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A144786
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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.
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10
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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, 27, 28, 29, 5, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 1, 43, 44, 45, 46, 47, 48, 49, 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
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OFFSET
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1,3
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COMMENTS
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As a motivation, consider the greedy decomposition of fractions 1/n into Egyptian fractions,
n=1: 2,3,7,43,1807,3263443,.. A000058
n=2: 3,7,43,1807,3263443,10650056950807,.. A000058
n=3: 4,13,157,24493,599882557,359859081592975693,.. A082732
n=4: 5,21,421,176821,31265489221,977530816197201697621,.. A144779
n=5: 6,31,931,865831,749662454731,561993796032558961827631,.. A144780
n=6: 7,43,1807,3263443,10650056950807,.. A000058
n=7: 8,57,3193,10192057,103878015699193,.. A144781
n=8: 9,73,5257,27630793,763460694178057,.. A144782
n=9: 10,91,8191,67084291,4500302031888391,.. A144783
n=10: 11,111,12211,149096311,22229709804712411,.. A144784
n=11: 12,133,17557,308230693,95006159799029557,.. A144785
n=12: 13,157,24493,599882557,.. A082732
k=13: 14,183,33307,1109322943,..
where the first few denominators of 1/n = 1/b(1)+1/b(2)+... have been tabulated.
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.
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.
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.
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).
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.
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LINKS
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FORMULA
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EXAMPLE
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n=2 = A000058(2), so a(2)=1 because there is 1 oblong number <=2 and >0.
n=6 = A000058(3), so a(6)=a(2) because there are 2 oblong numbers <=6 and >0.
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CROSSREFS
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Cf. A000058, A002378, A078358, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786.
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KEYWORD
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nonn,easy,frac
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AUTHOR
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EXTENSIONS
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a(57)=57 inserted, a(61)=61 corrected and better definition provided by Omar E. Pol, Dec 29 2008
I did some further editing of this entry, but many of the lines are still obscure. - N. J. A. Sloane, Dec 29 2008
Comments that connect to Egyptian fractions rephrased by R. J. Mathar, Oct 01 2009
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STATUS
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approved
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