

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



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
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OFFSET

1,2


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, 19381964, MAA, 1980, pages 398399.


LINKS

Table of n, a(n) for n=1..35.
John Scholes, 14th Putnam Mathematical Competition, 1954, Problem B6, after Gleason, Greenwood & Kelly.
Index entries for sequences related to Egyptian fractions


FORMULA

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


MATHEMATICA

x0=4Sum[1/k, {k, 1, 30}];
Nm=10;
j=0;
While[x0>0j==Nm, a0=Ceiling[1/x0];
x0=x01/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 *)


PROG

(Python)
from sympy import egyptian_fraction
print(egyptian_fraction(4)) # Pontus von BrÃ¶mssen, Feb 10 2019


CROSSREFS

Cf. A140335, A306349.
Sequence in context: A230034 A269331 A246103 * A273888 A192218 A070915
Adjacent sequences: A281870 A281871 A281872 * A281874 A281875 A281876


KEYWORD

nonn,fini,full,changed


AUTHOR

Yuriy Sibirmovsky, Jan 31 2017


STATUS

approved



