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A251364
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Difference between average of two consecutive odd primes and the sum of all prime factors of the average.
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1
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0, 1, 3, 5, 7, 10, 11, 11, 20, 15, 23, 30, 34, 38, 43, 48, 52, 43, 60, 53, 69, 41, 59, 82, 80, 90, 95, 71, 106, 83, 65, 110, 130, 135, 134, 145, 146, 146, 157, 165, 150, 177, 174, 179, 159, 179, 209, 202, 210, 173, 224, 200, 125, 238, 238, 254
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OFFSET
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1,3
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COMMENTS
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Sequence starts with the 2nd prime because the average of the first two primes is not an integer.
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LINKS
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FORMULA
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a(n) = ((prime(n+1) + prime(n+2))/2) - (sopfr((prime(n+1) + prime(n+2))/2)), where sopfr is A001414, the sum of primes dividing n (with repetition).
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EXAMPLE
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For n = 1, the average of prime(2) and prime(3) is 4. The prime factors of 4 are 2 and 2. 4 - (2 + 2) = 0.
For n = 2, the average of prime(3) and prime(4) is 6. The prime factors of 6 are 2 and 3. 6 - (2 + 3) = 1.
For n = 3, the average of prime(4) and prime(5) is 9. The prime factors of 9 are 3 and 3. 9 - (3 + 3) = 3.
For n = 4, the average of prime(5) and prime(6) is 12. The prime factors of 12 are 2, 2, and 3. 12 - (2 + 2 + 3) = 5.
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MATHEMATICA
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f[{a_, b_}] := Table[a, {b}]; g[n_] := Block[{d = (Prime[n + 1] + Prime[n])/2}, d - Plus @@ Flatten[f /@ FactorInteger@ d]]; Table[g@ n, {n, 2, 57}] (* Michael De Vlieger, Mar 25 2015 *)
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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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STATUS
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approved
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