A063740 Number of integers k such that cototient(k) = n.
1, 1, 2, 1, 1, 2, 3, 2, 0, 2, 3, 2, 1, 2, 3, 3, 1, 3, 1, 3, 1, 4, 4, 3, 0, 4, 1, 4, 3, 3, 4, 3, 0, 5, 2, 2, 1, 4, 1, 5, 1, 4, 2, 4, 2, 6, 5, 5, 0, 3, 0, 6, 2, 4, 2, 5, 0, 7, 4, 3, 1, 8, 4, 6, 1, 3, 1, 5, 2, 7, 3, 5, 1, 7, 1, 8, 1, 5, 2, 6, 1, 9, 2, 6, 0, 4, 2, 10, 2, 4, 2, 5, 2, 7, 5, 4, 1, 8, 0, 9, 1, 6, 1, 7
Offset: 2
Keywords
Examples
Cototient(x) = 101 for x in {485, 1157, 1577, 1817, 2117, 2201, 2501, 2537, 10201}, with a(101) = 8 terms; e.g. 485 - phi(485) = 485 - 384 = 101. Cototient(x) = 102 only for x = 202 so a(102) = 1.
Links
- T. D. Noe, Table of n, a(n) for n = 2..10000
Crossrefs
Programs
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Mathematica
Table[Count[Range[n^2], k_ /; k - EulerPhi@ k == n], {n, 2, 105}] (* Michael De Vlieger, Mar 17 2017 *)
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PARI
first(n)=my(v=vector(n),t); forcomposite(k=4,n^2, t=k-eulerphi(k); if(t<=n, v[t]++)); v[2..n] \\ Charles R Greathouse IV, Mar 17 2017
Formula
From Amiram Eldar, Apr 08 2023 (Start)
a(A005278(n)) = 0.
a(A131825(n)) = 1.
a(A063741(n)) = n. (End)
Extensions
Name edited by Charles R Greathouse IV, Mar 17 2017
Comments