

A238733


Number of primes p < n such that floor((np)/3) = (q1)*(q3)/8 for some prime q.


1



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
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OFFSET

1,4


COMMENTS

Conjecture: (i) For any integers m > 2 and n > 2, there is a prime p < n such that floor((np)/m) has the form (q1)*(q3)/8 with q an odd prime.
(ii) If m > 2 and n > m + 1, then there is a prime p < n such that floor((np)/m) has the form (q^2  1)/8 with q an odd prime, except for the case m = 3 and n = 19.
Note that (q1)*(q3)/8 = r*(r+1)/2 with r = (q3)/2. It seems that a(n) = 1 only for n = 3, 19, 25, 29, 31, 67, 79, 95, 96, 331, 373, 409.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), no.1, 6576. arXiv:0803.3737.
ZhiWei Sun, Problems on combinatorial properties of primes, 1402.6641, 2014.


EXAMPLE

a(25) = 1 since floor((2523)/3) = 0 = (31)*(33)/8 with 23 and 3 both prime.
a(96) = 1 since floor((9611)/3) = 28 = (171)*(173)/8 with 11 and 17 both prime.
a(409) = 1 since floor((409379)/3) = 10 = (111)*(113)/8 with 379 and 11 both prime.


MATHEMATICA

TQ[n_]:=PrimeQ[Sqrt[8n+1]+2]
t[n_, k_]:=TQ[Floor[(nPrime[k])/3]]
a[n_]:=Sum[If[t[n, k], 1, 0], {k, 1, PrimePi[n1]}]
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, n1, s+=has((np)\3)); s \\ Charles R Greathouse IV, Mar 03 2014


CROSSREFS

Cf. A000040, A000217, A132399, A144590, A238732.
Sequence in context: A046918 A238732 A060287 * A165001 A059253 A108133
Adjacent sequences: A238730 A238731 A238732 * A238734 A238735 A238736


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 03 2014


STATUS

approved



