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 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. 3
 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 (list; graph; refs; listen; history; text; internal format)
 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, 1938-1964, MAA, 1980, pages 398-399. LINKS John Scholes, 14th Putnam Mathematical Competition, 1954, Problem B6, after Gleason, Greenwood & Kelly. FORMULA Sum_{k=1..35} 1/a(k) = 4. 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 *) 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 AUTHOR Yuriy Sibirmovsky, Jan 31 2017 STATUS approved

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Last modified October 18 12:18 EDT 2019. Contains 328160 sequences. (Running on oeis4.)