OFFSET
0,1
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.
From Zhuorui He, Aug 07 2025: (Start)
This sequence has A266203(11)+1 terms and a(A266203(11))=0 is the last term of this sequence. The maximum term in this sequence is a((A266203(11)-1)/2)=(A266203(11)+1)/2. 10^^8 < A266203(11) < 10^^9.
More precisely, 10^(10^(10^(10^(10^(10^(10^619.29937)))))) < A266203(11) < 10^(10^(10^(10^(10^(10^(10^619.299371)))))). (End)
REFERENCES
K. Hrbacek and T. Jech, Introduction to Set Theory, Taylor & Francis Group, 1999, pp. 125-127.
LINKS
Zhuorui He, Table of n, a(n) for n = 0..10000 (Terms 0..998 from Harvey P. Dale)
FORMULA
To obtain a(n + 1), write a(n) in base n + 2, increase the base to n + 3 and subtract 1.
EXAMPLE
a(0) = 11 = 2^3 + 2^1 + 2^0
a(1) = 3^3 + 3^1 + 3^0 - 1 = 30
a(2) = 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 *)
PROG
(PARI) a(n, m=11) = { my(wn = m); for (k=2, n+1, wn = fromdigits(digits(wn, k), k+1) - 1); wn; } \\ Zhuorui He, Aug 08 2025
CROSSREFS
KEYWORD
nonn,fini
AUTHOR
Nicholas Matteo (kundor(AT)kundor.org), Apr 15 2008
EXTENSIONS
Offset changed to 0 by Zhuorui He, Aug 07 2025
STATUS
approved
