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A251364
Difference between average of two consecutive odd primes and the sum of all prime factors of the average.
1
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
OFFSET
1,3
COMMENTS
Sequence starts with the 2nd prime because the average of the first two primes is not an integer.
LINKS
FORMULA
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).
EXAMPLE
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.
MATHEMATICA
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 *)
CROSSREFS
Cf. A000040 (primes), A001414 (sopfr).
Sequence in context: A319585 A133452 A356665 * A229791 A189293 A140607
KEYWORD
nonn,easy
AUTHOR
Conner L. Delahanty, Mar 20 2015
STATUS
approved