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A099153 Iterated heptagonal numbers (A000566), starting at 7. +0
7
1, 7, 112, 31192, 2432305372, 14790273553001687902, 546880479431552932161867875823030372157, 747695646958212974238278880467821187888728169501525193422768463793490256523387 (list; graph; listen)
OFFSET

0,2

COMMENT

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

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 (mathar(AT)strw.leidenuniv.nl), 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: A128576 A147631 A010795 this_sequence A079296 A081531 A142537

Adjacent sequences: A099150 A099151 A099152 this_sequence A099154 A099155 A099156

KEYWORD

easy,nonn

AUTHOR

Jonathan Vos Post (jvospost3(AT)gmail.com), Nov 15 2004

EXTENSIONS

Corrected and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 09 2008

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Last modified November 30 13:13 EST 2009. Contains 167758 sequences.


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