OFFSET
0,2
COMMENTS
When A225401(k) = 0, i.e. k is a term of A353003, then a(k) > a(k+1); 1st example is for k = 3 with a(3) = 1753 > a(4) = 753; otherwise, a(n) < a(n+1).
When n <> k, a(n) coincides with the 'backward concatenation' of A225401(n-1) up to A225401(0), where A225401 is the 10-adic integer x such that x^3 = -7/9 (see table in Example section); when n= k, a(k) must be calculated directly with the definition.
Without "exactly" in the name, terms a'(n) should be: 1, 3, 53, 753, 753, 60753, 660753, ...
There are similar sequences when cubes end with 1, 3, 8 or 9; but there's no similar sequence for squares, because when a square ends in more than three identical digits, these digits are necessarily 0.
EXAMPLE
a(1) = 3 because 3^3 = 27;
a(2) = 53 because 53^2 = 148877;
a(3) = 1753 because 1753^3 = 5386984777;
a(4) = 753 because 753^2 = 426957777;
a(5) = 60753 because 60753^3 = 224234888577777.
Table with a(n) and A225401(n-1)
---------------------------------------------------------------------------
| | a(n) | a'(n) | A225401(n-1) | concatenation |
| n | with "exactly" | without "exactly" | = b(n-1) | b(n-1)...b(0) |
---------------------------------------------------------------------------
0 1 1
1 3 3 3 ...3
2 53 53 5 ...53
3 1753 753 7 ...753
4 753 753 0 ...0753
5 60753 60753 6 ...60753
6 660753 660753 6 ...660753
7 9660753 9660753 9 ...9660753
..........................................................................
Also, as A225401(23) = 0, we have from a(21) up to a(25):
a(21) = 559631497233899660753;
a(22) = 3559631497233899660753;
a(23) = 193559631497233899660753, found by Marius A. Burtea;
a(24) = 93559631497233899660753;
a(25) = 2093559631497233899660753.
PROG
(Python)
def a(n):
k, s, target = 1, "1", "7"*n
while s.rstrip("7") + target != s: k += 1; s = str(k**3)
return k
print([a(n) for n in range(8)]) # Michael S. Branicky, Apr 14 2022
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Apr 14 2022
EXTENSIONS
a(8)-a(9) from Marius A. Burtea, Apr 14 2022
STATUS
approved