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Question: can the (nonzero) value of any terms be proved?
a(n)=my(best=0); if(n==14, 76, for(k=1, max(9, 30094\sqrt(log(n)), ), if(isA043096(n^k), best=k)); best ) \\ (conjectural) _Charles R Greathouse IV_, Sep 17 2012
\\ (conjectural) Charles R Greathouse IV, Sep 17 2012
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a(n)=my(best=0); for(k=1, max(9, 147300\log(n)), if(isA043096(n^k), best=k)); best
\\ _(conjectural) _Charles R Greathouse IV_, Sep 17 2012
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Contribution from _Charles R Greathouse IV, _, Sep 17 2012: (Start)
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Contribution from ~~~~Charles R Greathouse IV, Sep 17 2012: (Start)
(Start)
A naive heuristic suggests that there are infinitely many n such that a(n) = 6 but only finitely many a(n) such that a(n) > 6. This suggests a weaker conjecture: this sequence is bounded. (end)
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Contribution from ~~~~:
(Start)
a(n) = 0 for infinitely many n; such n have positive density in this sequence. Question: are such n of density 1?
Question: can the (nonzero) value of any terms be proved?
A naive heuristic suggests that there are infinitely many n such that a(n) = 6 but only finitely many a(n) such that a(n) > 6. This suggests a weaker conjecture: this sequence is bounded.
(end)
(PARI) isA043096(n)=my(v=digits(n)); for(i=2, #v, if(v[i]==v[i-1], return(0))); 1
a(n)=my(best=0); for(k=1, max(9, 147\log(n)), if(isA043096(n^k), best=k)); best
\\ Charles R Greathouse IV, Sep 17 2012
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