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%I #29 Apr 17 2024 08:34:22
%S 6,18,42,54,66,78,102,114,126,138,162,174,186,198,222,234,246,258,282,
%T 294,306,318,342,354,366,378,402,414,426,438,462,474,486,498,522,534,
%U 546,558,582,594,606,618,642,654,666,678,702,714,726,738,762,774,786
%N Numbers k such that the penultimate 3 divisors of k sum to k.
%C It seems that only numbers that are 6 mod 12 are present except for multiples of 30.
%C From _David A. Corneth_, Jun 18 2021: (Start)
%C 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).
%C 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)
%H Michael S. Branicky, <a href="/A074837/b074837.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Harvey P. Dale)
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).
%F a(n) = a(n-4) + 60. - _David A. Corneth_, Jun 18 2021
%F From _Chai Wah Wu_, Apr 16 2024: (Start)
%F a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
%F G.f.: x*(6*x^4 + 12*x^3 + 24*x^2 + 12*x + 6)/(x^5 - x^4 - x + 1). (End)
%e 18 has the divisors 1,2,3,6,9,18. The penultimate 3 are 3,6,9, which sum to 18.
%t Select[Range[1000],Length[Divisors[ # ]]>3 && Sum[Divisors[ # ][[ -i]],{i,2,4}]==# &] (* _Stefan Steinerberger_, Aug 01 2007 *)
%t 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 *)
%o (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, ", "))))
%o (PARI) is(n) = n%12 == 6 && n % 5 != 0 \\ _David A. Corneth_, Jun 18 2021
%o (Python)
%o from sympy import divisors
%o def ok(n): d = divisors(n); return False if len(d)<4 else n==sum(d[-4:-1])
%o print(list(filter(ok, range(800)))) # _Michael S. Branicky_, Jun 18 2021
%K nonn,easy
%O 1,1
%A _Jon Perry_, Sep 09 2002
%E More terms from _Stefan Steinerberger_, Aug 01 2007