A265432 a(n) = smallest k with concat(1,k) and concat(n,k) both square numbers.
225, 6, 1025, 6, 225, 9937257544619140625, 80625, 225, 19025, 14797831640625, 5625, 89450791534674072265625, 96, 69, 44, 21, 1993672119140625, 2002541101386962890625, 225, 6, 8734765625, 99758030615478515625, 5625, 863225, 80625, 6, 40625, 225, 890625, 158764150390625
Offset: 0
Examples
a(0) = 225 because 1225 is a square as is (0)225. (In other words, 225 is the first term in A272672 that is itself a square). - _N. J. A. Sloane_, May 21 2016 a(2) = 1025 because concat(1,1025) = 11025 = 105^2 and concat(2,1025) = 21025 = 145^2.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 0..20000, May 25 2016 [First 10000 terms from David W. Wilson, Dec 08 2015]
- N. J. A. Sloane, Table of n, a(n), sqrt(concat(n,a(n))), sqrt(concat(1,a(n))) (based on _David W. Wilson_ 10000-term b-file)
- David W. Wilson, Table of n, a(n) for n = 0..10000, May 25 2016 [The graph of the first 10000 terms looks rather different from the graph of the first 20000 terms, so this file is worth preserving. - _N. J. A. Sloane_, May 25 2016]
Crossrefs
Programs
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Mathematica
<< Combinatorica` A265432[0] = 225; A265432[1] = 6; A265432[n_] := Block[{x = {-1, 1, 0, 1}[[Mod[n, 4, 1]]], d = Infinity, l, i}, While[d > Sqrt[10.0^(x - 1)] (Sqrt[10.0 n + 1] - Sqrt[11.0]), x++; d = Infinity; l = Divisors[((n - 1) 10^x)/4]; i = BinarySearch[l, 0.5 Sqrt[(n + 1) 10.0^x - 1] - 0.5 Sqrt[2*10.0^x - 1]]; If[i <= Length@l, d = 2*l[[i + 1/2]]]]; (((n - 1) 10^x - d^2)/(2 d))^2 - 10^x] (* Davin Park, Apr 11 2017 *)
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