OFFSET
1,1
FORMULA
a(n) = 9*10^(2n-1) - 10^n - Sum_{k=10^(n-1)+1..10^n-1} ceiling(10^(2n-1)/k).
a(n) ~ (9-log(10))*10^(2n-1).
EXAMPLE
a(1)=58 since we get the following pairs: (2, 5), ..., (2, 9), (3, 4), ..., (3, 9), (4, 3), ..., (4, 9), (5, 2), ..., (5, 9), (6, 2), ..., (6, 9), (7, 2), ..., (7, 9), (8, 2), ..., (8, 9), (9, 2), ..., (9, 9).
PROG
(Python)
import math
ende = 1
for i in range(1, 10):
anz = 0
for a in range(ende, 10*ende):
z = math.ceil((ende*ende*10)/a)
if z < ende*10:
anz = anz + ende*10 - z
ende = ende*10
print(i, anz)
(PARI) a(n) = 9*10^(2*n-1) - 10^n - sum(k=10^(n-1)+1, 10^n-1, ceil(10^(2*n-1)/k)); \\ Michel Marcus, Jun 25 2019
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Reiner Moewald, May 23 2019
STATUS
approved