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.

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

Offset: 1

Author

Felice Russo

Keywords

• nonn
• nice

Comments

Least solutions for 'Difference between two squares is a repunit of length n'.

Examples

			For n=2, 6^2 - 5^2 = 11.
		

References

• David Wells, "Curious and Interesting Numbers", Revised Ed. 1997, Penguin Books, p. 119. ISBN 0-14-026149-4.

Links

• H. Havermann, Repunit Square Differences (gives many more terms)

Crossrefs

Cf. A048611, A000042, A002275, A033676, A033677.

Programs

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 *)

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

Formula

a(n) = (A033677((10^n-1)/9)-A033676((10^n-1)/9))/2. - Chai Wah Wu, Apr 05 2021

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