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A304096 Number of Lucas numbers larger than 3 (4, 7, 11, 18, ...) that divide n.

%I A304096 #15 Dec 31 2023 03:40:47
%S A304096 0,0,0,1,0,0,1,1,0,0,1,1,0,1,0,1,0,1,0,1,1,1,0,1,0,0,0,2,1,0,0,1,1,0,
%T A304096 1,2,0,0,0,1,0,1,0,2,0,0,1,1,1,0,0,1,0,1,1,2,0,1,0,1,0,0,1,1,0,1,0,1,
%U A304096 0,1,0,2,0,0,0,2,2,0,0,1,0,0,0,2,0,0,1,2,0,1,1,1,0,1,0,1,0,1,1,1,0,0,0,1,1,0,0,2,0,1,0,2,0,0,0,2,0,0,1,1
%N A304096 Number of Lucas numbers larger than 3 (4, 7, 11, 18, ...) that divide n.
%C A304096 a(n) is the number of the divisors d of n that are of the form d = A000045(k-1) + A000045(k+1), for k >= 3.
%H A304096 Antti Karttunen, <a href="/A304096/b304096.txt">Table of n, a(n) for n = 1..65537</a>
%F A304096 a(n) = Sum_{d|n, d>3} A102460(d).
%F A304096 a(n) = A304094(n) - A079978(n) - 1.
%F A304096 a(n) = A304092(n) - A059841(n) - A079978(n) - 1.
%F A304096 a(n) = A007949(A304104(n)).
%F A304096 Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A093540 - 4/3 = 0.629524... . - _Amiram Eldar_, Dec 31 2023
%e A304096 The divisors of 4 are 1, 2 and 4. Of these only 4 is a Lucas number larger than 3, thus a(4) = 1.
%e A304096 The divisors of 28 are 1, 2, 4, 7, 14 and 28. Of these 4 and 7 are Lucas numbers (A000032) larger than 3, thus a(28) = 2.
%o A304096 (PARI)
%o A304096 A102460(n) = { my(u1=1,u2=3,old_u1); if(n<=2,sign(n),while(n>u2,old_u1=u1;u1=u2;u2=old_u1+u2);(u2==n)); };
%o A304096 A304096(n) = sumdiv(n,d,(d>3)*A102460(d));
%Y A304096 Cf. A000032, A000045, A000204, A093540, A102460, A304092, A304094, A304095, A304104.
%Y A304096 Cf. also A005086, A300837.
%K A304096 nonn
%O A304096 1,28
%A A304096 _Antti Karttunen_, May 13 2018