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A102863
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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.
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3
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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
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1,1
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COMMENTS
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a(n) = 0 if and only if n is in A013916. - Robert Israel, Jan 04 2017
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LINKS
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Robert Israel, Table of n, a(n) for n = 1..10000
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EXAMPLE
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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.
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MAPLE
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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
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MATHEMATICA
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a[n_] := Module[{pp = Prime[Range[n]], t}, t = Total[pp]; Boole[AnyTrue[pp, Divisible[t, #]&]]];
Array[a, 100] (* Jean-François Alcover, Jun 16 2020 *)
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CROSSREFS
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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
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KEYWORD
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easy,nonn
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AUTHOR
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Giovanni Teofilatto, Mar 01 2005
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EXTENSIONS
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Edited and extended by Emeric Deutsch, Apr 19 2005
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STATUS
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approved
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