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A099153
Iterated heptagonal numbers (A000566), starting at 7.
7
1, 7, 112, 31192, 2432305372, 14790273553001687902, 546880479431552932161867875823030372157, 747695646958212974238278880467821187888728169501525193422768463793490256523387
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
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
Sequence in context: A010795 A377639 A293456 * A270121 A079296 A081531
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Nov 15 2004
EXTENSIONS
Corrected and extended by R. J. Mathar, Jun 09 2008
STATUS
approved