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 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified June 6 11:08 EDT 2020. Contains 334830 sequences. (Running on oeis4.)