A117805 Start with 3. Square the previous term and subtract it.

3, 6, 30, 870, 756030, 571580604870, 326704387862983487112030, 106735757048926752040856495274871386126283608870, 11392521832807516835658052968328096177131218666695418950023483907701862019030266123104859068030

Offset: 0

Author

Jacob Vecht, Apr 29 2006

Keywords

• easy
• nonn

Comments

The next term is too large to include.

a(n) = A005267(n+1)+1. - R. J. Mathar, Apr 22 2007. This is true by induction. - M. F. Hasler, May 04 2007<

For any a(0) > 2, the sequence a(n) = a(n-1) * (a(n-1) - 1) gives a constructive proof that there exists integers with at least n + 1 distinct prime factors, e.g., a(n). As a corollary, this gives a constructive proof of Euclid's theorem stating that there are an infinity of primes. - Daniel Forgues, Mar 03 2017

Examples

			Start with 3; 3^2 - 3 = 6; 6^2 - 6 = 30; etc.
		

Crossrefs

Cf. A007018.

Programs

Maple

f:=proc(n) option remember; if n=0 then RETURN(3); else RETURN(f(n-1)^2-f(n-1)); fi; end;

Mathematica

k=3;lst={k};Do[k=k^2-k;AppendTo[lst,k],{n,9}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
RecurrenceTable[{a[0]==3, a[n]==a[n-1]*(a[n-1] - 1)}, a, {n, 0, 10}] (* Vaclav Kotesovec, Dec 17 2014 *)
NestList[#^2-#&,3,10] (* Harvey P. Dale, Oct 11 2023 *)

Formula

a(0) = 3, a(n) = (a(n-1))^2 - a(n-1).

a(n) ~ c^(2^n), where c = 2.330283023986140936420341573975137247354077600883596774023675490739568138... . - Vaclav Kotesovec, Dec 17 2014