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A340865
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Primes p such that (p^2 + 1)/2 and 2*p^2 - 1 are also prime.
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1
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3, 11, 59, 181, 199, 379, 409, 571, 739, 1039, 1439, 2239, 2269, 2351, 2381, 2671, 2719, 2789, 3049, 3529, 4021, 4201, 4721, 4999, 5431, 5531, 5839, 6329, 6619, 8329, 9241, 9419, 9631, 9689, 10151, 11329, 11551, 12071, 12421, 13339, 14489, 15091, 17419, 18301
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OFFSET
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1,1
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COMMENTS
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How many triangular numbers with 6 divisors (A292989) can be divisible by the same squared prime p^2?
The k-th triangular number T(k) = A000217(k) = k*(k+1)/2 can be written as the product of two coprime factors A and B where A=k and B=(k+1)/2 for odd k, A=k/2 and B=k+1 for even k. If a triangular number has 6 divisors, then it is of the form p^2*q where p and q are distinct primes. We can identify four cases:
Case 1: A = k = p^2 and B = (k+1)/2 = q, so q = (p^2 + 1)/2; solutions occur at primes p in A048161.
Case 2: A = k = q and B = (k+1)/2 = p^2, so 2*p^2 - 1 = q; solutions occur at primes p in A106483.
Case 3: A = k/2 = p^2 and B = k+1 = q. In this case, 2*p^2 + 1 = q. For p = 2, we would get q = 9 (nonprime), so p must be odd. If prime p > 3 (so q > 19), we have p^2 == 1 (mod 3), so q == 0 (mod 3), hence nonprime. So the only solution for this case occurs at p=3, q=19, t = 3^2*19 = 171.
Case 4: A = k/2 = q and B = k+1 = p^2. In this case, 2*q + 1 = p^2, so p is odd, but then p^2 == 1 (mod 8), so q == 0 (mod 4), hence q is not prime: no solutions exist.
Since Case 4 has no solutions, at most three triangular numbers with 6 divisors can be divisible by the same squared prime p^2; Case 3 has a solution only at p=3 and, in fact, there are three triangular numbers with 6 divisors that are divisible by 3^2: t = 3^2*5 = 45 = T(9), t = 3^2*17 = 153 = T(17), and 3^2*19 = 171 = T(18).
For all primes p > 3, then, at most two triangular numbers with 6 divisors are divisible by p^2; this sequence (after the initial term, 3) lists the primes p such that p^2 divides exactly two triangular numbers that have 6 divisors.
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LINKS
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EXAMPLE
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Both (3^2 + 1)/2 = 5 and 2*3^2 - 1 = 17 are prime, so 3 is in the sequence.
(5^2 + 1)/2 = 13 is prime, but 2*5^2 - 1 = 49 = 7^2 is not prime, so 5 is not in the sequence.
(7^2 + 1)/2 = 25 is not prime, so even though 2*7^2 - 1 = 97 is prime, 7 is not in the sequence.
Neither (23^2 + 1)/2 = 265 = 5*53 nor 2*23^2 - 1 = 1057 = 7*151 is prime, so 23 is not in the sequence.
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PROG
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(PARI) isok(p) = (p>2) && isprime(p) && isprime((p^2+1)/2) && isprime(2*p^2-1); \\ Michel Marcus, Jan 25 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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