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%I #110 Jan 23 2022 12:10:37
%S 2,2,2,1,2,2,2,2,3,2,2,2,2,2,2,1,2,3,2,1,2,2,2,3,2,2,2,1,2,2,2,1,2,2,
%T 2,1,2,2,2,2,2,2,2,1,3,2,2,1,2,2,2,1,2,2,2,2,2,2,2,2,2,2,3,1,2,2,2,1,
%U 2,2,2,1,2,2,2,1,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,1,2,2,2,2,2,2,3,1
%N a(n) is the number of positive k (can be greater than n) such that A000005(n)/n = A000005(k)/k.
%C The first 1 occurs at a(4). The corresponding k value is 4.
%C The first 2 occurs at a(1). The corresponding k values are 1, 2.
%C The first 3 occurs at a(9). The corresponding k values are 9, 18, 24.
%C The first 4 occurs at a(243). The corresponding k values are 243, 405, 486, 810.
%C Does every number appear in this sequence?
%C The first 5 is at a(3189375) with k-values {3189375, 5467500, 6378750, 6561000, 8748000} while the first 6 is at 3176523 with k-values {3176523, 4084101, 6353046, 6806835, 8168202, 13613670}. If 7 appears it does so past 10^8. - _Charles R Greathouse IV_, Oct 05 2021
%H David A. Corneth, <a href="/A348172/b348172.txt">Table of n, a(n) for n = 1..10000</a>
%H Charles R Greathouse IV, <a href="/A348172/a348172_1.gp.txt">PARI/GP script</a>
%e For n = 9, the k such that A000005(9)/9 = 1/3 = A000005(k)/k are 9, 18, and 24. Therefore, a(9) = 3.
%t Array[Function[r, Count[Range[Ceiling[4/r^2]], _?(DivisorSigma[0, #]/# == r &)]][DivisorSigma[0, #]/#] &, 105] (* or *)
%t Block[{nn = 7, m, s}, m = 2^(2 nn); s = KeySort@ PositionIndex[Array[DivisorSigma[0, #]/# &, m]]; s = Reverse@ KeyDrop[s, TakeWhile[Keys@ s, 4/#^2 > m &]]; Length /@ Array[Lookup[s, DivisorSigma[0, #]/#] &, 2^nn]] (* _Michael De Vlieger_, Oct 04 2021 *)
%o (PARI) a(n) = {my(q=numdiv(n)/n); sum(i=1, 4/q^2, numdiv(i)/i == q);} \\ _Michel Marcus_, Oct 04 2021
%o (PARI) a(n) = my(q=numdiv(n)/n, s=denominator(q), res = 0); forstep(i=s, 4/q^2, s, if(numdiv(i) == q * i, res++)); res \\ _David A. Corneth_, Oct 07 2021
%o (PARI) See Greathouse link
%Y Cf. A000005, A090387, A090395, A005382, A005383, A348184.
%K nonn
%O 1,1
%A _Tejo Vrush_, Oct 04 2021