A277521 Numbers k such that number of divisors of k and sum of divisors of k divides product of divisors of k and the average of the divisors of k is an integer.
1, 6, 30, 66, 102, 210, 270, 318, 330, 420, 462, 510, 546, 570, 642, 672, 690, 714, 840, 870, 924, 930, 966, 1122, 1320, 1410, 1428, 1518, 1590, 1638, 1722, 1770, 1890, 1932, 2130, 2226, 2280, 2310, 2346, 2370, 2670, 2730, 2760, 2838, 2970, 2982, 3102, 3162, 3210, 3360, 3444, 3486, 3498, 3570, 3720, 3780, 3948, 3990
Offset: 1
Keywords
Examples
a(2) = 6 because 6 has 4 divisors {1,2,3,6}, 1*2*3*6/4 = 9, 1*2*3*6/(1 + 2 + 3 + 6) = 3 and (1 + 2 + 3 + 6)/4 = 3 are integer.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Vaclav Kotesovec, Plot of a(n)/n^(3/2) for n = 1..20000
- Eric Weisstein's World of Mathematics, Divisor Product.
- Eric Weisstein's World of Mathematics, Divisor Function.
- Wikipedia, Arithmetic number.
Crossrefs
Programs
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Maple
with(numtheory): P:=proc(q) local a,b,k,n;for n from 1 to q do a:=divisors(n); b:=mul(a[k],k=1..nops(a)); if type(sigma(n)/tau(n),integer) and type(b/sigma(n),integer) and type(b/tau(n),integer) then print(n); fi; od; end: P(10^5); # Paolo P. Lava, Oct 20 2016
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Mathematica
Select[Range[4000], Divisible[Sqrt[#1]^DivisorSigma[0, #1], DivisorSigma[1, #1]] && Divisible[Sqrt[#1]^DivisorSigma[0, #1], DivisorSigma[0, #1]] && Divisible[DivisorSigma[1, #1], DivisorSigma[0, #1]] & ]
Comments