A048611 -id:A048611 - cp's OEIS frontend

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

Felice Russo

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

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

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

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

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

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 *)

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

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

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

cp's OEIS Frontend

A039668 Replaced by the pair of sequences A048611 and A048612.

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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.

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