Conner L. Delahanty - cp's OEIS frontend

User: Conner L. Delahanty

Conner L. Delahanty has authored 2 sequences.

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

Conner L. Delahanty, Mar 20 2015

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

			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.

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

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

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 *)

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).

A233468 The digital root of prime(n+1) minus the digital root of prime(n).

Original entry on oeis.org

1, 2, 2, -5, 2, 4, -7, 4, -3, 2, -3, 4, 2, -5, 6, -3, 2, -3, 4, -7, 6, -5, 6, -1, -5, 2, 4, -7, 4, -4, 4, -3, 2, 1, 2, -3, -3, 4, -3, 6, -7, 1, 2, 4, -7, 3, 3, -5, 2, 4, -3, 2, 1, -3, -3, 6, -7, 6, -5, 2, 1, -4, 4, 2, -5, 5, -3, 1, 2, -5, 6

Offset: 1

Conner L. Delahanty, Apr 18 2014

			For n = 1, (prime(2) mod 9) - (prime(1) mod 9) =  3 (mod 9) - 2 (mod 9) = 3-2 = 1.
For n = 2, (prime(3) mod 9) - (prime(2) mod 9) =  5 (mod 9) - 3 (mod 9) = 5-3 = 2.
For n = 3, (prime(4) mod 9) - (prime(3) mod 9) =  7 (mod 9) - 5 (mod 9) = 7-5 = 2.
For n = 4, (prime(5) mod 9) - (prime(4) mod 9) = 11 (mod 9) - 7 (mod 9) = 2-7 = -5.

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

Cf. A000040, A010888.

A233468:=n->(ithprime(n+1) mod 9) - (ithprime(n) mod 9); seq(A233468(n), n=1..100); # Wesley Ivan Hurt, Apr 19 2014

Table[Mod[Prime[n + 1], 9] - Mod[Prime[n], 9], {n, 100}] (* Wesley Ivan Hurt, Apr 19 2014 *)

dd=[]
def prim(end):
    num=3
    primes=[2, 3]
    while (len(primes)<=end):
        num+=1
        prime=False
        length=len(primes)
        for y in range(0, length):
            if (num % primes[y]!=0):
                prime=True
            else:
                prime=False
                break
        if (prime):
            primes.append(num)
    for x in range(len(primes)-1):
        dd.append((primes[x+1]%9) - (primes[x]%9))
    return dd

a(n) = (prime(n+1) mod 9) - (prime(n) mod 9).
a(n) = prime(n + 1) - 9*floor((prime(n + 1) - 1)/9) - prime(n) + 9*floor((prime(n) - 1)/9). - Wesley Ivan Hurt, Apr 19 2014
a(n) = A010888(A000040(n+1)) - A010888(A000040(n)). - Michel Marcus, Apr 19 2014

cp's OEIS Frontend

User: Conner L. Delahanty

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula