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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 03 2014
STATUS
approved