OFFSET
1,2
COMMENTS
Conjecture: For every n there exists a k different from n (possibly k > n) such that n*pi(k) = k*pi(n).
From David A. Corneth, Nov 15 2019: (Start)
If n*pi(k) = k*pi(n) then n/pi(n) = k/pi(k). So to find terms, one can make a list of pairs (k/pi(k), k) and sort them.
Then if for two such pairs (m/pi(m), m) and (k/pi(k), k), m > k have the same first element, i.e., m/pi(m) = k/pi(k) then a(m) = k for the least k with that ratio.
Amarnath Murthy's conjecture above is false. For n = 3 we have pi(n)/n = 2/3. For no other k we have pi(k)/k = 2/3. Therefore the conjecture is false. (End)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10007
EXAMPLE
a(15) = 10 as 15*pi(10) = 15*4 = 60 = 10*pi(15) = 10*6.
For k in {2, 4, 6, 8} we have pi(k)/k = 1/2 and for no k < 2 this holds. So for all these values a(k) = 2. - David A. Corneth, Nov 15 2019
MATHEMATICA
Table[SelectFirst[Range[n], n PrimePi[#] == # PrimePi[n] &], {n, 72}] (* Michael De Vlieger, Dec 14 2019 *)
PROG
(PARI) {m=72; pi=vector(m, n, omega(n!)); for(n=1, m, k=1; while(n*pi[k]!=k*pi[n], k++); print1(k, ", "))}
(PARI) first(n) = {n = nextprime(n); my(v = vector(n), t = -1, q = 1, res = vector(n), m); v[1] = [0, 1]; v[2] = [1/2, 2]; forprime(p = 2, n, t++; for(c = q, p - 1, v[c] = [t/c, c]; ); q = p ); v[n] = [t/n, n]; v = vecsort(v); res[1] = 1; for(i = 2, #v, if(v[i-1][1] != v[i][1], m = v[i][2]; ); res[v[i][2]] = m ); res } \\ David A. Corneth, Nov 15 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amarnath Murthy, May 30 2004
EXTENSIONS
Edited and extended by Klaus Brockhaus, Jun 01 2004
STATUS
approved