A237721 Number of primes p <= n with floor( sqrt(n-p) ) a square.

0, 1, 2, 2, 3, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 4, 4, 4, 4, 5, 5, 5, 5, 4, 3, 5, 4, 5, 4, 4, 4, 4, 3, 4, 3, 4, 4, 4, 3, 4, 3, 3, 3, 4, 3, 3, 3, 2, 2, 4, 3, 3, 2, 2, 2, 4, 4, 5, 4, 4, 4, 3, 2, 3, 2, 3, 3

Offset

1,3

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 9, 10, 11, 12, 15, 16, 17.

We have verified this for n up to 10^6.

See also A237705, A237706 and A237720 for similar conjectures.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(2) = 1 since 2 is prime with floor(sqrt(2-2)) = 0^2.
a(3) = 2 since 2 is prime with floor(sqrt(3-2)) = 1^2, and 3 is prime with floor(sqrt(3-3)) = 0^2.
a(9) = a(10) = 1 since 7 is prime with floor(sqrt(9-7)) = floor(sqrt(10-7)) = 1^2.
a(11) = 1 since 11 is prime with floor(sqrt(11-11)) = 0^2.
a(12) = 1 since 11 is prime with floor(sqrt(12-11)) = 1^2.
a(15) = a(16) = 1 since 13 is prime with floor(sqrt(15-13)) = floor(sqrt(16-13)) = 1^2.
a(17) = 1 since 17 is prime with floor(sqrt(17-17)) = 0^2.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
q[n_]:=SQ[Floor[Sqrt[n]]]
a[n_]:=Sum[If[q[n-Prime[k]], 1, 0], {k, 1, PrimePi[n]}]
Table[a[n], {n, 1, 70}]

Crossrefs

Cf. A000040, A000290, A237705, A237706, A237710, A237720.

Sequence in context: A259977 A341434 A115312 * A254296 A248371 A237768

Adjacent sequences: A237718 A237719 A237720 * A237722 A237723 A237724

Keyword

nonn

Author

Zhi-Wei Sun, Feb 12 2014

Status

approved