A283556 Digital root of the sum of the first n primes.

0, 2, 5, 1, 8, 1, 5, 4, 5, 1, 3, 7, 8, 4, 2, 4, 3, 8, 6, 1, 9, 1, 8, 1, 9, 7, 9, 4, 3, 4, 9, 1, 6, 8, 3, 8, 6, 1, 2, 7, 9, 8, 9, 2, 6, 5, 6, 1, 8, 1, 5, 4, 9, 7, 6, 2, 4, 3, 4, 2, 4, 8, 4, 5, 1, 8, 1, 8, 3, 8, 6, 8, 7, 5, 9, 1, 6, 8, 9, 5, 9, 5, 3, 2, 3, 1, 3, 2, 9, 2, 6, 5, 7, 8, 4, 8, 7, 3, 2, 3

Offset: 0

Author

Dimitris Valianatos, Mar 10 2017

Keywords

• nonn
• base

Examples

			For n=3, a(3)=1 because the sum of the first 3 primes is 10 and the sum of digits of 10 is 1.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A007504, A010888, A051351.

Programs

Maple

0, op(subs(0=9, ListTools:-PartialSums(select(isprime, [2,seq(i,i=3..1000,2)])) mod 9)); # Robert Israel, Mar 30 2017

Mathematica

With[{nn = 78}, {0}~Join~Table[NestWhile[Total@ IntegerDigits@ # &, #, # >= 10 &] &@ Total@ Take[#, n], {n, nn}] &@ Array[Prime, nn]] (* Michael De Vlieger, Mar 15 2017 *)

PARI

{
p=0;print1(p", ");
forprime(n=2,1000,
         p+=n;
         while(p>9,p=sumdigits(p))
         ;print1(p", ")
        )
}

Python

from sympy import primerange
from itertools import accumulate
prime_sum = [0] + list(accumulate(primerange(2, 1000)))
def dig_root(n): return 1+(n-1)%9
def a(n):
    return 0 if n<1 else dig_root(prime_sum[n])
print([a(n) for n in range(101)]) # Indranil Ghosh, Mar 30 2017

Formula

a(n) = ((A007504(n) - 1) mod 9) + 1.

a(n) = ((A051351(n) - 1) mod 9) + 1.

a(n) = A010888(A007504(n)). - Michel Marcus, Mar 26 2017

Extensions

Corrected by Robert Israel, Mar 30 2017