A048250 Sum of the squarefree divisors of n.

1, 3, 4, 3, 6, 12, 8, 3, 4, 18, 12, 12, 14, 24, 24, 3, 18, 12, 20, 18, 32, 36, 24, 12, 6, 42, 4, 24, 30, 72, 32, 3, 48, 54, 48, 12, 38, 60, 56, 18, 42, 96, 44, 36, 24, 72, 48, 12, 8, 18, 72, 42, 54, 12, 72, 24, 80, 90, 60, 72, 62, 96, 32, 3, 84, 144, 68, 54, 96, 144, 72, 12, 74

Offset

1,2

Comments

Also sum of divisors of the squarefree kernel of n: a(n) = A000203(A007947(n)). - Reinhard Zumkeller, Jul 19 2002

The absolute values of the Dirichlet inverse of A001615. - R. J. Mathar, Dec 22 2010

Row sums of the triangle in A206778. - Reinhard Zumkeller, Feb 12 2012

Inverse Möbius transform of n * mu(n)^2 = |A055615(n)|. - Wesley Ivan Hurt, Jun 08 2023

References

D. Suryanarayana, On the core of an integer, Indian J. Math. 14 (1972) 65-74.

Links

T. D. Noe, Table of n, a(n) for n = 1..10000

Steven R. Finch, Unitarism and infinitarism.

Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]

Index entries for sequences related to sums of squares

Formula

If n = Product p_i^e_i, a(n) = Product (p_i + 1). - Vladeta Jovovic, Apr 19 2001

Dirichlet g.f.: zeta(s)*zeta(s-1)/zeta(2*s-2). - Michael Somos, Sep 08 2002

a(n) = Sum_{d|n} mu(d)^2*d = Sum_{d|n} |A055615(d)|. - Benoit Cloitre, Dec 09 2002

Pieter Moree (moree(AT)mpim-bonn.mpg.de), Feb 20 2004 can show that Sum_{n <= x} a(n) = x^2/2 + O(x*sqrt{x}) and adds: "As S. R. Finch pointed out to me, in Suryanarayana's paper this is proved under the Riemann hypothesis with error term O(x^{7/5+epsilon})".

a(n) = psi(rad(n)) = A001615(A007947(n)). - Enrique Pérez Herrero, Aug 24 2010

a(n) = rad(n)*psi(n)/n = A001615(n)*A007947(n)/n. - Enrique Pérez Herrero, Aug 31 2010

G.f.: Sum_{k>=1} mu(k)^2*k*x^k/(1 - x^k). - Ilya Gutkovskiy, Jan 03 2017

Lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k = 1. - Amiram Eldar, Jun 10 2020

a(n) = Sum_{d divides n} mu(d)^2*core(d), where core(n) = A007913(n). - Peter Bala, Jan 24 2024

Example

For n=1000, out of the 16 divisors, four are squarefree: {1,2,5,10}. Their sum is 18. Or, 1000 = 2^3*5^3 hence a(1000) = (2+1)*(5+1) = 18.

Maple

A048250 := proc(n) local ans, i:ans := 1: for i from 1 to nops(ifactors(n)[ 2 ]) do ans := ans*(1+ifactors(n)[ 2 ][ i ] [ 1 ]): od: RETURN(ans) end:
# alternative:
seq(mul(1+p, p = numtheory:-factorset(n)), n=1..1000); # Robert Israel, Mar 18 2015

Mathematica

sumOfSquareFreeDivisors[ n_ ] := Plus @@ Select[ Divisors[ n ], MoebiusMu[ # ] != 0 & ]; Table[ sumOfSquareFreeDivisors[ i ], {i, 85} ]
Table[Total[Select[Divisors[n], SquareFreeQ]], {n, 80}] (* Harvey P. Dale, Jan 25 2013 *)
a[1] = 1; a[n_] := Times@@(1 + FactorInteger[n][[;; , 1]]); Array[a, 100] (* Amiram Eldar, Dec 19 2018 *)

Prog

(PARI) a(n)=if(n<1, 0, sumdiv(n, d, if(core(d)==d, d)))
(PARI) a(n)=if(n<1, 0, direuler(p=2, n, (1+p*X)/(1-X))[n])
(PARI) a(n)=sumdiv(n, d, moebius(d)^2*d); \\ Joerg Arndt, Jul 06 2011
(PARI) a(n)=my(f=factor(n)); for(i=1, #f~, f[i, 2]=1); sigma(f) \\ Charles R Greathouse IV, Sep 09 2014
(Haskell)
a034448 = sum . a206778_row  -- Reinhard Zumkeller, Feb 12 2012
(Sage)
def A048250(n): return mul(map(lambda p: p+1, prime_divisors(n)))
[A048250(n) for n in (1..73)]  # Peter Luschny, May 23 2013
(Python)
from math import prod
from sympy import primefactors
def A048250(n): return prod(p+1 for p in primefactors(n)) # Chai Wah Wu, Apr 20 2023

Crossrefs

Cf. A003557, A007913, A007947, A023900, A034448, A206787, A309192.

Sum of the k-th powers of the squarefree divisors of n for k=0..10: A034444 (k=0), this sequence (k=1), A351265 (k=2), A351266 (k=3), A351267 (k=4), A351268 (k=5), A351269 (k=6), A351270 (k=7), A351271 (k=8), A351272 (k=9), A351273 (k=10).

Cf. A240976 (tenth of Dgf at s=3).

Sequence in context: A324335 A238162 A367503 * A323363 A073181 A183100

Adjacent sequences: A048247 A048248 A048249 * A048251 A048252 A048253

Keyword

nonn,easy,nice,mult

Author

Labos Elemer

Status

approved