

A108582


n appears n^3 times.


1



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
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OFFSET

1,2


COMMENTS

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 A108582(A000537(n)) = A108582((A000217(n))^2) = n and thus A108582(1+A000217(n)) = A108582(1+(A000217(n))^2) = n+1. The current sequence is, loosely, the inverse function of the square of the triangular number sequence. See also: A000537 Sum of first n cubes. See also: A002024 n appears n times. See also: A074279 n appears n^2 times.  Jonathan Vos Post, Mar 18 2006


LINKS

Table of n, a(n) for n=1..107.


MATHEMATICA

Flatten @ Table[ Table[k, {k^3}], {k, 5}] (* Giovanni Resta, Jun 17 2016 *)


CROSSREFS

Cf. A000027, A000578, A002024, A072649, A072649, A074279.
Cf. A000217, A000330, A000537, A002024, A006331, A050446, A050447, A000537, A006003, A005900, A074279.
Sequence in context: A069903 A331003 A086007 * A071840 A000193 A059939
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



