1,1

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)

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).

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)

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

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

Sequence in context: A191829 A023620 A386276 * A286308 A015942 A009945

Adjacent sequences: A074834 A074835 A074836 * A074838 A074839 A074840

nonn,easy

Jon Perry, Sep 09 2002

More terms from Stefan Steinerberger, Aug 01 2007

approved