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 A099153 Iterated heptagonal numbers (A000566), starting at 7. 7
 1, 7, 112, 31192, 2432305372, 14790273553001687902, 546880479431552932161867875823030372157, 747695646958212974238278880467821187888728169501525193422768463793490256523387 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The number of digits approximately doubles moving to the next member in the sequence; therefore a(8) onwards are not shown. - R. J. Mathar, Jun 09 2008 REFERENCES J. V. Post, "Iterated Polygonal Numbers", preprint. J. V. Post, "Iterated Triangular Numbers", preprint. LINKS Eric Weisstein's World of Mathematics, "Heptagonal Number." FORMULA 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, Jun 09 2008 EXAMPLE 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. CROSSREFS Cf. A007501, A000566. Sequence in context: A147631 A010795 A293456 * A270121 A079296 A081531 Adjacent sequences:  A099150 A099151 A099152 * A099154 A099155 A099156 KEYWORD easy,nonn AUTHOR Jonathan Vos Post, Nov 15 2004 EXTENSIONS Corrected and extended by R. J. Mathar, Jun 09 2008 STATUS approved

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Last modified November 18 10:08 EST 2019. Contains 329261 sequences. (Running on oeis4.)