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Periodic part of continued fraction for sqrt(n), encoded by raising successive primes to the terms. If sqrt(n)=c0+[c1,c2,c3...] then a(n)=2^c1*3^c2*5^c3*...

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`%I #6 Oct 14 2022 17:57:58
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`%S 1,4,18,1,16,324,72030,162,1,64,5832,2916,372027810,10588410,1458,1,
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`%T 256,104976,1036385881030500,26244,9421689827550,4946387159463750,
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`%U 1556496270,13122,1,1024,1889568,2542277241000,76256028326940,236196
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`%N Periodic part of continued fraction for sqrt(n), encoded by raising successive primes to the terms. If sqrt(n)=c0+[c1,c2,c3...] then a(n)=2^c1*3^c2*5^c3*...
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`%C Could be made less gigantic by omitting final terms in continued fraction, which are always 2*c0.
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`%F a(n) = A059900(A059904(n)).
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`%e sqrt(14) = 3+[1,2,1,6] so a(14) = 2^1*3^2*5^1*7^6 = 10588410.
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`%Y Cf. A059904, A059900, A003285.
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`%K easy,nonn
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`%O 1,2
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`%A _Marc LeBrun_, Feb 07 2001
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`%E Offset corrected by _Sean A. Irvine_, Oct 14 2022
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