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%I #4 Oct 01 2013 17:58:07
%S 1,2,2,1,2,2,1,2,1,2,2,2,2,2,2,2,2,2,1,2,2,2,2,2,2,2,2,1,2,3,2,3,2,2,
%T 2,3,1,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,3,1,2,2,2,1,3,2,3,
%U 2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,3,2,2,2,2,3,2,1,3,2,2,2,2,2,2,2,2,2,3,2,2,2
%N Number of distinct divisors of n+1 where n and n+1 are composite or twin composite numbers.
%C It is interesting to note that the first such consecutive pair of composite numbers is 8 and 9 which are perfect powers: 2^3 and 3^2. Conjecture: 8 and 9 are the only 2 consecutive composite numbers that are both perfect powers. Or, if x>2, x^m+1 != y^n for all m,n,x,y. Now if we relax the condition that 0 and 1 are not composite, we have 0^m+1 = 1^n for all m,n an infinity of solutions.
%e For n=8 n+1 = 9 = 3*3 or 1 distinct divisor.
%o (PARI) f(n) = for(x=1,n,y=composite(x)+1;if(!isprime(y),print1(omega(y)","))) composite(n) =\The n-th composite number. 1 is def as not prime nor composite. { local(c,x); c=1; x=1; while(c <= n, x++; if(!isprime(x),c++); ); return(x) }
%K easy,nonn
%O 1,2
%A _Cino Hilliard_, Feb 19 2005
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