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A067338 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 number of elements in the n-th set gives a(n).

%I A067338 #41 Sep 08 2022 08:45:05
%S A067338 2,3,12,138,11937,73102188,2672848933402062,
%T A067338 3572060905817696883164853272313,
%U A067338 6379809557435582128907282471156933713351634534272773703460187
%N A067338 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 number of elements in the n-th set gives a(n).
%C A067338 The sets begin {1,2}, {3,4,5}, {6,7,8,...,17}, ...
%C A067338 Starting with {1}, one would get {1}, {2}, {3,4}, {5,6,7, 8,9,10, 11} ... with sums (1,2,7,56, 2212 ...) = A002658. - _M. F. Hasler_, Jan 21 2015
%F A067338 a(n)= a(n-1) *( 1 +2*[a(1)+a(2)+...+a(n-2)] +a(n-1) )/2. - Corrected by _R. J. Mathar_, Jan 22 2015
%F A067338 a(n) = a(n-1)*(2*a(n-1) + a(n-2)*a(n-1) + a(n-2)^2)/(2*a(n-2)). - _David W. Wilson_, Jan 22 2015
%F A067338 a(n+1) = a(n)*(2*A067339(n)-a(n)+1)/2. - _M. F. Hasler_, Jan 23 2015
%F A067338 a(n) ~ 2 * c^(2^n), where c = 1.312718001584962838462131787518361199185077166417566246117... . - _Vaclav Kotesovec_, Dec 09 2015
%p A067338 # Return [start,number,sum] of n-th group
%p A067338 A067338aux := proc(n)
%p A067338     local StrNumSu,Strt,Num,Su ;
%p A067338     option remember;
%p A067338     if n = 1 then
%p A067338         return [1,2,3] ;
%p A067338     else
%p A067338         strNumSu := procname(n-1) ;
%p A067338         Strt := strNumSu[1]+strNumSu[2] ;
%p A067338         Num := strNumSu[3] ;
%p A067338         Su := Num*(Num+2*Strt-1)/2 ;
%p A067338         return [Strt,Num,Su] ;
%p A067338     end if;
%p A067338 end proc:
%p A067338 A067338 := proc(n)
%p A067338     A067338aux(n)[2] ;
%p A067338 end proc:
%p A067338 seq(A067338(n),n=1..10) ; # _R. J. Mathar_, Jan 21 2015
%t A067338 RecurrenceTable[{a[n] == a[n-1]*(2*a[n-1] + a[n-2]*a[n-1] + a[n-2]^2)/(2*a[n-2]), a[1]==2, a[2]==3}, a, {n, 1, 10}] (* _Vaclav Kotesovec_, Dec 09 2015 *)
%o A067338 (PARI) print1(a=n=2);for(i=2,9,print1(","n=n*(a+a-n+1)/2);a+=n) \\ _M. F. Hasler_, Jan 21 2015
%o A067338 (Magma) I:=[2,3]; [n le 2 select I[n] else Self(n-1)*(2*Self(n-1) + Self(n-2)*Self(n-1) + Self(n-2)^2)/(2*Self(n-2)): n in [1..10]]; // _Vincenzo Librandi_, Jan 23 2015
%Y A067338 Cf. A067339 (partial sums).
%K A067338 easy,nonn
%O A067338 1,1
%A A067338 _Floor van Lamoen_, Jan 16 2002
%E A067338 Corrected and extended by _Harvey P. Dale_ and _M. F. Hasler_, Jan 21 2015