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A113320
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a(1)=1 and a(n) for n>1 has the smallest positive value such that Sum_{i=1..n} a(i)^a(n-i+1) is prime.
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8
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1, 1, 1, 2, 2, 4, 4, 4, 6, 2, 6, 4, 18, 6, 4, 20, 6, 30, 4, 40, 30, 8, 18, 16, 40, 128, 24, 40, 58, 194, 78, 84, 56, 56, 72, 112, 98, 300, 444, 54, 978, 1938, 120, 126, 6, 1750
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OFFSET
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1,4
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COMMENTS
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Previous name was: Least integers so ascending descending base exponent transforms all prime.
This is the first 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 infinite, but it is hard to characterize the asymptotic cost of adding an n-th term. 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.
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LINKS
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FORMULA
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a(1) = 1. For n>1, a(n) = min {k>0: a(1)^k + k^a(1) + Sum_{i=2..n-1} a(i)^a(n-i+1) is prime}.
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EXAMPLE
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a(1) = 1 by definition.
a(2) = 1 because 1 is the min such that 1^a(2) + a(2)^1 is prime (p=2).
a(3) = 1 because 1 is the min such that 1^a(3) + 1^1 + a(3)^1 is prime (p=5).
a(4) = 2 because 2 is the min such that 1^a(4) + 1^1 + 3^1 + a(4)^1 is prime (p=7).
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MATHEMATICA
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inve[w_] := Total[w^Reverse[w]]; a[1] = 1; a[n_] := a[n] = Block[{k = 0}, While[! PrimeQ[ inve@ Append[Array[a, n-1], ++k]]]; k]; Array[a, 46] (* Giovanni Resta, Jun 13 2016 *)
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PROG
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(PARI) lista(n)={my(a=vector(n)); a[1]=1; print1(1, ", "); for(n=2, #a, my(t=sum(i=2, n-1, a[i]^a[n-i+1])); my(k=1); while(!ispseudoprime(t+1+k), k++); a[n]=k; print1(k, ", "))} \\ Andrew Howroyd, Jan 03 2020
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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