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a(n) = smallest k > 1 such that k-1 and k+1 together have n prime divisors.
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%I #5 Nov 02 2013 05:21:37

%S 2,4,3,5,7,15,17,31,65,129,127,449,511,2561,1025,7937,12799,20481,

%T 8191,28673,65537,131071,458751,360449,966655,524287,4194303,2097151,

%U 29360129,34865153,67108865,134217729,33554431,608174081,268435457,536870911,4831838207

%N a(n) = smallest k > 1 such that k-1 and k+1 together have n prime divisors.

%C Prime divisors are counted with multiplicity.

%C Similar to A155800, where k is restricted to primes.

%C Terms of the form 2^m-1 or 2^m+1 seem to occur frequently.

%e Adjacent to 2 are the numbers 1 and 3 which together have one prime divisor, hence a(1) = 2. Adjacent to 3 are 2 and 4; together they have three prime divisors, hence a(3) = 3. Adjacent to 4 are the primes 3 and 5, each having one prime divisor; hence a(2) = 4.

%e For k = 129, k-1 = 128 = 2*2*2*2*2*2*2 and k+1 = 130 = 2*5*13 together have ten prime divisors. For all numbers k < 129 the adjacent numbers k-1 and k+1 together have fewer or more than ten (for 127 there are eleven) prime divisors, hence a(10) = 129.

%o (PARI) {for(n=2, 150000000, s=bigomega(n-1)+bigomega(n+1); if(v[s]==0, v[s]=n)); v}

%Y Cf. A001222 (number of prime divisors of n), A155800, A154704, A154598.

%K nonn

%O 1,1

%A _Klaus Brockhaus_, Jan 28 2009, Jan 31 2009

%E a(34)-a(37) from _Donovan Johnson_, Nov 02 2013