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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 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

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Last modified September 22 08:40 EDT 2023. Contains 365519 sequences. (Running on oeis4.)