A238733 Number of primes p < n such that floor((n-p)/3) = (q-1)*(q-3)/8 for some prime q.

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

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) has the form (q-1)*(q-3)/8 with q an odd prime.

(ii) If m > 2 and n > m + 1, then there is a prime p < n such that floor((n-p)/m) has the form (q^2 - 1)/8 with q an odd prime, except for the case m = 3 and n = 19.

Note that (q-1)*(q-3)/8 = r*(r+1)/2 with r = (q-3)/2. It seems that a(n) = 1 only for n = 3, 19, 25, 29, 31, 67, 79, 95, 96, 331, 373, 409.

Links

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

Zhi-Wei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), no.1, 65-76. arXiv:0803.3737.

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

Example

a(25) = 1 since floor((25-23)/3) = 0 = (3-1)*(3-3)/8 with 23 and 3 both prime.
a(96) = 1 since floor((96-11)/3) = 28 = (17-1)*(17-3)/8 with 11 and 17 both prime.
a(409) = 1 since floor((409-379)/3) = 10 = (11-1)*(11-3)/8 with 379 and 11 both prime.

Mathematica

TQ[n_]:=PrimeQ[Sqrt[8n+1]+2]
t[n_, k_]:=TQ[Floor[(n-Prime[k])/3]]
a[n_]:=Sum[If[t[n, k], 1, 0], {k, 1, PrimePi[n-1]}]
Table[a[n], {n, 1, 80}]

Prog

(PARI) has(x)=issquare(8*x+1, &x) && isprime(x+2)
a(n)=my(s); forprime(p=2, n-1, s+=has((n-p)\3)); s \\ Charles R Greathouse IV, Mar 03 2014

Crossrefs

Cf. A000040, A000217, A132399, A144590, A238732.

Sequence in context: A046918 A238732 A060287 * A165001 A364200 A059253

Adjacent sequences: A238730 A238731 A238732 * A238734 A238735 A238736

Keyword

nonn

Author

Zhi-Wei Sun, Mar 03 2014

Status

approved