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Largest denominator of greedy Egyptian fraction sum for M/N.
2

%I #16 Sep 21 2024 15:09:13

%S 2,6,4,20,6,231,24,45,20,4070,12,2145,231,120,240,3039345,45,2359420,

%T 180,1428,4070,1019084,120,53307975,2145,1350,1428,1003066152,120,

%U 1127619917796295,16800,26796,3039345,1104740,72,884004,2359420,1288092

%N Largest denominator of greedy Egyptian fraction sum for M/N.

%C Each term gives the largest of the N-1 terms in A050210 corresponding to the fractions with denominator N.

%D Guy, R. K. "Egyptian Fractions." section D11 in "Unsolved Problems in Number Theory", 2nd ed. New York: Springer-Verlag, pp. 158-166, 1994.

%H Seiichi Manyama, <a href="/A100140/b100140.txt">Table of n, a(n) for n = 2..198</a>

%H Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-a100140.html">Largest Denominator of Greedy Egyptian Fraction Sum for M/N</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fractions</a>.

%e Consider a(5). There are 4 fractions with 5 in the denominator: 1/5=1/5, 2/5=1/3+1/15, 3/5=1/2+1/10 and 4/5=1/2+1/4+1/20. Of these, the largest denominator is 20, so a(5)=20.

%o (Maxima) /* MACSYMA or maxima */ egypt(x) := block([i,n,d,p,e, on, od], ( n : num(x), d : n/x, on : n, od : d, p : 0, e : [], for i:1 while x>0 do ( n : num(x), d : n/x, p : fix((d+n-1)/n), x : x - 1/p, e : append(e, [p]) ), return(p) ) ); for b:2 step 1 through 100 do ( max:2, for a:2 step 1 through b-1 do ( if gcd(a,b)=1 then ( m : egypt(a/b), if m>max then max : m ) ), print("a[", b, "]=", max) ), t$

%Y Cf. A050210, A098853, A260618.

%K nonn

%O 2,1

%A _Robert Munafo_, Nov 06 2004

%E a(6) corrected by _Seiichi Manyama_, Sep 18 2022