OFFSET
1,5
COMMENTS
For any positive integer q, clearly q + 1 - pi(q+1) - (q - pi(q)) = 1 + pi(q) - pi(q+1) is 0 or 1. Thus {q - pi(q): q = 1, ..., n} = {1, 2, ..., n-pi(n)}. Note that n - pi(n) > n/2 for n > 8. If p is at least sqrt(n/2), then n - p^2 <= n/2 and hence n - p^2 = q - pi(q) for some q = 1, ..., n. So it can be proved that a(n) > 0 for all n > 4. - Li-Lu Zhao and Zhi-Wei Sun, Feb 26 2014
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(9) = 1 since 9 = 2^2 + 9 - pi(9) with 2 prime and pi(9) = 4.
a(40) = 1 since 40 = 5^2 + 24 - pi(24) with 5 prime and pi(24) = 9.
a(120) = 1 since 120 = 7^2 + 95 - pi(95) with 7 prime and pi(95) = 24.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]]&&PrimeQ[Sqrt[n]]
a[n_]:=Sum[If[SQ[n-q+PrimePi[q]], 1, 0], {q, 1, n}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 26 2014
STATUS
approved