A258252 Sequence of distinct positive integers having lowest possible denominators of sums of 1/a(n).

1, 2, 6, 3, 4, 12, 15, 10, 14, 35, 5, 30, 42, 7, 8, 24, 18, 9, 33, 88, 40, 60, 84, 63, 99, 22, 26, 143, 11, 154, 238, 51, 21, 28, 20, 55, 66, 78, 91, 56, 72, 90, 110, 132, 156, 13

Offset: 1

Author

Ivan Neretin, May 24 2015

Keywords

• nonn

Comments

a(n) is chosen among the unused positive integers so that the denominator of Sum_{i=1..n} 1/a(i), is as low as possible.

Presumably a permutation of positive integers.

Primes do not always occur in natural order.

The numbers which retain their natural positions (that is, a(n)=n) are 1, 2, 48, 80601...

Inverse (A258253): 1, 2, 4, 5, 11, 3, 14, 15, 18, 8, 29, 6, 46, 9, 7, 47, 73, 17, 134, 35, 33, 26, 153, 16, 96, ..., . - Robert G. Wilson v, Jun 18 2015

Records: 1, 2, 6, 12, 15, 35, 42, 88, 99, 143, 154, 238, 260, 460, 544, 840, 1645, 1666, 2109, 2622, 3876, 4599, 5644, 6565, 6734, 8701, 9492, 10272, ..., . - Robert G. Wilson v, Jun 18 2015

Examples

			After 2 terms, the partial sum of 1/a(i) reaches 3/2. Adding 1 or 1/2 is impossible, since 1 and 2 are already taken. The rest of positive integers lead to the following sums: +1/3 -> 11/6, +1/4 -> 7/4, +1/5 -> 17/10, +1/6 -> 5/3 with denominator 3 which is the lowest we can get. Hence a(3)=6.
For this specific term, the fractions that are encountered are 3/2 + 1/k with k>2. The resulting sequence of denominators are: 6, 4, 10, 3, 14, 8, 18, 5, 22, 12, ... (see A145979) within which the smallest term is indeed 3. - _Michel Marcus_, Jun 04 2015
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..25000
• Ivan Neretin, C++ program generating this sequence and A258254

Crossrefs

Cf. A157248 (another reordering of the harmonic series), A258253 (putative inverse), A258254 (denominators of partial sums of 1/a(n)), A258255 (positions where partial sums reach integers).

Programs

Mathematica

f[lst_] := Block[{c = 0, d, dk, k, mk, mn = Infinity, t = Total[1/lst]}, d = Denominator@ t; k = d; While[c < 101, If[ !MemberQ[lst, k], c++; dk = Denominator[t + 1/k]; If[dk < mn, mn = dk; mk = k]]; k += d]; Append[lst, mk]]; Nest[f, {}, 60] (* Robert G. Wilson v, Jun 18 2015 *)