A100140 Largest denominator of greedy Egyptian fraction sum for M/N.

2, 6, 4, 20, 6, 231, 24, 45, 20, 4070, 12, 2145, 231, 120, 240, 3039345, 45, 2359420, 180, 1428, 4070, 1019084, 120, 53307975, 2145, 1350, 1428, 1003066152, 120, 1127619917796295, 16800, 26796, 3039345, 1104740, 72, 884004, 2359420, 1288092

Offset: 2

Author

Robert Munafo, Nov 06 2004

Keywords

• nonn

Comments

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

Examples

			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.
		

References

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

Links

• Seiichi Manyama, Table of n, a(n) for n = 2..198
• Robert Munafo, Largest Denominator of Greedy Egyptian Fraction Sum for M/N
• Eric Weisstein's World of Mathematics, Egyptian Fractions.

Crossrefs

Cf. A050210, A098853, A260618.

Programs

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 thru 100 do ( max:2, for a:2 step 1 thru b-1 do ( if gcd(a,b)=1 then ( m : egypt(a/b), if m>max then max : m ) ), print("a[", b, "]=", max) ), t$

Extensions

a(6) corrected by Seiichi Manyama, Sep 18 2022