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A216244
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a(n) = (prime(n)^2 - 1)/2 for n >= 2.
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6
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4, 12, 24, 60, 84, 144, 180, 264, 420, 480, 684, 840, 924, 1104, 1404, 1740, 1860, 2244, 2520, 2664, 3120, 3444, 3960, 4704, 5100, 5304, 5724, 5940, 6384, 8064, 8580, 9384, 9660, 11100, 11400, 12324, 13284, 13944, 14964, 16020, 16380, 18240, 18624, 19404, 19800
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OFFSET
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2,1
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COMMENTS
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Subsequence of A055523 restricted to the case of the other (shorter) leg of the triangle equal to a prime.
There is only one value of a(n) for each prime(n). (This is not necessarily true if the shorter leg is not a prime.)
Note that a(1) is nonexistent since there is no solution with prime = 2.
All terms are divisible by 4.
The values of m (the length of the hypotenuse) always equals a(n) + 1.
a(n) = (prime(n)^2 - 1)/2 for all n > 1.
This follows algebraically given m = a(n) + 1 (or vice versa).
The same two relationships apply when the shorter leg is an odd nonprime, but for only those results corresponding to the longest possible leg of the triangle.
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LINKS
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FORMULA
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a(n) = (prime(n)^2 - 1)/2 for n >= 2.
a(n) = (prime(n)-1)*(prime(n)+1)/2 = lcm(prime(n)+1, prime(n)-1) for n > 1 because one of prime(n)+1 or prime(n)-1 is even and the other is divisible by 4. Say prime(n)-1 is divisible by 4; then (prime(n)+1)/2 and (prime(n)-1)/4 must be coprime. - Frank M Jackson, Dec 11 2018
Product_{n>=2} (1 + 1/a(n)) = 3/2. - Amiram Eldar, Jun 03 2022
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EXAMPLE
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24^2 + 7^2 = 625 = 25^2 = (24 +1)^2 and a(4) = (prime(4)^2 -1)/2 = (49 - 1)/2 = 24.
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MAPLE
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MATHEMATICA
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PROG
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(PARI) vector(50, n, n++; (prime(n)^2 -1)/2) \\ G. C. Greubel, Dec 14 2018
(Magma) [(NthPrime(n)^2 - 1)/2: n in [2..50]]; // G. C. Greubel, Dec 14 2018
(Sage) [(nth_prime(n)^2 -1)/2 for n in (2..50)] # G. C. Greubel, Dec 14 2018
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CROSSREFS
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Equals A084921 excluding its first term.
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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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