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A094757
Least positive k <= n such that n*pi(k) = k*pi(n), where pi(n) is the number of primes <= n (A000720).
4
1, 2, 3, 2, 5, 2, 7, 2, 9, 10, 11, 12, 13, 14, 10, 16, 17, 18, 19, 10, 21, 22, 23, 16, 25, 26, 27, 28, 29, 27, 31, 32, 27, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 40, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 56, 64, 65, 66, 67, 68, 69, 70, 71, 72
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
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
Cf. A095299 for n such that a(n) < n.
Sequence in context: A007387 A105222 A280503 * A095171 A343004 A270443
KEYWORD
nonn,easy
AUTHOR
Amarnath Murthy, May 30 2004
EXTENSIONS
Edited and extended by Klaus Brockhaus, Jun 01 2004
STATUS
approved