|
|
A115439
|
|
Numbers m such that the square of m is the concatenation of two numbers k and k+5.
|
|
13
|
|
|
4, 7, 45, 56, 38163, 61838, 83618, 346980, 653021, 950051, 8647555, 9534265, 8167822283, 9007920992, 9209900792, 9950000501, 4737445289221, 4990568257187, 5009431742814, 5262554710780, 8373808925585, 8626931893551, 34323166122692, 34532758615690, 49625657225895, 49835249718893
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
EXAMPLE
|
38163^2 = 14564_14569.
|
|
CROSSREFS
|
Cf. A030467, A057934, A106497, A115428, A115427, A115438, A115440, A115441, A115442, A115443, A115444, A115445, A115446, A115447.
|
|
KEYWORD
|
base,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|