login
Numbers n such that sum_{k=1..n} (-1)^(n-k) *bigomega(k) is prime.
0

%I #4 Mar 30 2012 18:35:54

%S 4,6,8,10,12,24,26,28,30,32,34,46,52,70,78,82,102,126,128,132,134,136,

%T 138,168,186,190,192,222,234,274,280,312,316,322,336,378,418,424,426,

%U 440,472,484,492,504,532,540,558,570,574,584,592,602,604,606,650,652

%N Numbers n such that sum_{k=1..n} (-1)^(n-k) *bigomega(k) is prime.

%C The partial alternating sum over bigomega(.)=A001222(.) in the definition starts at n=1

%C as 0, 1, 0, 2, -1, 3, -2, 5, -3, 5, -4, 7, -6, 8 ...

%C The first primes in this signed sequence are

%C 2, 3, 5, 5, 7, 17, 17, 17, 19, 23, 23, 31, 37, 53, 59, 61, 79, 97, 103, 107, 107, 107, 109,...

%C occurring at positions 4, 6, 8, 10 etc, which define the sequence.

%e 6 is in the sequence because sum_{k=1..6}(-1)^ (6-k)*bigomega(k) =

%e ((-1)^5)*0 + ((-1)^4)*1 + ((-1)^3)*1 + ((-1)^2)*2 + ((-1)^1)*1 + ((-1)^0)*2 =

%e 0 + 1 -1 + 2 -1 + 2 = 3 is prime.

%p with(numtheory):for n from 1 to 1000 do: s:=0: for k from 1 to n do :s:=s+((-1)^(n-k))*bigomega(k):od: if type(s,prime)=true then printf(`%d, `, n):else fi:od:

%Y Cf. A001222.

%K nonn

%O 1,1

%A _Michel Lagneau_, Oct 01 2010

%E Comment slightly extended - _R. J. Mathar_, Oct 03 2010