1,3

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

n=13: 14,183,33307,1109322943,.. A179090

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.

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

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.

(Python)

from math import isqrt

from functools import lru_cache

from sympy.ntheory.primetest import is_square

@lru_cache(maxsize=None)

def A144786(n): return A144786(isqrt(m)-1>>1) if is_square(m:=(n<<2)|1) else n # Chai Wah Wu, Feb 12 2026

Cf. A000058, A002378, A078358, A082732, A144779, A144780, A144781, A144782, A144783, A144784, A144785, A179090.

Sequence in context: A135347 A370116 A345708 * A247369 A009389 A385046

Adjacent sequences: A144783 A144784 A144785 * A144787 A144788 A144789

nonn,easy

Artur Jasinski, Sep 22 2008, Sep 26 2008

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

approved