This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A238402 Number of ways to write n = p^2 + q - pi(q) with p prime and q among 1, ..., n, where pi(.) is given by A000720. 2
 0, 0, 0, 0, 3, 2, 2, 1, 1, 4, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 4, 3, 3, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 3, 3, 2, 2, 2, 3, 4, 3, 2, 2, 2, 3, 3, 2, 3, 4, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 2, 3, 3, 3, 2, 2, 2, 2, 4, 2, 2, 3 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A000040, A000290, A000720, A232443, A232463, A238386. Sequence in context: A054546 A065310 A226352 * A016558 A154395 A214609 Adjacent sequences:  A238399 A238400 A238401 * A238403 A238404 A238405 KEYWORD nonn AUTHOR Zhi-Wei Sun, Feb 26 2014 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified September 20 10:10 EDT 2019. Contains 327229 sequences. (Running on oeis4.)