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A281873
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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.
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3
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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
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1,2
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COMMENTS
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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.
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REFERENCES
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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.
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LINKS
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FORMULA
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Sum_{k=1..35} 1/a(k) = 4.
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MATHEMATICA
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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 *)
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PROG
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(Python)
from sympy import egyptian_fraction
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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STATUS
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approved
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