A255245 Numbers that divide the average of the squares of their aliquot parts.

10, 65, 140, 420, 2100, 2210, 20737, 32045, 200725, 207370, 1204350, 1347905, 1762645, 16502850, 31427800, 37741340, 107671200, 130643100, 200728169, 239719720, 357491225, 417225900, 430085380, 766750575, 1088692500, 1132409168, 1328204850, 1788379460

Offset

1,1

Comments

Ratio: 1, 1, 5, 10, 78, 1, 109, 565,...

If the ratio is equal to 1 we have 10, 65, 20737 (A140362).

Links

Giovanni Resta, Table of n, a(n) for n = 1..59 (terms < 10^11)

Example

Aliquot parts of 10 are 1, 2, 5. The average of their squares is (1^2 + 2^2 + 5^2) / 3 = (1 + 4 + 25) / 3 = 30 / 3 = 10 and 10 / 10 = 1.

Maple

with(numtheory); P:=proc(q) local a, b, k, n;
for n from 2 to q do a:=sort([op(divisors(n))]);
b:=add(a[k]^2, k=1..nops(a)-1)/(nops(a)-1);
if type(b/n, integer) then lprint(n);
fi; od; end: P(10^6);

Mathematica

Select[Range[10^6], Mod[Mean[Most[Divisors[#]^2]], #]==0&] (* Ivan N. Ianakiev, Mar 03 2015 *)

Prog

(PARI) isok(n) = (q=(sumdiv(n, d, (d!=n)*d^2)/(numdiv(n)-1))) && (type(q)=="t_INT") && ((q % n) == 0); \\ Michel Marcus, Feb 20 2015
(Python)
from __future__ import division
from sympy import factorint
A255245_list = []
for n in range(2, 10**9):
....s0 = s2 = 1
....for p, e in factorint(n).items():
........s0 *= e+1
........s2 *= (p**(2*(e+1))-1)//(p**2-1)
....q, r = divmod(s2-n**2, s0-1)
....if not (r or q % n):
........A255245_list.append(n) # Chai Wah Wu, Mar 08 2015

Crossrefs

Cf. A001065, A255244.

Sequence in context: A033863 A233246 A229996 * A210369 A058920 A263472

Adjacent sequences: A255242 A255243 A255244 * A255246 A255247 A255248

Keyword

nonn

Author

Paolo P. Lava, Feb 20 2015

Extensions

More terms from Michel Marcus, Feb 20 2015

a(17)-a(24) from Chai Wah Wu, Mar 08 2015

a(25)-a(28) from Giovanni Resta, May 30 2016

Status

approved