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A113336 Least integers, starting with 2, so ascending descending base exponent transforms all prime. +0
9
2, 1, 6, 6, 18, 12, 18, 42 (list; graph; listen)
OFFSET

1,1

COMMENT

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.

FORMULA

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

Adjacent sequences: A113333 A113334 A113335 this_sequence A113337 A113338 A113339

Sequence in context: A117753 A145883 A062820 this_sequence A113979 A053442 A019082

KEYWORD

easy,nonn

AUTHOR

Jonathan Vos Post (jvospost3(AT)gmail.com), Jan 07 2006

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Last modified November 8 20:39 EST 2009. Contains 166234 sequences.


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