|
|
A048612
|
|
Find smallest pair (x,y) such that x^2-y^2 = 11...1 (n times) = (10^n-1)/9; sequence gives value of y.
|
|
4
|
|
|
0, 5, 17, 45, 115, 67, 2205, 2933, 166667, 44445, 245795, 6667, 132683733, 4444445, 2012917, 23767083, 2680575317, 666667, 555555555555555555, 83053525, 3263104267, 12488376483, 5555555555555555555555, 66666667, 2952525627555
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Least solutions for 'Difference between two squares is a repunit of length n'.
|
|
REFERENCES
|
David Wells, "Curious and Interesting Numbers", Revised Ed. 1997, Penguin Books, p. 119. ISBN 0-14-026149-4.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
For n=2, 6^2 - 5^2 = 11.
|
|
MATHEMATICA
|
s = Flatten[Table[r = (10^i - 1)/9; d = Divisors[r]; p = d[[Length[d]/2]]; Solve[{x - y == p, x + y == r/p}, {y, x}], {i, 2, 56}]]; Prepend[Cases[s, Rule[y, n_] -> n], 0]
Join[{0}, Table[y/.Solve[{x>0, y>0, x^2-y^2==FromDigits[PadRight[{}, n, 1]]}, {x, y}, Integers][[1]], {n, 2, 30}]](* Harvey P. Dale, Jun 12 2018 *)
|
|
PROG
|
(Python)
from sympy import divisors
d = divisors((10**n-1)//9)
l = len(d)
return (d[l//2]-d[(l-1)//2])//2 # Chai Wah Wu, Apr 05 2021
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,nice
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|