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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).
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%I #41 Sep 08 2022 08:45:05

%S 2,3,12,138,11937,73102188,2672848933402062,

%T 3572060905817696883164853272313,

%U 6379809557435582128907282471156933713351634534272773703460187

%N 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 The sets begin {1,2}, {3,4,5}, {6,7,8,...,17}, ...

%C 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 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 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 a(n+1) = a(n)*(2*A067339(n)-a(n)+1)/2. - _M. F. Hasler_, Jan 23 2015

%F a(n) ~ 2 * c^(2^n), where c = 1.312718001584962838462131787518361199185077166417566246117... . - _Vaclav Kotesovec_, Dec 09 2015

%p # Return [start,number,sum] of n-th group

%p A067338aux := proc(n)

%p local StrNumSu,Strt,Num,Su ;

%p option remember;

%p if n = 1 then

%p return [1,2,3] ;

%p else

%p strNumSu := procname(n-1) ;

%p Strt := strNumSu[1]+strNumSu[2] ;

%p Num := strNumSu[3] ;

%p Su := Num*(Num+2*Strt-1)/2 ;

%p return [Strt,Num,Su] ;

%p end if;

%p end proc:

%p A067338 := proc(n)

%p A067338aux(n)[2] ;

%p end proc:

%p seq(A067338(n),n=1..10) ; # _R. J. Mathar_, Jan 21 2015

%t 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 (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 (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 Cf. A067339 (partial sums).

%K easy,nonn

%O 1,1

%A _Floor van Lamoen_, Jan 16 2002

%E Corrected and extended by _Harvey P. Dale_ and _M. F. Hasler_, Jan 21 2015