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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
a(n) = (A033677((10^n-1)/9)-A033676((10^n-1)/9))/2. - Chai Wah Wu, Apr 05 2021
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
def A048612(n):
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
Sequence in context: A299335 A247618 A269962 * A320554 A218135 A271122
KEYWORD
nonn,nice
AUTHOR
EXTENSIONS
Corrected and extended by Patrick De Geest, Jun 15 1999
More terms from Hans Havermann, Jul 02 2000
Offset corrected by Chai Wah Wu, Apr 05 2021
STATUS
approved

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Last modified March 19 06:25 EDT 2024. Contains 370953 sequences. (Running on oeis4.)