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a(1) = 4, a(n) = smallest number > a(n-1) such that the concatenation of a(n-1) and a(n) is a square.
8

%I #10 Oct 13 2017 03:31:12

%S 4,9,61,504,4516,47504,382025,3975209,33057329,80214016,454665681,

%T 4507966404,44168848384,69005350809,163894140625,784386132324,

%U 5954843762641,7954794246144,53996843222416,69176076458289,379510987739761,1641640879622564,7593632535763529,31733339799107600

%N a(1) = 4, a(n) = smallest number > a(n-1) such that the concatenation of a(n-1) and a(n) is a square.

%e a(3) is 61 since it is the least number greater than a(2)=9 which concatenated with 9 forms a perfect square, i.e., 961 = 31^2.

%t f[n_] := Block[{x = n, d = 1 + Floor@ Log10@ n}, q = (Floor@ Sqrt[(10^d + 1) x] + 1)^2; If[q < (10^d) (x + 1), Mod[q, 10^d], Mod[(Floor@ Sqrt[(10^d)(10x + 1) - 1] + 1)^2, 10^(d + 1)] ]]; NestList[f, 4, 23] (* after the algorithm of _David W. Wilson_ in A090566 *)

%Y Cf. A090566, A265147, A265149, A265150, A265151, A265152, A265153, A265154, A265155.

%K nonn,base

%O 1,1

%A _Robert G. Wilson v_, Dec 02 2015