387, cf. A374595 (positions of 1's).
Cf. A000188, A003987, A048833 (counts prime signatures that are represented), A059897, A329329.
Subsequences: A006881 (semiprime terms), A030229 (squarefree terms), A238748 (differs first by missing a(115) = 360 and lists more subsequences).
Subsequences for prime signatures not within A238748: A163569, A190111, A190468.
Subsequence of A000379, A028260. Differs from their intersection, A374472, by omitting 64, 144, 324 etc.
Related to A262675 via A225546.
Ordered odd bisection of A334205.

Programs

  • Mathematica
    Select[Range[200], # == 1 || BitXor @@ Last /@ FactorInteger[#] == 0 &] (* Amiram Eldar, Nov 27 2020 *)

Formula

From Peter Munn, Oct 30 2019: (Start)
For k >= 0, prime p_1, prime p_2, {m : m = A059897(p_1^k, a(n)), n >= 1} = {m : m = A059897(p_2^k, a(n)), n >= 1}.
For n >= 1, k >= 0, prime p, A268387(A059897(p^k, a(n))) = k.
(End)
From Peter Munn, Nov 24 2021: (Start)
{a(n) : n >= 1} = {A000188(a(n)) : n >= 1}.
{a(n) : n >= 1} = {A225546(A262675(n)) : n >= 1}.
{A059897(a(n), A262675(m)) : n >= 1, m >= 1} = {A000379(k) : k >= 1}.
(End)

Extensions

New name from Peter Munn, Jul 15 2024

A055076 Multiplicity of Max{gcd(d, n/d)} when d runs over divisors of n.

Original entry on oeis.org

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

Views

Author

Labos Elemer, Jun 13 2000

Keywords

Comments

Number of distinct values of gcd(d, n!/d) if d runs over divisors of n! seems to be A046951(n).
a(n) = 1 iff n is a square. - Bernard Schott, Oct 22 2019
a(n) is the number of the unitary divisors (cf. A077610) of n that are exponentially odd (A268335). - Amiram Eldar, Nov 11 2022
The number of infinitary divisors of n that are squarefree (A005117). - Amiram Eldar, Jan 09 2024

Examples

			n=120, the set of gcd(d, 120/d) values for the 16 divisors of 120 is {1,2,1,2,1,2,1,2,2,1,2,1,2,1,2,1}. The max is 2 and it occurs 8 times, so a(120)=8. This sequence seems to consist of powers of 2.
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    a:= n->(p->coeff(p, x, degree(p)))(add(x^igcd(d, n/d), d=divisors(n))):
    seq(a(n), n=1..105);  # Alois P. Heinz, Jul 21 2015
  • Mathematica
    a[n_] := With[{g = GCD[#, n/#]& /@ Divisors[n]}, Count[g, Max[g]]];
    Array[a, 105] (* Jean-François Alcover, Mar 28 2017 *)
    f[p_, e_] := 2^Mod[e, 2]; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    A055076(n) = if(1==n,n,my(es=factor(n)[,2]~); prod(i=1,#es,2^(es[i]%2))); \\ Antti Karttunen, Apr 05 2021
    
  • Scheme
    ;; With memoization-macro definec.
    (definec (A055076 n) (if (= 1 n) n (* (+ 1 (A000035 (A067029 n))) (A055076 (A028234 n))))) ;; Antti Karttunen, Dec 02 2017

Formula

Multiplicative with a(p^e) = 2^(e mod 2). - Vladeta Jovovic, Dec 13 2002
a(n) = 2^A162642(n). - Antti Karttunen, Dec 02 2017
a(n) = A034444(A007913(n)). [Found by LODA miner, see C. Krause link. Essentially the same formula as the above ones] - Antti Karttunen, Apr 05 2021
From Amiram Eldar, Sep 09 2023: (Start)
a(n) = A034444(A350389(n)).
Dirichlet g.f.: zeta(2*s) * Product_{p prime} (1 + 2/p^s). (End)
From Vaclav Kotesovec, Sep 09 2023: (Start)
Let f(s) = Product_{p prime} (1 - 3/p^(2*s) + 2/p^(3*s)).
Dirichlet g.f.: zeta(s)^2 * zeta(2*s) * f(s).
Sum_{k=1..n} a(k) ~ (Pi^2 * f(1) * n / 6) * (log(n) + 2*gamma - 1 + 12*zeta'(2)/Pi^2 + f'(1)/f(1)), where
f(1) = A065473 = Product_{primes p} (1 - 3/p^2 + 2/p^3) = 0.286747428434478734107892712789838446434331844097056995641477859336652243...,
f'(1) = f(1) * Sum_{primes p} 6*log(p) / (p^2 + p - 2) = f(1) * 2.798014228561519243358371276385174449737670294137200281334256087932048625...
and gamma is the Euler-Mascheroni constant A001620. (End)

A132188 Number of 3-term geometric progressions with no term exceeding n.

Original entry on oeis.org

1, 2, 3, 6, 7, 8, 9, 12, 17, 18, 19, 22, 23, 24, 25, 32, 33, 38, 39, 42, 43, 44, 45, 48, 57, 58, 63, 66, 67, 68, 69, 76, 77, 78, 79, 90, 91, 92, 93, 96, 97, 98, 99, 102, 107, 108, 109, 116, 129, 138, 139, 142, 143, 148, 149, 152, 153, 154, 155, 158
Offset: 1

Views

Author

Gerry Myerson, Nov 21 2007

Keywords

Comments

a(n) = number of pairs (i,j) in [1..n] X [1..n] with integral geometric mean sqrt(i*j). Cf. A000982, A362931. - N. J. A. Sloane, Aug 28 2023
Also the number of 2 X 2 symmetric singular matrices with entries from {1, ..., n} - cf. A064368.
Rephrased: Number of ordered triples (w,x,y) with all terms in {1,...,n} and w^2=x*y. See A211422. - Clark Kimberling, Apr 14 2012

Examples

			a(4) counts these six (w,x,y) - triples: (1,1,1), (2,1,4), (2,4,1), (2,2,2), (3,3,3), (4,4,4). - _Clark Kimberling_, Apr 14 2012
		

Crossrefs

Programs

  • Haskell
    a132188 0 = 0
    a132188 n = a132345 n + (a120486 $ fromInteger n)
    -- Reinhard Zumkeller, Apr 21 2012
    
  • Maple
    a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+
          1+2*add(`if`(issqr(i*n), 1, 0), i=1..n-1))
        end:
    seq(a(n), n=1..60);  # Alois P. Heinz, Aug 28 2023
  • Mathematica
    t[n_] := t[n] = Flatten[Table[w^2 - x*y, {w, 1, n}, {x, 1, n}, {y, 1, n}]]
    c[n_] := Count[t[n], 0]
    t = Table[c[n], {n, 0, 80}]  (* Clark Kimberling, Apr 14 2012 *)
  • Python
    from sympy.ntheory.primetest import is_square
    def A132188(n): return n+(sum(1 for x in range(1,n+1) for y in range(1,x) if is_square(x*y))<<1) # Chai Wah Wu, Aug 28 2023

Formula

a(n) = Sum [sqrt(n/k)]^2, where the sum is over all squarefree k not exceeding n.
If we call A120486, this sequence and A132189 F(n), P(n) and S(n), respectively, then P(n) = 2 F(n) - n = S(n) + n. The Finch-Sebah paper cited at A000188 proves that F(n) is asymptotic to (3 / pi^2) n log n. In the reference, we prove that F(n) = (3 / pi^2) n log n + O(n), from which it follows that P(n) = (6 / pi^2) n log n + O(n) and similarly for S(n).
a(n) = Sum_{1 <=x,y <=n} A010052(x*y). - Clark Kimberling, Apr 14 2012
a(n) = n+2*Sum_{1<=xA010052(x*y). - Chai Wah Wu, Aug 28 2023

A053149 Smallest cube divisible by n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 8, 27, 1000, 1331, 216, 2197, 2744, 3375, 64, 4913, 216, 6859, 1000, 9261, 10648, 12167, 216, 125, 17576, 27, 2744, 24389, 27000, 29791, 64, 35937, 39304, 42875, 216, 50653, 54872, 59319, 1000, 68921, 74088, 79507, 10648
Offset: 1

Views

Author

Henry Bottomley, Feb 28 2000

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := For[k = 1, True, k++, If[ Divisible[c = k^3, n], Return[c]]]; Table[a[n], {n, 1, 44}] (* Jean-François Alcover, Sep 03 2012 *)
    f[p_, e_] := p^(e + Mod[3 - e, 3]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
    scdn[n_]:=Module[{c=Ceiling[Surd[n,3]]},While[!Divisible[c^3,n],c++];c^3]; Array[scdn,50] (* Harvey P. Dale, Jun 13 2020 *)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(f[i,2] + (3-f[i,2])%3));} \\ Amiram Eldar, Oct 27 2022

Formula

a(n) = (n/A000189(n))^3 = A008834(n)*A019554(A050985(n))^3 = n*A050985(n)^2/A000188(A050985(n))^3.
a(n) = n * A048798(n). - Franklin T. Adams-Watters, Apr 08 2009
From Amiram Eldar, Jul 29 2022: (Start)
Multiplicative with a(p^e) = p^(e + ((3-e) mod 3)).
Sum_{n>=1} 1/a(n) = Product_{p prime} ((p^3+2)/(p^3-1)) = 1.655234386560802506... . (End)
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(9)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013667*A330596/(4*A002117) = 0.1559906... . - Amiram Eldar, Oct 27 2022

A055155 a(n) = Sum_{d|n} gcd(d, n/d).

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 6, 5, 4, 2, 8, 2, 4, 4, 10, 2, 10, 2, 8, 4, 4, 2, 12, 7, 4, 8, 8, 2, 8, 2, 14, 4, 4, 4, 20, 2, 4, 4, 12, 2, 8, 2, 8, 10, 4, 2, 20, 9, 14, 4, 8, 2, 16, 4, 12, 4, 4, 2, 16, 2, 4, 10, 22, 4, 8, 2, 8, 4, 8, 2, 30, 2, 4, 14, 8, 4, 8, 2, 20, 17, 4, 2, 16, 4, 4, 4, 12, 2, 20, 4, 8, 4, 4
Offset: 1

Views

Author

Leroy Quet, Jul 02 2000

Keywords

Comments

a(n) is odd iff n is odd square. - Vladeta Jovovic, Aug 27 2002
From Robert Israel, Dec 26 2015: (Start)
a(n) >= A000005(n), with equality iff n is squarefree (i.e., is in A005117).
a(n) = 2 iff n is prime. (End)

Examples

			a(9) = gcd(1,9) + gcd(3,3) + gcd(9,1) = 5, since 1, 3, 9 are the positive divisors of 9.
		

Crossrefs

Programs

  • Maple
    N:= 1000: # to get a(1) to a(N)
    V:= Vector(N):
    for k from 1 to N do
       for j from 1 to floor(N/k) do
         V[k*j]:= V[k*j]+igcd(k,j)
       od
    od:
    convert(V,list); # Robert Israel, Dec 26 2015
  • Mathematica
    Table[DivisorSum[n, GCD[#, n/#] &], {n, 94}] (* Michael De Vlieger, Sep 23 2017 *)
    f[p_, e_] := If[EvenQ[e], (p^(e/2)*(p+1)-2)/(p-1), 2*(p^((e+1)/2)-1)/(p-1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 30 2020 *)
  • PARI
    a(n) = sumdiv(n, d, gcd(d, n/d)); \\ Michel Marcus, Aug 03 2016
    
  • Python
    from sympy import divisors, gcd
    def A055155(n): return sum(gcd(d,n//d) for d in divisors(n,generator=True)) # Chai Wah Wu, Aug 19 2021

Formula

Multiplicative with a(p^e) = (p^(e/2)*(p+1)-2)/(p-1) for even e and a(p^e) = 2*(p^((e+1)/2)-1)/(p-1) for odd e. - Vladeta Jovovic, Nov 01 2001
Dirichlet g.f.: (zeta(s))^2*zeta(2s-1)/zeta(2s); inverse Mobius transform of A000188. - R. J. Mathar, Feb 16 2011
Dirichlet convolution of A069290 and A008966. - R. J. Mathar, Oct 31 2011
Sum_{k=1..n} a(k) ~ 3*n / (2*Pi^6) * (Pi^4 * log(n)^2 + ((8*g - 2)*Pi^4 - 24 * Pi^2 * z1) * log(n) + 2*Pi^4 * (1 - 4*g + 5*g^2 - 6*sg1) + 288 * z1^2 - 24 * Pi^2 * (-z1 + 4*g*z1 + z2)), where g is the Euler-Mascheroni constant A001620, sg1 is the first Stieltjes constant A082633, z1 = Zeta'(2) = A073002, z2 = Zeta''(2) = A201994. - Vaclav Kotesovec, Feb 01 2019
a(n) = (1/n)*Sum_{i=1..n} sigma(gcd(n,i^2)). - Ridouane Oudra, Dec 30 2020
a(n) = Sum_{k=1..n} gcd(gcd(n,k),n/gcd(n,k))/phi(n/gcd(n,k)), where phi = A000010. - Richard L. Ollerton, May 09 2021

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

A097246 Replace factors of n that are squares of a prime with the prime succeeding this prime.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 6, 5, 10, 11, 9, 13, 14, 15, 9, 17, 10, 19, 15, 21, 22, 23, 18, 7, 26, 15, 21, 29, 30, 31, 18, 33, 34, 35, 15, 37, 38, 39, 30, 41, 42, 43, 33, 25, 46, 47, 27, 11, 14, 51, 39, 53, 30, 55, 42, 57, 58, 59, 45, 61, 62, 35, 27, 65, 66, 67, 51, 69, 70, 71, 30, 73
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 03 2004

Keywords

Crossrefs

Cf. A097247, A097248 (fixed points of iteration), A097249 (number of iterations needed to reach them for each n), A277886, A277899.

Programs

Formula

Multiplicative with p^e -> NextPrime(p)^floor(e/2) * p^(e mod 2), where p prime and NextPrime(p)=A000040(A049084(p)+1).
a(n) <= n; a(n) = n iff n is squarefree: a(A005117(n)) = A005117(n);
a(m*n) <= a(m)*a(n); a(m*n) = a(m)*a(n) iff m and n are coprime;
a(A000040(k)^n) = A000040(k+1)^floor(n/2)*A000040(k)^(n mod 2); a(2^n) = 3^floor(n/2) * (1 + n mod 2);
a(A000040(k)*A002110(n)/A002110(k-1)) = A000040(k+1)*A002110(n)/A002110(k) for k <= n, see also A097250.
From Antti Karttunen, Nov 15 2016: (Start)
a(1) = 1; for n > 1, a(n) = 2^A000035(A007814(n)) * 3^A004526(A007814(n)) * A003961(a(A064989(n))).
a(n) = A003961(A000188(n)) * A007913(n).
A048675(a(n)) = A048675(n).
(End)
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (p^4-p^2)/(p^4-nextprime(p)) = 0.4059779303..., where nextprime is A151800. - Amiram Eldar, Nov 29 2022
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