cp's OEIS Frontend

A113336 Least integers, starting with 2, so ascending descending base exponent transforms all prime.

%I A113336 #10 Dec 17 2019 05:59:19
%S A113336 2,1,6,6,18,12,18,42,288,108,180,1122,1458,660
%N A113336 Least integers, starting with 2, so ascending descending base exponent transforms all prime.
%C A113336 This is the second sequence submitted as a solution to an "ascending descending base exponent transform inverse problem" where the sequence is iteratively defined such that the transform meets a constraint. The sequence is probably infinite, but it is hard to characterize the asymptotic cost of adding an n-th term (the 9th terms is at least 250). A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154.
%F A113336 a(1) = 2. For n>1: a(n) = min {n>0: Sum_{i=1..n} a(i)^a(n-i+1) is prime}.
%e A113336 a(1) = 2 by definition.
%e A113336 a(2) = 1 because 1 is the min such that 2^a(2) + a(2)^2 is prime (p=3).
%e A113336 a(3) = 6 because 6 is the min such that 2^a(3) + 1^1 + a(3)^2 is prime (2^6 + 1^1 + 6^1 = 101).
%e A113336 a(4) = 6 because 2^6 + 1^6 + 6^1 + 6^2 = 107 is prime.
%e A113336 a(5) = 18 because 2^18 + 1^6 + 6^6 + 6^1 + 18^2 = 309131 is prime.
%e A113336 a(6) = 12 because 2^12 + 1^18 + 6^6 + 6^6 + 18^1 + 12^2 = 97571 is prime.
%e A113336 a(7) = 18 because 2^18 + 1^12 + 6^18 + 6^6 + 18^6 + 12^1 + 18^2 = 101559990989777 is prime.
%e A113336 a(8) = 42 because 2^42 + 1^18 + 6^12 + 6^18 + 18^6 + 12^6 + 18^1 + 42^2 = 105960216961847 is prime.
%e A113336 a(9) > 250.
%Y A113336 Cf. A000040, A005408, A113122, A113153, A113154.
%K A113336 easy,nonn
%O A113336 1,1
%A A113336 _Jonathan Vos Post_, Jan 07 2006
%E A113336 a(9)-a(14) from _Giovanni Resta_, Jun 13 2016