A108582 n appears n^3 times.

1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5

Offset

1,2

Comments

From Jonathan Vos Post, Mar 18 2006: (Start)

The key to this sequence is: 1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+3+...+n)^2.

Since the last occurrence of n comes one before the first occurrence of n+1 and the former is at Sum_{i=0..n} i^3 = A000537(n) = (A000217(n))^2 = (n*(n+1)/2)^2 = (C(n+1,2))^2, have a(A000537(n)) = a((A000217(n))^2) = n and thus a(1+A000537(n)) = a(1+(A000217(n))^2) = n+1.

The current sequence is, loosely, the inverse function of the square of the triangular number sequence. (End)

Links

Boris Putievskiy, Table of n, a(n) for n = 1..8281

Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.

Formula

a(n) = ceiling((1/2)*(sqrt(8*sqrt(n) + 1) - 1)). - Boris Putievskiy, Jun 19 2024

Mathematica

Flatten @ Table[ Table[k, {k^3}], {k, 5}] (* Giovanni Resta, Jun 17 2016 *)
a[n_]:=Ceiling[1/2 (Sqrt[8 Sqrt[n]+1]-1)]
Nmax=225; Table[a[n], {n, 1, Nmax}] (* Boris Putievskiy, Jun 19 2024 *)

Crossrefs

Cf. A000027, A000578, A002024, A072649, A074279, A000217.

Cf. A000330, A000537, A006331, A050446, A050447, A006003, A005900.

Sequence in context: A069903 A331003 A086007 * A345378 A071840 A000193

Adjacent sequences: A108579 A108580 A108581 * A108583 A108584 A108585

Keyword

easy,nonn

Author

Jonathan Vos Post, Jul 25 2005

Extensions

Two missing terms from Giovanni Resta, Jun 17 2016

Status

approved