A074837 Numbers k such that the penultimate 3 divisors of k sum to k.

6, 18, 42, 54, 66, 78, 102, 114, 126, 138, 162, 174, 186, 198, 222, 234, 246, 258, 282, 294, 306, 318, 342, 354, 366, 378, 402, 414, 426, 438, 462, 474, 486, 498, 522, 534, 546, 558, 582, 594, 606, 618, 642, 654, 666, 678, 702, 714, 726, 738, 762, 774, 786

Offset: 1

Author

Jon Perry, Sep 09 2002

Keywords

• nonn
• easy

Comments

It seems that only numbers that are 6 mod 12 are present except for multiples of 30.

From David A. Corneth, Jun 18 2021: (Start)

The above is true. Proof: Suppose a(n) = k. Then the penultimate three divisors of k are k/d1, k/d2 and k/d3 for some divisors d1, d2 and d3 where 1 < d1 < d2 < d3 of k. We have k = k/d1 + k/d2 + k/d3 = k * (1/d1 + 1/d2 + 1/d3) i.e. 1 = 1/d1 + 1/d2 + 1/d3. The only triplet satisfying this (including the inequalities) is (d1, d2, d3) = (2, 3, 6).

So k cannot be divisible by 4 and not by 5 but must be divisible by lcm(2, 3, 6) = 6. The only numbers satisfying this are numbers of the form 6 mod 12 that are not multiples of 5. (End)

Examples

			18 has the divisors 1,2,3,6,9,18. The penultimate 3 are 3,6,9, which sum to 18.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Programs

Mathematica

Select[Range[1000],Length[Divisors[ # ]]>3 && Sum[Divisors[ # ][[ -i]],{i,2,4}]==# &] (* Stefan Steinerberger, Aug 01 2007 *)
p3dQ[n_]:=Module[{d=Divisors[n]},Length[d]>3&&Total[Take[Most[d],-3]] == n]; Select[Range[800],p3dQ] (* Harvey P. Dale, Dec 06 2012 *)

PARI

for (n=1,800,dn=divisors(n); dnl=length(dn); if (dnl>3,if (n==dn[dnl-1]+dn[dnl-2]+dn[dnl-3],print1(n, ", "))))

PARI

is(n) = n%12 == 6 && n % 5 != 0 \\ David A. Corneth, Jun 18 2021

Python

from sympy import divisors
def ok(n): d = divisors(n); return False if len(d)<4 else n==sum(d[-4:-1])
print(list(filter(ok, range(800)))) # Michael S. Branicky, Jun 18 2021

Formula

a(n) = a(n-4) + 60. - David A. Corneth, Jun 18 2021

From Chai Wah Wu, Apr 16 2024: (Start)

a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.

G.f.: x*(6*x^4 + 12*x^3 + 24*x^2 + 12*x + 6)/(x^5 - x^4 - x + 1). (End)

Extensions

More terms from Stefan Steinerberger, Aug 01 2007