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Indices n where prime(n) + 2*prime(n+1) and 2*prime(n) + prime(n+1) have the same number of prime divisors counted with multiplicity.
1

%I #12 Mar 20 2017 12:08:28

%S 2,3,6,11,12,15,16,17,19,20,23,25,27,30,33,34,37,38,47,48,51,53,56,57,

%T 58,60,66,68,75,76,77,78,79,86,87,89,90,93,94,99,100,101,107,110,123,

%U 124,128,133,137,138,139,141,143,145,147,151

%N Indices n where prime(n) + 2*prime(n+1) and 2*prime(n) + prime(n+1) have the same number of prime divisors counted with multiplicity.

%H Charles R Greathouse IV, <a href="/A284091/b284091.txt">Table of n, a(n) for n = 1..10000</a>

%F n such that A001222(A000040(n)+2*A000040(n+1))=A001222(2*A000040(n)+A000040(n+1)). - _Robert Israel_, Mar 20 2017

%e n = 15, prime(n) = 47, prime(n+1) = 53, both 2*47 + 53 = 147 = 3*7^2 and 47 + 2*53 = 153 = 3^2*17 are products of 3 primes.

%p select(t -> numtheory:-bigomega(2*ithprime(t)+ithprime(t+1)) = numtheory:-bigomega(ithprime(t)+2*ithprime(t+1)), [$1..1000]); # _Robert Israel_, Mar 20 2017

%t Select[Range[1000],PrimeOmega[{2,1}.{(p=Prime[#]),(q=Prime[#+1])}]==PrimeOmega[{1,2}.{p,q}]&]

%o (PARI) list(lim)=my(v=List(),p=2,n); forprime(q=3,, if(n++>lim, break); if(bigomega(p+2*q)==bigomega(2*p+q), listput(v,n)); p=q); Vec(v) \\ _Charles R Greathouse IV_, Mar 20 2017

%Y A241945 is a subsequence.

%Y Cf. A000040, A001222.

%K nonn

%O 1,1

%A _Zak Seidov_, Mar 19 2017