A253905 -id:A253905 - cp's OEIS frontend

A366762 Numbers whose canonical prime factorization contains only exponents which are congruent to 1 modulo 3.

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 48, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 102

Offset: 1

Amiram Eldar, Oct 21 2023

First differs from A274034 at n = 42, and from A197680 and A361177 at n = 84.

The asymptotic density of this sequence is zeta(3) * Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = A002117 * A330523 = A253905 * A065465 = 0.644177671086029533405... .

Amiram Eldar, Table of n, a(n) for n = 1..10000

Similar sequences with exponents of a given form: A000290 (2*k), A268335 (2*k+1), A000578 (3*k), A182120 (3*k+2).
Cf. A002117, A065465, A253905, A330523.
Cf. A197680, A274034, A361177, A366761.

q[n_] := AllTrue[FactorInteger[n][[;; , 2]], Mod[#, 3] == 1 &]; Select[Range[120], q]

is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 2]%3 != 1, return(0))); 1;}

Sum_{n>=1} 1/a(n)^s = zeta(3*s) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s)), for s > 1.

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

Peter Kagey, Feb 06 2015

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

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

Peter Kagey, Table of n, a(n) for n = 1..5000

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

Cf. A253905.

a254732 n = head [k | k <- [n + 1 ..], mod (k ^ 2) n == 0]
-- Reinhard Zumkeller, Feb 07 2015

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 *)

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

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

def A254732(n):
    k = n + 1
    while pow(k,2,n):
        k += 1
    return k # Chai Wah Wu, Feb 15 2015

def a(n)
  (n+1..2*n).find { |k| k**2 % n == 0 }
end

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

A038388 Let f(n) be the smallest number such that the arithmetic mean (A) and geometric mean (G) of n and f(n) are both integers; sequence gives G values.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 4, 3, 10, 11, 12, 13, 14, 15, 8, 17, 6, 19, 20, 21, 22, 23, 12, 5, 26, 9, 28, 29, 30, 31, 8, 33, 34, 35, 12, 37, 38, 39, 20, 41, 42, 43, 44, 15, 46, 47, 24, 7, 10, 51, 52, 53, 18, 55, 28, 57, 58, 59, 60, 61, 62, 21, 16, 65, 66, 67, 68, 69, 70, 71, 12, 73, 74, 15

Offset: 1

N. J. A. Sloane

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A038387, A038389.

Cf. A053626, A053627, A253905.

Table[k = 1; While[Nand @@ IntegerQ /@ {(n + k)/2, g = Sqrt[n*k]}, k++]; g, {n, 75}] (* Jayanta Basu, Jul 14 2013 *)
f[p_, e_] := If[OddQ[e], p^((e + 1)/2), If[p == 2, 2^(e/2 + 1), p^(e/2)]]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 09 2020*)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%2, f[i,1]^((f[i,2]+1)/2), if(f[i,1]==2, 2^(f[i,2]/2+1), f[i,1]^(f[i,2]/2))));} \\ Amiram Eldar, Oct 27 2022

Multiplicative with a(p^e) = p^((e+1)/2) if e is odd, 2^(e/2+1) if p=2 and e is even, p^(e/2) if p>2 and e is even. - Vladeta Jovovic, May 15 2003

Sum_{k=1..n} a(k) ~ c * n^2, where c = (13/24)*zeta(3)/zeta(2) = 0.395829... . - Amiram Eldar, Oct 27 2022

More terms from Vladeta Jovovic, May 15 2003

A367407 a(n) = sqrt(A367406(n)).

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 4, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 12, 26, 9, 29, 30, 31, 8, 33, 34, 35, 37, 38, 39, 20, 41, 42, 43, 46, 47, 51, 53, 18, 55, 28, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 44, 89, 91, 93, 94, 95, 24, 97

Offset: 1

Amiram Eldar, Nov 17 2023

A permutation of the positive integers.

Cf. A064549, A268335, A367406, A367417, A367419.

Cf. A065463, A065465, A253905.

s[n_] := Sqrt[n * Times @@ FactorInteger[n][[;;, 1]]]; s /@ Select[Range[100], AllTrue[FactorInteger[#][[;; , 2]], OddQ] &]

b(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%2, f[i,1]^(f[i,2]+1), 0));}
lista(kmax) = {my(b1); for(k = 1, kmax, b1 = b(k); if(b1 > 0, print1(sqrtint(b1), ", ")));}

a(n) = sqrt(A064549(A268335(n))).
a(n) = sqrt(A268335(n)*A367417(n)).
a(n) = A268335(n)/A367419(n).
Sum_{k=1..n} a(k) = c * n^2 / 2, where c = (zeta(3)/(zeta(2)*d^2)) * Product_{p prime} (1 - 1/(p^2*(p+1))) = A253905 * A065465 / d^3 = 1.29812028442810841122..., and d = A065463 is the asymptotic density of the exponentially odd numbers (A268335).

A335005 Decimal expansion of Pi^2/(12*zeta(3)).

Original entry on oeis.org

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

Offset: 0

Amiram Eldar, May 19 2020

			0.68421638881010293786838292699239597056540695732620...

Eckford Cohen, Arithmetical functions associated with the unitary divisors of an integer, Mathematische Zeitschrift, Vol. 74, No. 1 (1960), pp. 66-80.
R. Sitaramachandrarao and D. Suryanarayana, On Sigma_{n<=x} sigma*(n) and Sigma_{n<=x} phi*(n), Proceedings of the American Mathematical Society, Vol. 41, No. 1 (1973), pp. 61-66.

Cf. A002117(zeta(3)), A013661 (zeta(2)), A034448, A064609, A072691 (Pi^2/12), A253905 (zeta(3)/zeta(2)).

RealDigits[Pi^2/12/Zeta[3], 10, 100][[1]]

Pi^2/(12*zeta(3)) \\ Michel Marcus, May 19 2020

Equals lim_{k->oo} A064609(k)/k^2, where A064609(k) is the partial sums of A034448, the sum of unitary divisors from 1 to k.

Equals zeta(2)/(2*zeta(3)) = A013661/(2*A002117) = A072691/A002117 = 1/(2*A253905).

A060367 Average order of an element in a cyclic group of order n rounded down.

Original entry on oeis.org

1, 1, 2, 2, 4, 3, 6, 5, 6, 6, 10, 6, 12, 9, 9, 10, 16, 10, 18, 11, 14, 15, 22, 12, 20, 18, 20, 16, 28, 14, 30, 21, 23, 24, 25, 18, 36, 27, 28, 22, 40, 21, 42, 27, 28, 33, 46, 24, 42, 31, 37, 33, 52, 30, 42, 33, 42, 42, 58, 26, 60, 45, 41, 42, 50, 35, 66, 44, 51

Offset: 1

Avi Peretz (njk(AT)netvision.net.il), Apr 01 2001

Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Cf. A000010, A000203, A001157, A018804, A057660, A065764, A065827, A253905.

seq(floor(numtheory:-sigma[2](n^2)/numtheory:-sigma(n^2)/n), n=1..1000); # Robert Israel, Mar 24 2015

f[n_] := Block[{i, j, k}, Reap@ For[j = 1, j <= n, j++, Sow[Floor[Sum[1/GCD[j, k], {k, 1, j}]]]]] // Flatten // Rest; f@ 49 (* Michael De Vlieger, Mar 24 2015 *)
a[n_] := Floor[DivisorSigma[2, n^2]/DivisorSigma[1, n^2]/n]; Array[a, 100] (* Amiram Eldar, Jul 25 2025 *)

a(n) = {my(f = factor(n^2)); floor(sigma(f, 2)/(n * sigma(f)));} \\ Amiram Eldar, Jul 25 2025

[floor(sum([1/gcd(n,k) for k in range(1,n+1)])) for n in range(1,50)] # Danny Rorabaugh, Mar 24 2015

Sequence A057660 gives the sum of the orders of the elements in a cyclic group with n elements so a(n) = floor(A057660(n) / n) = floor(Sum_{k=1..n} 1/GCD(n, k)) = floor(Sum of 1/d times phi(n/d)) for all divisors d of n, where phi is Euler's phi function. This sum may also be expressed as the product of (p^(2*e(p)+1)+1)/((p+1)*p^e(p)) over all prime divisors p of n where the canonical factorization of n is the product of p^e(p), the e(p) being the exponents of the power of p in the factorization.
From Amiram Eldar, Jul 25 2025: (Start)
a(n) = floor(sigma_2(n^2)/(n*sigma(n))) = floor(A001157(n^2)/(n*A000203(n^2))) = floor(A065827(n)/(n*A065764(n))).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(3)/zeta(2) (A253905). (End)

Offset corrected and terms a(18)-a(50) added by Danny Rorabaugh, Mar 24 2015

A084302 Remainder of tau(n) modulo 6.

Original entry on oeis.org

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

Offset: 1

Labos Elemer, Jun 02 2003

The sums of the first 10^k terms, for k = 1, 2, ..., are 27, 236, 2275, 22166, 220070, 2195376, 21933228, 219259514, 2192385128, 21923168052, ... . Conjecture: the asymptotic mean of this sequence is 3*zeta(3)/zeta(2) = 3 * A253905 = 2.192288... . The conjecture is true if A211337 and A211338 have an equal asymptotic density (see also A059269). - Amiram Eldar, Jul 11 2024

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n.

Cf. A000005, A054763, A084299, A084300, A084301.

Cf. A059269, A211337, A211338

Mod[DivisorSigma[0,Range[110]],6] (* Harvey P. Dale, Sep 04 2020 *)

A084302(n) = (numdiv(n)%6); \\ Antti Karttunen, Jul 07 2017

a(n) = A000005(n) modulo 6.

cp's OEIS Frontend

A366762 Numbers whose canonical prime factorization contains only exponents which are congruent to 1 modulo 3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

PARI

Python

Ruby

Formula

A038388 Let f(n) be the smallest number such that the arithmetic mean (A) and geometric mean (G) of n and f(n) are both integers; sequence gives G values.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A367407 a(n) = sqrt(A367406(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A335005 Decimal expansion of Pi^2/(12*zeta(3)).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A060367 Average order of an element in a cyclic group of order n rounded down.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Sage

Formula

Extensions

A084302 Remainder of tau(n) modulo 6.

Views

Author

Keywords

Comments

Links