cp's OEIS Frontend

Also numbers k such that k concatenated with k+8 gives the product of two numbers which differ by 2.

Also numbers k such that k concatenated with k+5 gives the product of two numbers which differ by 4.

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. All numbers of the form f(n)=3(n).4.6(n).5.3(n+1) are in the sequence because if k(n)=1(n).2.0(n+1).8(n).5 then f(n)^2= k(n).(k(n)+4). For example f(3)=333466653333; k(3)=111200008885 and f(3)^2=333466653333^2=k(3).(k(3)+4)=111200008885.111200008889.

2. All numbers of the form g(n)=6(n).5.3(n).4.6(n).8 are in the sequence because g(0)=548 is in the sequence(548^2=300.304) and for n>0 if h(n)=4(n).2.6(n-1).70.2(n).0 then g(n)^2=h(n).(h(n)+4). For example g(5)=666665333334666668; h(5)=444442666670222220 and g(5)^2=h(5).(h(5)+4)=444442666670222220.444442666670222224. (End)