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A048611 Find smallest pair (x,y) such that x^2 - y^2 = 11...1 (n times) = (10^n-1)/9; sequence gives value of x. 3

%I #24 Apr 05 2021 13:33:03

%S 1,6,20,56,156,340,2444,4440,167000,55556,267444,333400,132687920,

%T 5555556,10731400,40938800,2682647040,333334000,555555555555555556,

%U 3334367856,11034444280,35595935980,5555555555555555555556

%N Find smallest pair (x,y) such that x^2 - y^2 = 11...1 (n times) = (10^n-1)/9; sequence gives value of x.

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

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

%H H. Havermann, <a href="http://chesswanks.com/pxp/RSD.html">Repunit Square Differences (gives many more terms)</a>

%F a(n) = (A033677((10^n-1)/9)+A033676((10^n-1)/9))/2. - _Chai Wah Wu_, Apr 05 2021

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

%t 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[x, n_] -> n], 1]

%o (Python)

%o from sympy import divisors

%o def A048611(n):

%o d = divisors((10**n-1)//9)

%o l = len(d)

%o return (d[l//2]+d[(l-1)//2])//2 # _Chai Wah Wu_, Apr 05 2021

%Y Cf. A048612, A000042, A002275, A033676, A033677.

%K nonn,nice

%O 1,2

%A _Felice Russo_

%E Corrected and extended by _Patrick De Geest_, Jun 15 1999

%E More terms from _Hans Havermann_, Jul 02 2000

%E Offset corrected by _Chai Wah Wu_, Apr 05 2021

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Last modified March 28 08:22 EDT 2024. Contains 371236 sequences. (Running on oeis4.)