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A225461
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a(n) = (prime(k) + prime(k+1))/2 where k = A098015(n).
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3
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9, 15, 21, 26, 34, 39, 45, 50, 56, 64, 69, 76, 81, 86, 99, 105, 111, 120, 129, 134, 144, 154, 160, 165, 170, 176, 186, 195, 225, 231, 236, 246, 254, 260, 266, 274, 279, 288, 300, 309, 315, 324, 334, 342, 351, 356, 370, 376, 381, 386, 399
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OFFSET
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1,1
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COMMENTS
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Former name: Numbers n such that n^2 = prime(i)*prime(i+1) + prime(j)^2, for some i, j > 0 and such that prime(i+1) = prime(i) + 2*prime(j).
The solution values for n = prime(i) + prime(j), when restricted by the condition prime(i+1) = prime(i) + 2*prime(j). Rather than being overly restrictive, the condition applies to the most prevalent type of solution to the equation above for n^2.
The density of solutions, as a percentage of prime(i) values, is 51.33% in the first 10000 primes, and 49.29% in the second 10000 primes.
Values of prime(j) for these solutions range from the prime 2 to the prime 31 in the first 10000 values of prime(i). In the second 10000 values of prime(i) the solutions shift somewhat to higher p(j) values. This trend is expected with increases in the average spacing of primes. This has the effect of slowing the decline in density of solutions at higher p(i) relative to the decline in density of primes themselves.
There is an infinite family of similar equations when "1" is replaced by "d" as in the following: n^2 = prime(i)*prime(i+d) + prime(j)^2, for any i,j, or d > 0.
In this general case, the condition becomes: prime(i+d) = prime(i) + 2*prime(j).
The solution sequence of n values, of course, differs for each d, as does the sequence of primes(i) that create them.
For this generalized version of the equation (and the condition) it is also true that the solution values of n = prime(i) + prime(j), for all d.
[This can be shown with basic algebra as follows, where p() = prime(): n^2 = (p(i) + p(j))^2 = p(i)^2 + 2*p(i)*p(j) + p(j)^2.
Now replace the first p(j) above with (p(i+d) - p(i))/2, and get: n^2 = p(i)^2 + 2*p(i)*(p(i+d) - p(i))/2 + p(j)^2 = p(i)*p(i+d) + p(j)^2.]
This also proves that for each p(i) there is, at most, one solution when the condition is applied.
There are other solutions to the equation above for n^2 with primes that do NOT obey the condition above. These are sparse. For d=1 (this case) they include 8, 18, 72, 450, and 882. All four of these are found at low values of i (i <= 13). These solutions, for all values of d, appear to have the form n = p(j) + 1, if they exist at all for given d. It can be shown, however, that for any given d NOT all p(i) have a solution, but they can occur with the same p(i) as the "algebraic" solutions above.
Heuristically, A001223(n)/2 should be prime with probability ~ 1/log log n, so (prime(n) + prime(n+1))/2 is in the sequence with probability ~ 1/log log n. - Robert Israel, Nov 30 2015
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LINKS
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EXAMPLE
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n = 9 is in the sequence (with i = 4 and j = 1), because 9^2 = 7*11 + 2^2 = prime(4)*prime(4+1) + 2*prime(1) and prime(4+1) = 11 = 7 + 2*2 = prime(4) + 2*prime(1). Moreover, 9 = 7 + 2 = prime(4) + prime(1).
Whereas n = 8 (not in the sequence) satisfies the first condition with i = 2 and j = 4 (8^2 = 3*5 + 7^2), it does NOT meet the second condition: p(i+1) = 5 != p(i) + 2*p(j) = 3 + 2*7 = 17. Moreover, 8 != 3 + 7.
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MAPLE
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P:= select(isprime, [2, seq(2*i+1, i=1..10^4)]):
G2:= (P[2..-1] - P[1..-2])/2:
inds:= select(i -> isprime(G2[i]), [$2..nops(G2)]):
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MATHEMATICA
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Reap[For[p = 2, p < 400, p = NextPrime[p], p1 = NextPrime[p]; If[PrimeQ[pj = (p1 - p)/2] && IntegerQ[n = Sqrt[p*p1 + pj^2]], Sow[n]]]][[2, 1]](* Jean-François Alcover, May 17 2013 *)
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CROSSREFS
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Cf. A136798 and A136799, first and last values for runs of adjacent composites.
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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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