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a(1) = 11, a(n) = smallest number > a(n-1) such that the concatenation of a(n-1) and a(n) is a square.
8

%I #9 Oct 13 2017 03:31:32

%S 11,56,169,744,769,5076,5625,43524,390625,1827776,2562500,8273225,

%T 37136225,38371001,43037561,258421444,792669636,2928667041,

%U 38512058944,260125180889,405701529401,688085041025,5890084946609,22508111494025,64017148660004,537387232526336

%N a(1) = 11, 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 169 since it is the least number greater than a(2)=56 which concatenated with 56 forms a perfect square, i.e., 56169 = 237^2.

%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, 11, 25] (* after the algorithm of _David W. Wilson_ in A090566 *)

%Y Cf. A090566, A265147, A265148, A265149, A265150, A265152, A265153, A265154, A265155.

%K nonn,base

%O 1,1

%A _Robert G. Wilson v_, Dec 02 2015