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A348172
a(n) is the number of positive k (can be greater than n) such that A000005(n)/n = A000005(k)/k.
3
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, 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, 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
OFFSET
1,1
COMMENTS
The first 1 occurs at a(4). The corresponding k value is 4.
The first 2 occurs at a(1). The corresponding k values are 1, 2.
The first 3 occurs at a(9). The corresponding k values are 9, 18, 24.
The first 4 occurs at a(243). The corresponding k values are 243, 405, 486, 810.
Does every number appear in this sequence?
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
LINKS
Charles R Greathouse IV, PARI/GP script
EXAMPLE
For n = 9, the k such that A000005(9)/9 = 1/3 = A000005(k)/k are 9, 18, and 24. Therefore, a(9) = 3.
MATHEMATICA
Array[Function[r, Count[Range[Ceiling[4/r^2]], _?(DivisorSigma[0, #]/# == r &)]][DivisorSigma[0, #]/#] &, 105] (* or *)
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 *)
PROG
(PARI) a(n) = {my(q=numdiv(n)/n); sum(i=1, 4/q^2, numdiv(i)/i == q); } \\ Michel Marcus, Oct 04 2021
(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
(PARI) See Greathouse link
CROSSREFS
KEYWORD
nonn
AUTHOR
Tejo Vrush, Oct 04 2021
STATUS
approved