

A113336


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


8



2, 1, 6, 6, 18, 12, 18, 42, 288, 108, 180, 1122, 1458, 660
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OFFSET

1,1


COMMENTS

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 nth 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.


LINKS

Table of n, a(n) for n=1..14.


FORMULA

a(1) = 2. For n>1: a(n) = min {n>0: SUM[from i = 1 to n] (a(i))^(a(ni+1)) is prime}.


EXAMPLE

a(1) = 2 by definition.
a(2) = 1 because 1 is the min such that 2^a(2) + a(2)^2 is prime (p=3).
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).
a(4) = 6 because 2^6 + 1^6 + 6^1 + 6^2 = 107 is prime.
a(5) = 18 because 2^18 + 1^6 + 6^6 + 6^1 + 18^2 = 309131 is prime.
a(6) = 12 because 2^12 + 1^18 + 6^6 + 6^6 + 18^1 + 12^2 = 97571 is prime.
a(7) = 18 because 2^18 + 1^12 + 6^18 + 6^6 + 18^6 + 12^1 + 18^2 = 101559990989777 is prime.
a(8) = 42 because 2^42 + 1^18 + 6^12 + 6^18 + 18^6 + 12^6 + 18^1 + 42^2 = 105960216961847 is prime.
a(9) > 250.


CROSSREFS

Cf. A000040, A005408, A113122, A113153, A113154.
Sequence in context: A117753 A145883 A062820 * A113979 A053442 A019082
Adjacent sequences: A113333 A113334 A113335 * A113337 A113338 A113339


KEYWORD

easy,nonn


AUTHOR

Jonathan Vos Post, Jan 07 2006


EXTENSIONS

a(9)a(14) from Giovanni Resta, Jun 13 2016


STATUS

approved



