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

%I #11 Mar 04 2014 13:16:56

%S 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,

%T 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,

%U 4,4,3,4,3,4,4,5,4,4,1,2

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

%C 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.

%C (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.

%C 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.

%H Zhi-Wei Sun, <a href="/A238733/b238733.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), no.1, 65-76. <a href="http://arxiv.org/abs/0803.3737">arXiv:0803.3737</a>.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, 1402.6641, 2014.

%e a(25) = 1 since floor((25-23)/3) = 0 = (3-1)*(3-3)/8 with 23 and 3 both prime.

%e a(96) = 1 since floor((96-11)/3) = 28 = (17-1)*(17-3)/8 with 11 and 17 both prime.

%e a(409) = 1 since floor((409-379)/3) = 10 = (11-1)*(11-3)/8 with 379 and 11 both prime.

%t TQ[n_]:=PrimeQ[Sqrt[8n+1]+2]

%t t[n_,k_]:=TQ[Floor[(n-Prime[k])/3]]

%t a[n_]:=Sum[If[t[n,k],1,0],{k,1,PrimePi[n-1]}]

%t Table[a[n],{n,1,80}]

%o (PARI) has(x)=issquare(8*x+1,&x) && isprime(x+2)

%o a(n)=my(s); forprime(p=2,n-1,s+=has((n-p)\3)); s \\ _Charles R Greathouse IV_, Mar 03 2014

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

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Mar 03 2014

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)