|
| |
|
|
A306686
|
|
Values of n such that 9^n ends in n, or expomorphic numbers relative to "base" 9.
|
|
6
|
|
|
|
9, 89, 289, 5289, 45289, 745289, 2745289, 92745289, 392745289, 7392745289, 97392745289, 597392745289, 7597392745289, 87597392745289, 8087597392745289, 48087597392745289, 748087597392745289, 10748087597392745289, 610748087597392745289, 5610748087597392745289
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 9) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]
For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.
Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).
a(n) is the backward concatenation of A133619(0) through A133619(n-1). So, a(1) = 9, a(2) = 89 and so on, with recognition of the former comments about the OEIS and terms beginning with 0. - Davis Smith, Mar 07 2019
|
|
|
LINKS
|
Charles W. Trigg, Problem 559, Crux Mathematicorum, page 192, Vol. 7, Jun. 81.
Emil Vaughan, Problem 226.8 - 999 nines, M500 Magazine of the Open University, number 226, February 2009, page 21; and Tony Forbes, Solution 226.8 - 999 nines, M500 Magazine of the Open University, number 232, February 2010, pages 8-9, calculating a(9) = 392745289.
|
|
|
EXAMPLE
|
9^9 = 387420489 ends in 9, so 9 is a term; 9^89 = .....289 ends in 89, so 89 is another term.
|
|
|
PROG
|
(PARI) tetrmod(b, n, m)=my(t=b); for(i=1, n, if(i>1, t=lift(Mod(b, m)^t), t)); t
for(n=1, 21, if(tetrmod(9, n, 10^n)!=tetrmod(9, n-1, 10^(n-1)), print1(tetrmod(9, n, 10^(n-1)), ", "))) \\ Davis Smith, Mar 09 2019
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,base
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
