OFFSET
1,1
COMMENTS
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
LINKS
David A. Corneth, Table of n, a(n) for n = 1..3733
Dario A. Alpern, Generic two integer variable equation solver
EXAMPLE
38163^2 = 14564_14569.
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Giovanni Resta, Jan 24 2006
EXTENSIONS
More terms from David A. Corneth, Jan 02 2021
STATUS
approved