A143782 Let b(0) = 3 and b(n) = b(n-1)^n - n^b(n-1) for n >= 1. a(n) = trunc(b(n)), where trunc(x) is floor(x) for x >= 0, ceiling(x) for x < 0.

3, 2, 0, -1, 0, -3, 898

Offset

0,1

Comments

The next number is too large to include.

Links

Table of n, a(n) for n=0..6.

Maple

b[0]:= 3:
for n from 1 to 7 do b[n]:= b[n-1]^n - n^b[n-1] od:
seq(trunc(b[i]), i=0..7); # Robert Israel, Nov 13 2024

Mathematica

lst={}; k=3; Do[k=k^n-n^k; AppendTo[lst, IntegerPart[k]], {n, 7}]; lst

Crossrefs

Sequence in context: A131290 A138741 A116604 * A303913 A117406 A290216

Adjacent sequences: A143779 A143780 A143781 * A143783 A143784 A143785

Keyword

sign

Author

Vladimir Joseph Stephan Orlovsky, Aug 31 2008

Extensions

Name clarified and a(0) inserted by Robert Israel, Nov 13 2024

Status

approved