A251364 Difference between average of two consecutive odd primes and the sum of all prime factors of the average.

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

Author

Conner L. Delahanty, Mar 20 2015

Keywords

• nonn
• easy

Comments

Sequence starts with the 2nd prime because the average of the first two primes is not an integer.

Examples

			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.
		

Links

• Conner L. Delahanty, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A000040 (primes), A001414 (sopfr).

Programs

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 *)

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).