A281873 a(n+1) is the smallest number greater than a(n) such that Sum_{j=1..n+1} 1/a(j) <= 4, a(1) = 1.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 200, 77706, 16532869712, 3230579689970657935732, 36802906522516375115639735990520502954652700

Offset: 1

Author

Yuriy Sibirmovsky, Jan 31 2017

Keywords

• nonn
• fini
• full

Comments

The method for any number A is to find the largest harmonic number H(n) smaller than A, then use the greedy algorithm to expand the difference A - H(n).

A140335 is the same sequence for 3. The sequence for 5 consists of 99 terms, the largest of which has 142548 digits.

References

• A. M. Gleason, R. E. Greenwood, and L. M. Kelly, The William Lowell Putnam Mathematical Competition, Problems and Solutions, 1938-1964, MAA, 1980, pages 398-399.

Links

• John Scholes, 14th Putnam Mathematical Competition, 1954, Problem B6, after Gleason, Greenwood & Kelly.
• Index entries for sequences related to Egyptian fractions

Crossrefs

Cf. A140335, A306349.

Programs

Mathematica

x0=4-Sum[1/k,{k,1,30}];
Nm=10;
j=0;
While[x0>0||j==Nm,a0=Ceiling[1/x0];
x0=x0-1/a0;
Print[a0];j++]
f[s_List, n_] := Block[{t = Total[1/s]}, Append[s, Max[ s[[-1]] +1, Ceiling[1/(n - t)]]]]; Nest[f[#, 4] &, {1}, 34] (* Robert G. Wilson v, Feb 05 2017 *)

Python

from sympy import egyptian_fraction
print(egyptian_fraction((4, 1))) # Pontus von Brömssen, Feb 10 2019

Formula

Sum_{k=1..35} 1/a(k) = 4.