A083205 - OEIS

%I #8 Aug 25 2014 11:50:38

%S 1,2,4,31,26129,466202136816810031,

%T 10584868011442581563053546190350068005080430521324963528734180259722815036145

%N a(1) = 1, then smallest number not included earlier such that a(n)*a(n+1) + 1 is an n-th power.

%C The sequence is infinite and a(n+1) <= ([a(n)+1]^n - 1)/a(n) when n is even, or a(n+1) <= ([a(n)-1]^n - 1)/a(n) when n is odd.

%C To find a(6), we need an x such that x^5 = 1 (mod a(5)); then a(6) = (x^5 - 1)/a(5). The multiplicative group mod a(5) has order phi(a(5)) = 23296, which is not divisible by 5. So the only 5th root of 1 in this group is 1. x = 1 would give a(6) = 0, this is not allowed, so we take x to be the next representative of 1 mod a(5), i.e. a(5)+1. So a(6) = [(a(5)+1)^5 - 1]/a(5). - _David Wasserman_, Mar 02 2004

%C Next term is approximately 8.1*10^451. - _David Wasserman_, May 26 2004

%e a(4) = 31, a(5) = 26129, 31*26129 + 1 = 810000 = 30^4.

%Y Cf. A083203, A083204.

%K nonn

%O 1,2

%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Apr 28 2003

%E More terms from _David Wasserman_, Mar 02 2004