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Search: id:A108582
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| 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, 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; listen)
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OFFSET
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1,2
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COMMENT
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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) = (A0000217(n))^2 = (n*(n+1)/2)^2 = (C(n+1,2))^2, have A108582(A000537(n)) = A108582((A0000217(n))^2) = n, and thus A108582(1+A0000217(n)) = A108582(1+(A0000217(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 (jvospost2(AT)yahoo.com), Mar 18 2006
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CROSSREFS
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Cf. A000027, A000578, A002024, A072649, A072649, A074279.
Cf. A000217, A000330, A000537, A002024, A006331, A050446, A050447, A000537, A006003, A005900, A074279.
Sequence in context: A104230 A069903 A086007 this_sequence A071840 A000193 A059939
Adjacent sequences: A108579 A108580 A108581 this_sequence A108583 A108584 A108585
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KEYWORD
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easy,nonn
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AUTHOR
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Jonathan Vos Post (jvospost2(AT)yahoo.com), Jul 25 2005
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