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A066888
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Number of primes p between triangular numbers T(n) < p <= T(n+1).
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12
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0, 2, 1, 1, 2, 2, 1, 2, 3, 2, 2, 3, 3, 3, 3, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 4, 5, 3, 6, 6, 7, 5, 5, 6, 4, 8, 5, 6, 6, 8, 6, 8, 5, 7, 5, 11, 4, 6, 9, 7, 8, 9, 8, 7, 7, 9, 7, 8, 7, 12, 5, 9, 9, 11, 9, 7, 7, 12, 10, 10, 9, 9, 9, 6, 11, 10, 11, 9, 12, 11, 12, 9, 10, 11, 12, 10, 13, 9, 11, 10, 12
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OFFSET
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0,2
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COMMENTS
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If the above conjecture is true, then for any k > 1 there is a prime p > k such that p <= (n+1)(n+2)/2, where n = floor(sqrt(2k)+1/2). Ignoring the floor function we can obtain a looser (but nicer) lower bound of p <= 1 + k + 2*sqrt(2k). - Dmitry Kamenetsky, Nov 26 2016
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LINKS
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FORMULA
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a(n) = pi(n*(n+1)/2) - pi(n*(n-1)/2).
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EXAMPLE
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Write the numbers 1, 2, ... in a triangle with n terms in the n-th row; a(n) = number of primes in n-th row.
Triangle begins
1 (0 primes)
2 3 (2 primes)
4 5 6 (1 prime)
7 8 9 10 (1 prime)
11 12 13 14 15 (2 primes)
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MATHEMATICA
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Table[PrimePi[(n^2 + n)/2] - PrimePi[(n^2 - n)/2], {n, 96}] (* Alonso del Arte, Sep 03 2011 *)
PrimePi[#[[2]]]-PrimePi[#[[1]]]&/@Partition[Accumulate[Range[0, 100]], 2, 1] (* Harvey P. Dale, Jun 04 2019 *)
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PROG
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(PARI) { tp(m)=local(r, t); r=1; for(n=1, m, t=0; for(k=r, n+r-1, if(isprime(k), t++)); print1(t", "); r=n+r; ) }
(PARI) {tpf(m)=local(r, t); r=1; for(n=1, m, t=0; for(k=r, n+r-1, if(isprime(k), t++); print1(k" ")); print1(" ("t" prime)"); print(); r=n+r; ) }
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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