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%I #22 Mar 30 2020 06:40:59
%S 1,1,3,4,5,1,7,8,9,10,11,3,13,14,15,16,17,18,19,4,21,22,23,24,25,26,
%T 27,28,29,5,31,32,33,34,35,36,37,38,39,40,41,1,43,44,45,46,47,48,49,
%U 50,51,52,53,54,55,7,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,8,73,74,75,76
%N If n is an oblong number A002378, then a(n)=a(j) where j is the number of oblong numbers in (0,n], otherwise a(n)=n.
%C As a motivation, consider the greedy decomposition of fractions 1/n into Egyptian fractions,
%C n=1: 2,3,7,43,1807,3263443,.. A000058
%C n=2: 3,7,43,1807,3263443,10650056950807,.. A000058
%C n=3: 4,13,157,24493,599882557,359859081592975693,.. A082732
%C n=4: 5,21,421,176821,31265489221,977530816197201697621,.. A144779
%C n=5: 6,31,931,865831,749662454731,561993796032558961827631,.. A144780
%C n=6: 7,43,1807,3263443,10650056950807,.. A000058
%C n=7: 8,57,3193,10192057,103878015699193,.. A144781
%C n=8: 9,73,5257,27630793,763460694178057,.. A144782
%C n=9: 10,91,8191,67084291,4500302031888391,.. A144783
%C n=10: 11,111,12211,149096311,22229709804712411,.. A144784
%C n=11: 12,133,17557,308230693,95006159799029557,.. A144785
%C n=12: 13,157,24493,599882557,.. A082732
%C k=13: 14,183,33307,1109322943,..
%C where the first few denominators of 1/n = 1/b(1)+1/b(2)+... have been tabulated.
%C For some sets of n, the list b(i) of denominators is essentially the same: consider for example A000058, which represents primarily n=1, then in truncated form also n=2, and then n=6, n=42 etc. Or consider A082732 which represents n=3, then in truncated form n=12, n=156 etc.
%C The OEIS sequence assigns the primary n to a(n). The interpretation of a(n) with ascending n is: n=1 is primary, a(1)=1.
%C Decomposition of n=2 is equivalent to n=1, a(2)=1. Cases n=3 to 5 are primary ("original", "new"), and a(n)=n in these cases. n=6 is not new but essentially the same Egyptian series as seen for n=1, so a(6)=1. Cases n=7 to n=11 are "new" sequences, again a(n)=n in these cases, but then n=12 is represented by A082732 as already seen for n=3, so a(12)=3.
%C Because the first denominator for the decomposition of 1/n is 1/(n+1), n+1 belongs to the sequence of denominators of the expansion of 1/a(n).
%C The sequences b(.) have recurrences which are essentially 1+b(n-1)*(b(n-1)-1), looking up the oblong number at the position of the previous b(.). This is the reason why reverse look-up of the n via A000194 (number of oblong numbers up to n) as used in the definition is equivalent to the assignment described above.
%F a(n) = a(A000194(n+1)) if n in A002378. a(n) = n if n in A078358.
%e n=1 is not in A002378, so a(1)=1.
%e n=2 = A000058(2), so a(2)=1 because there is 1 oblong number <=2 and >0.
%e n=3 is not in A002378, so a(3)=3.
%e n=6 = A000058(3), so a(6)=a(2) because there are 2 oblong numbers <=6 and >0.
%Y Cf. A000058, A002378, A078358, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786.
%K nonn,easy,frac
%O 1,3
%A _Artur Jasinski_, Sep 22 2008, Sep 26 2008
%E a(57)=57 inserted, a(61)=61 corrected and better definition provided by _Omar E. Pol_, Dec 29 2008
%E I did some further editing of this entry, but many of the lines are still obscure. - _N. J. A. Sloane_, Dec 29 2008
%E Comments that connect to Egyptian fractions rephrased by _R. J. Mathar_, Oct 01 2009
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