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%I #32 Apr 23 2024 14:32:03
%S 1,2,8,12,20,24,48,60,68,72,92,96,120,140,144,188,192,200,212,216,240,
%T 288,300,332,336,360,384,428,432,440,452,480,500,548,576,600,648,660,
%U 668,672,680,692,696,720,768,780,788,812,864,908,932,960,1008,1028,1052
%N Numbers k such that the average of the first k Lucas numbers is an integer.
%C A111035 is the equivalent for Fibonacci numbers and has many elements in common with this sequence. _T. D. Noe_, who extended this sequence, noticed that, for some reason, 24 divides many of those k.
%C All terms are even except for the first term. - _Harvey P. Dale_, Apr 22 2024
%H Alois P. Heinz, <a href="/A111058/b111058.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Harvey P. Dale)
%F k such that (Sum_{i=1..k} A000204(i))/k is an integer.
%F { k : A027961(k) == 0 (mod k) }. - _Alois P. Heinz_, Apr 23 2024
%t Lucas[n_] := Fibonacci[n+1]+Fibonacci[n-1]; lst={}; s=0; Do[s=s+Lucas[n]; If[Mod[s, n]==0, AppendTo[lst, n]], {n, 1000}]; lst (* _T. D. Noe_ *)
%t Module[{nn=1000,ln},ln=LucasL[Range[nn]];Table[If[IntegerQ[Mean[Take[ln,n]]],n,Nothing],{n,nn}]] (* _Harvey P. Dale_, Apr 22 2024 *)
%Y Cf. A000032, A000204, A027961, A111035, A160665.
%K easy,nonn,changed
%O 1,2
%A _Jonathan Vos Post_, Oct 07 2005
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