

A144786


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.


10



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

1,3


COMMENTS

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(n1)*(b(n1)1), looking up the oblong number at the position of the previous b(.). This is the reason why reverse lookup of the n via A000194 (number of oblong numbers up to n) as used in the definition is equivalent to the assignment described above.


LINKS

Table of n, a(n) for n=1..76.


FORMULA

a(n) = a(A000194(n+1)) if n in A002378. a(n) = n if n in A078358.


EXAMPLE

n=1 is not in A002378, so a(1)=1.
n=2 = A000058(2), so a(2)=1 because there is 1 oblong number <=2 and >0.
n=3 is not in A002378, so a(3)=3.
n=6 = A000058(3), so a(6)=a(2) because there are 2 oblong numbers <=6 and >0.


CROSSREFS

Cf. A000058, A002378, A078358, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A144786.
Sequence in context: A280490 A135347 A345708 * A247369 A009389 A091828
Adjacent sequences: A144783 A144784 A144785 * A144787 A144788 A144789


KEYWORD

nonn,easy,frac


AUTHOR

Artur Jasinski, Sep 22 2008, Sep 26 2008


EXTENSIONS

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


STATUS

approved



