cp's OEIS Frontend

All numbers of the form f(n)=9(n).5.0(2n).5.0(n-1).1 where n>0 are in the sequence because if k(n)=9(n).0(n).25.0(n-1).9(n).6 then f(n)^2=k(n).(k(n)+5). For example f(2)=9950000501; k(2)=9900250996 and f(2)^2=9950000501^2=9900250996.9900251001 =k(2).(k(2)+5). - Farideh Firoozbakht, Nov 26 2006

m^2 = (k)|(k+5) = (k)|(k) + 5 = (10^q + 1)*k + 5 where | denotes concatenation and q is the number of digits of k gives a nonlinear equation that can be solved using the solver below. - David A. Corneth, Jan 02 2021

From Farideh Firoozbakht, Nov 26 2006: (Start)

1. a(n).(a(n)+4) = A115438^2 where "." denotes concatenation.

2. All numbers of the form f(j) = 4{j}.2.6{j-1}.70.2{j}.0 where each expression in braces denotes the multiplicity of the digit preceding the expression (e.g., "4{j}" means that the digit "4" appears j times consecutively) and where j > 0 are in the sequence because if k(j) = 6{j}.5.3{j}.4.6{j}.8 then k(j)^2 = f(j).(f(j)+4). For example, f(4) = 444426667022220, k(4) = 666653333466668, and k(4)^2 = 666653333466668^2 = f(4).(f(4)+4) = 444426667022220.444426667022224.

3. All numbers of the form f(j) = 1{j}.2.0{j+1}.8{j}.5 where j > -1 are in the sequence because if k(j) = 3{j}.4.6{j}.5.3{j+1} then k(j)^2 = f(j).(f(j)+4). For example, f(5) = 111112000000888885, k(5) = 333334666665333333, and k(5)^2 = 333334666665333333^2 = f(5).(f(5)+4) = 111112000000888885.111112000000888889. (End)