A238732 Number of primes p < n with floor((n-p)/3) a square.

0, 0, 1, 2, 2, 3, 3, 3, 2, 2, 1, 2, 1, 3, 4, 4, 3, 3, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 4, 5, 5, 4, 4, 3, 3, 1, 2, 2, 3, 3, 4, 3, 4, 4, 4, 2, 3, 2, 4, 5, 4, 3, 3, 5, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 4, 3, 3, 3, 3, 4, 5, 4, 4, 4, 3, 3, 4, 5, 6

Offset

1,4

Comments

Conjecture: (i) For any integers m > 2 and n > 2, there is a prime p < n such that floor((n-p)/m) is a square.

(ii) For any integer k > 2, if m and n are sufficiently large, then there is a prime p < n such that floor((n-p)/m) is a k-th power. In particular, for any integer n > 2, there is a prime p < n such that floor((n-p)/5) is a cube.

We have verified part (i) for m = 3, ..., 10 and n = 3, ..., 10^6.

Links

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

Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Example

a(3) = 1 since 2 is prime with floor((3-2)/3) = 0^2.
a(11) = 1 since 7 is prime with floor((11-7)/3) = 1^2.
a(13) = 1 since 13 is prime with floor((13-11)/3) = 0^2.
a(28) = 1 since 23 is prime with floor((28-23)/3) = 1^2.
a(37) = 1 since 23 is prime with floor((37-23)/3) = 2^2.
a(173) = 1 since 97 is prime with floor((173-97)/3) = 5^2.
It seems that a(n) = 1 only for n = 3, 11, 13, 28, 37, 173.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
s[n_, k_]:=SQ[Floor[(n-Prime[k])/3]]
a[n_]:=Sum[If[s[n, k], 1, 0], {k, 1, PrimePi[n-1]}]
Table[a[n], {n, 1, 80}]

Crossrefs

Cf. A000040, A000290, A237706, A238701, A238134.

Sequence in context: A220604 A331484 A046918 * A060287 A238733 A165001

Adjacent sequences: A238729 A238730 A238731 * A238733 A238734 A238735

Keyword

nonn

Author

Zhi-Wei Sun, Mar 03 2014

Status

approved