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%I #34 Aug 03 2020 05:36:14
%S 1,6,57960,36802906522516375115639735990520502954652700
%N a(n) is the last (and thus largest) denominator of an Egyptian fraction representing n, where each consecutive denominator is as small as possible.
%C Values grow extremely quickly, a(5) has 142548 decimal digits.
%C The denominators for n = 3 are given in A140335.
%C The denominators for n = 4 are given in A281873.
%C The number of terms in the representation of n is A306349(n).
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Egyptian_fraction">Egyptian fraction</a>
%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%e a(3) = 57960 because (1/1) + (1/2) + (1/3) + (1/4) + (1/5) + (1/6) + (1/7) + (1/8) + (1/9) + (1/10) + (1/15) + (1/230) + (1/57960) = 3 and the final and greatest denominator is 57960. (Sequence A140335 has the full list of denominators.)
%o (PARI) a(n)={my(s=n,k=1); while(s>1/k, s-=1/k; k++); while(s!=0, k=ceil(1/s); s-=1/k); k} \\ _Andrew Howroyd_, Sep 01 2019
%o (Python)
%o from sympy import egyptian_fraction
%o def A323725(n): return egyptian_fraction(n)[-1] # _Pontus von Brömssen_, Aug 03 2020
%Y A140335 and A281873 are the denominatorial sequences for 3 and 4, respectively.
%Y Cf. A306349.
%K nonn
%O 1,2
%A _AJ Tatum_, Aug 31 2019