|
|
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).
|
|
1
|
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
The sets begin {1, 2}, {3, 4, 5}, {6, 7, 8, ..., 17}, ...
|
|
LINKS
|
|
|
FORMULA
|
a(n)=a(n-1)*(a(n-1)+1)/2 + 2
a(n) ~ 2 * c^(2^n), where c = 1.312718001584962838462131787518361199185077166417566246117... . - Vaclav Kotesovec, Dec 09 2015
|
|
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 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|