|
%I #31 Sep 29 2023 19:42:23
%S 0,1,6,231,1186570,347357071281165,2076895351339769460477611370186681,
%T 143892868802856286225154411591351342616163027795335641150249224655238508171
%N Let T(n) be the n-th triangular number n*(n+1)/2; then a(n) = n-th iteration T(T(T(...(n)))).
%C The next term, a(8), has 162 digits. - _Harvey P. Dale_, May 29 2013
%H Alois P. Heinz, <a href="/A099129/b099129.txt">Table of n, a(n) for n = 0..10</a>
%F a(n) = A000217^n(n).
%F The sequence grows like O(n^2^n*1/2^n). This can be derived from the growth O(n^2*1/2) of the triangle sum by iteration. - _Hieronymus Fischer_, Jan 21 2006
%e a(3) = 231 because we can write the 3-time iterated expression on T(3), the triangular number sequence n*(n+1)/2, namely: T(T(T(3))) = 231.
%p a:= n-> (t-> (t@@n)(n))(j-> j*(j+1)/2):
%p seq(a(n), n=0..7); # _Alois P. Heinz_, Sep 29 2023
%t Table[Nest[(#(#+1))/2&,n,n],{n,8}] (* _Harvey P. Dale_, May 29 2013 *)
%o (PARI) a(n) = my(k = n); for (j=1, n, k = k*(k+1)/2;); k; \\ _Michel Marcus_, Jan 01 2017
%Y Cf. A000217, A007501, A058009 (analog with primes), A097547.
%K nonn,easy
%O 0,3
%A _Jonathan Vos Post_, Nov 14 2004
%E Offset changed to 1 by _Georg Fischer_, Jun 20 2022
%E a(0)=0 prepended by _Alois P. Heinz_, Sep 29 2023
| 