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A067339 Divide the natural numbers in sets of consecutive numbers, starting with {1,2}, each set with number of elements equal to the sum of elements of the preceding set. The final element of the n-th set gives a(n).

%I A067339 #14 Dec 15 2017 17:35:50
%S A067339 2,5,17,155,12092,73114280,2672849006516342,
%T A067339 3572060905817699556013859788655,
%U A067339 6379809557435582128907282471160505774257452233828787563248842
%N A067339 Divide the natural numbers in sets of consecutive numbers, starting with {1,2}, each set with number of elements equal to the sum of elements of the preceding set. The final element of the n-th set gives a(n).
%C A067339 The sets begin {1, 2}, {3, 4, 5}, {6, 7, 8, ..., 17}, ...
%F A067339 a(n)=a(n-1)*(a(n-1)+1)/2 + 2
%F A067339 a(n)=a(n-1)+A067338(n). - _M. F. Hasler_, Jan 23 2015
%F A067339 a(n) ~ 2 * c^(2^n), where c = 1.312718001584962838462131787518361199185077166417566246117... . - _Vaclav Kotesovec_, Dec 09 2015
%t A067339 RecurrenceTable[{a[n] == a[n-1]*(a[n-1]+1)/2 + 2, a[1]==2}, a, {n, 1, 10}] (* _Vaclav Kotesovec_, Dec 09 2015 *)
%t A067339 NestList[(#(#+1))/2+2&,2,10] (* _Harvey P. Dale_, Jun 17 2017 *)
%o A067339 (PARI) a(n) = if(n>1,a(n-1)*(a(n-1)+1)/2)+2 \\ Edited by _M. F. Hasler_, Jan 23 2015
%o A067339 (PARI) vector(10,i,if(i>1,n=n*(a+a-n+1)/2;a+=n,n=a=2)) \\ _M. F. Hasler_, Jan 23 2015
%Y A067339 Cf. A006894, A002658. Partial sums of A067338.
%K A067339 easy,nonn
%O A067339 1,1
%A A067339 _Floor van Lamoen_, Jan 16 2002
%E A067339 More terms from _Jason Earls_, Jan 16 2002