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Search: id:A099185
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| A099185 |
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Iterated octahedral numbers, starting at oct(2) = 6. |
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+0
1
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| 1, 6, 146, 2074806, 5954444157018557346, 140744820294208035204656447906095566299588102457814757606, 18586858963650566404526041827782437558782101283254936314363949423284878019246427\ 07284183892998994140529418479176238949509991689659112331788418314832322689820822\ 344586546
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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This need not start at oct(2) = 6. For example, if a(1) = oct(3) = 19, then a(4)= oct(19) = 4579; a(5) = oct(4579) = 64005999219; a(6) = oct(64005999219) = 174811816875659072517015216413379.
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REFERENCES
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Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, p. 50, 1996.
Dickson, L. E. History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Chelsea, 1952.
J. V. Post, "Iterated Triangular numbers", preprint.
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LINKS
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J. V. Post, Table of Polytope Numbers, Sorted, Through 1,000,000.
Eric Weisstein's World of Mathematics, "Octahedral Number."
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FORMULA
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Given the octahedral number formula oct(n) = (2*n^3 + n)/3, define: a(0, n) = 0; a(1, n) = oct(n); a(2, n) = oct(oct(n)); in general for k>0 a(k+1, n) = oct(a(k, n)); the octahedral number of the octahedral number of ... of n. a(n) = 2*a(n-1) + 3; generating function = 1/(exp(x)-1).
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EXAMPLE
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a(3) = 2074806 because a(1) = the 2nd octahedral number = oct(2) = 6; a(2) = oct(oct(2)) = the 6th octahedral number = oct(6) = (2*6^3 + 6)/3 = 146; a(3) = oct(oct(oct(2))) = the 146th octahedral number = oct(146) = (2*146^3 + 146)/3 = 2074806.
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CROSSREFS
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Cf. A007501, A005900.
Sequence in context: A166837 A166809 A063419 this_sequence A065986 A012791 A089480
Adjacent sequences: A099182 A099183 A099184 this_sequence A099186 A099187 A099188
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KEYWORD
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easy,nonn,uned
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AUTHOR
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Jonathan Vos Post (jvospost3(AT)gmail.com), Nov 15 2004
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