OFFSET
1,1
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)
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).
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)
EXAMPLE
18 has the divisors 1,2,3,6,9,18. The penultimate 3 are 3,6,9, which sum to 18.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jon Perry, Sep 09 2002
EXTENSIONS
More terms from Stefan Steinerberger, Aug 01 2007
STATUS
approved