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

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

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

Programs

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}]

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