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

1,2

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

Table of n, a(n) for n=1..25.

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

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

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

Sequence in context: A299335 A247618 A269962 * A320554 A218135 A271122

Adjacent sequences: A048609 A048610 A048611 * A048613 A048614 A048615

Keyword

nonn,nice

Author

Felice Russo

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