OFFSET
2,1
COMMENTS
The sequence eventually goes to zero, as can be seen by noting that multiples of the highest exponent (3 in this case) only go down; in fact the 8th term, a(8) = 7*8^2 + 7*8 + 7 = 511; after which the multiple of the square term will only go down, etc.
This sequence, for 11, grows beyond the quintillions of digits before going to zero.
REFERENCES
K. Hrbacek & T. Jech, Introduction to Set Theory, Taylor & Francis Group, 1999, pp. 125-127.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 2..1000
FORMULA
To obtain a(n + 1), write a(n) in base n, increase the base to n + 1 and subtract 1.
EXAMPLE
a(2) = 11 = 2^3 + 2^1 + 2^0
a(3) = 3^3 + 3^1 + 3^0 - 1 = 30
a(4) = 4^3 + 4^1 - 1 = 4^3 + 3*4^0 = 67
MATHEMATICA
nxt[{n_, a_}]:={n+1, FromDigits[IntegerDigits[a, n+1], n+2]-1}; Transpose[ NestList[ nxt, {1, 11}, 50]][[2]] (* Harvey P. Dale, Feb 09 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Nicholas Matteo (kundor(AT)kundor.org), Apr 15 2008
STATUS
approved