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A099179 Iterated tetrahedral numbers, starting at Tet(2) = 4. 6

%I

%S 2,4,20,1540,609896980,37811003218473324699257860,

%T 9009555207802177724984164589516456320805205201729086740415363658290866918420

%N Iterated tetrahedral numbers, starting at Tet(2) = 4.

%C The next term, a(8), has 228 digits. - _Harvey P. Dale_, Dec 18 2012

%D J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 83.

%D H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.

%D J. V. Post, "Iterated Triangular Numbers", preprint.

%H J. V. Post, <a href="http://magicdragon.com/poly.html">Table of Polytope Numbers, Sorted, Through 1,000,000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TetrahedralNumber.html">"Tetrahedral Number."</a>

%F Given the tetrahedral number formula Tet(n) = n*(n+1)*(n+2)/6, define a(1) = 2; a(2) = the 2nd tetrahedral number = 2*(2+1)*(2+2)/6 = 4; Define a(k+1) = Tet(a(k)) = a(k)*[a(k)+1]*[a(k)+2]/6.

%F a(n)= A000292(a(n-1)). - _R. J. Mathar_, Jun 09 2008

%e a(2) = Tet(2) = the 2nd tetrahedral number = 2*(2+1)*(2+2)/6 = 4;

%e a(3) = Tet(Tet(2)) = the 4th tetrahedral number = 4*(4+1)*(4+2)/6 = 20;

%e a(4) = Tet(Tet(Tet(2))) = the 20th tetrahedral number = 20*(20+1)*(20+2)/6 = 1540.

%t NestList[(#(#+1)(#+2))/6&,2,6] (* _Harvey P. Dale_, Dec 18 2012 *)

%Y Cf. A007501, A000292.

%K easy,nonn

%O 1,1

%A _Jonathan Vos Post_, Nov 15 2004

%E Corrected and extended by _R. J. Mathar_, Jun 09 2008

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Last modified December 16 01:32 EST 2019. Contains 330013 sequences. (Running on oeis4.)