A119734 - OEIS

%I #8 Oct 02 2013 15:12:51

%S 0,3,1,0,2,2,1,0,0,4,-1,4,2

%N a(n) = least k such that 2^k + n is a perfect power, or -1 if no such k exists.

%C Comments from Herman Jamke (hermanjamke(AT)fastmail.fm), May 05 2007: (Start) The first term for which k does not exist is a(10) since neither for k=0 nor 1 do we get a perfect power and for k>1 2^k+10=2*(2^(k-1)+5)=2*odd which can't be a perfect power since the exponent of 2 is not greater than 1.

%C For n=2*n' where n' is odd: if n+1 is a perfect power, a(n)=0; else if n+2 is a perfect power, a(n)=1. Otherwise a(n)=-1 because if we assume, for some k>1, that 2^k + n = 2*(2^(k-1) + n') is a perfect power m^e then, since 2^(k-1)+n' is odd, m must have its factor 2 raised to a multiple of e equal to 1 and so e=1, a contradiction. For example:

%C a(2*1) = 1 since n+2=2*1+2=4, a perfect power.

%C a(2*3) = 1 since n+2=2*3+2=8, a perfect power.

%C a(2*5) = -1 since n+1=11 and n+2=12 are not perfect powers.

%C a(2*7) = 1 since n+2=2*7+2=16, a perfect power.

%C a(2*9) = -1 since n+1=19 and n+2=20 are not perfect powers.

%C a(2*11) = -1 since n+1=23 and n+2=24 are not perfect powers.

%C a(2*13) = 0 since n+1=2*13+1=27, a perfect power.

%C a(2*15) = 1 since n+2=2*15+2=32, a perfect power.

%C a(2*17) = 1 since n+2=2*17+2=36, a perfect power. (End)

%e The least k such that 2^k + 5 is a perfect power is 2, since 2^2 + 5 = 9 = 3^2, so a(5) = 2.

%Y Cf. A001597.

%K hard,more,sign

%O 0,2

%A _Ryan Propper_, Jun 15 2006

%E a(10)-a(12) from Herman Jamke (hermanjamke(AT)fastmail.fm), May 05 2007