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Search: id:A099153
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| 1, 7, 112, 31192, 2432305372, 14790273553001687902, 546880479431552932161867875823030372157, 747695646958212974238278880467821187888728169501525193422768463793490256523387
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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The number of digits approximately doubles moving to the next member in the sequence; therefore a(8) onwards are not shown. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 09 2008
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REFERENCES
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J. V. Post, "Iterated Polygonal Numbers", preprint.
J. V. Post, "Iterated Triangular Numbers", preprint.
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LINKS
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Eric Weisstein's World of Mathematics, "Heptagonal Number."
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FORMULA
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a(0, n) = 1. a(1, n) = Hep(n) = the n-th heptagonal number = n*(5*n-3)/2. a(2, n) = Hep(Hep(n)) = the Hep(n)th heptagonal number = [n*(5*n-3)/2]*{5*n*(5*n-3)/2-3}/2 = (1/4)*{[Hep(n)]^2 - 3*Hep(n)}. a(3, n) = Hep(Hep(Hep(n)))) = (1/8)*{125*[Hep(n)]^4 - 90*[Hep(n)]^3 + 9*[Hep(n)]^2} = (1/8)*{78125*n^8 - 187500*n^7 + 150000*n^6 - 33750*n^5 - 9375*n^4 + 3150*n^3 + 315*n^2 - 27*n}. In general, a(k+1, n) = Hep[a(k, n)] = a(k, n)* [5*a(k, n)-3]/2.
a(n)= A000566(a(n-1)), n>1. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 09 2008
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EXAMPLE
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a(3) = 31192 because a(1) = the first heptagonal number = 7; a(2) = the 7th heptagonal number = 7*(5*7-3)/2 = 112; a(3) = the 112th heptagonal number = 112*(5*112-3)/2 = 31192.
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CROSSREFS
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Cf. A007501, A000566.
Sequence in context: A128576 A147631 A010795 this_sequence A079296 A081531 A142537
Adjacent sequences: A099150 A099151 A099152 this_sequence A099154 A099155 A099156
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KEYWORD
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easy,nonn
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AUTHOR
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Jonathan Vos Post (jvospost3(AT)gmail.com), Nov 15 2004
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EXTENSIONS
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Corrected and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 09 2008
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