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).

2, 5, 17, 155, 12092, 73114280, 2672849006516342, 3572060905817699556013859788655, 6379809557435582128907282471160505774257452233828787563248842

Offset: 1

Author

Floor van Lamoen, Jan 16 2002

Keywords

• easy
• nonn

Comments

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

Crossrefs

Cf. A006894, A002658. Partial sums of A067338.

Programs

Mathematica

RecurrenceTable[{a[n] == a[n-1]*(a[n-1]+1)/2 + 2, a[1]==2}, a, {n, 1, 10}] (* Vaclav Kotesovec, Dec 09 2015 *)
NestList[(#(#+1))/2+2&,2,10] (* Harvey P. Dale, Jun 17 2017 *)

PARI

a(n) = if(n>1,a(n-1)*(a(n-1)+1)/2)+2 \\ Edited by M. F. Hasler, Jan 23 2015

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

Formula

a(n)=a(n-1)*(a(n-1)+1)/2 + 2

a(n)=a(n-1)+A067338(n). - M. F. Hasler, Jan 23 2015

a(n) ~ 2 * c^(2^n), where c = 1.312718001584962838462131787518361199185077166417566246117... . - Vaclav Kotesovec, Dec 09 2015

Extensions

More terms from Jason Earls, Jan 16 2002