

A102863


a(n)=1 if at least one of the first n primes is a divisor of the sum of the first n primes; otherwise a(n)=0.


3



1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1
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OFFSET

1,1


COMMENTS

a(n) = 0 if and only if n is in A013916.  Robert Israel, Jan 04 2017


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000


EXAMPLE

a(2)=0 because none of the first 2 primes (2, 3) is a divisor of 2+3; a(5)=1 because among the first 5 primes (namely, 2,3,5,7,11) there are divisors of 2+3+5+7+11=28.


MAPLE

with(numtheory):
a:=proc(n)
if nops(factorset(sum(ithprime(k), k=1..n)) intersect {seq(ithprime(j), j=1..n)}) >0 then
1
else
0
fi
end:
seq(a(n), n=1..130); # Emeric Deutsch
# alternative:
N:= 500: # to get the first N terms
A:= Vector(N):
S:= 2: P:= 2: p:= 2: A[1]:= 1:
for n from 2 to N do
p:= nextprime(p);
S:= S+p; P:= P*p;
if igcd(S, P) > 1 then A[n]:= 1 fi
od:
convert(A, list); # Robert Israel, Jan 04 2017


CROSSREFS

A105783(n) gives number of primes among the first n primes that are divisors of the sum of the first n primes.
Cf. A013916, A136443.
Sequence in context: A195062 A229940 A179761 * A131483 A077052 A133566
Adjacent sequences: A102860 A102861 A102862 * A102864 A102865 A102866


KEYWORD

easy,nonn


AUTHOR

Giovanni Teofilatto, Mar 01 2005


EXTENSIONS

Edited and extended by Emeric Deutsch, Apr 19 2005


STATUS

approved



