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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
 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 (list; graph; refs; listen; history; text; internal format)
 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 A059253 A108133 Adjacent sequences:  A238730 A238731 A238732 * A238734 A238735 A238736 KEYWORD nonn AUTHOR Zhi-Wei Sun, Mar 03 2014 STATUS approved

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Last modified October 26 05:46 EDT 2021. Contains 348256 sequences. (Running on oeis4.)