OFFSET
1,2
COMMENTS
30 divides all terms of the sequence. For all nonnegative integers m and n all numbers of the form f1(m,n) = 660(10^(m + 2) - 1)*(10^((m + 4)*n) - 1)/(10^(m + 4) - 1) are in the sequence, in fact f1(m,n) = (65.(9)(m).34)(n).0 where dot between numbers means concatenation and "(r)(t)" means number of r's is t. With this definition a(1) = 0 = f1(0,0), a(2) = 65340 = f1(0,1), a(3) = 659340 = f1(1,1), a(4) = 6599340 = f1(2,1), a(5) = 65999340 = f1(3,1), a(6) = 653465340 = f1(0,2), a(7) = 659999340 = f1(4,1), a(9) = 6599999340 = f1(5,1), etc. f1(m,1) = 660(10^(m + 2) - 1) = 65.(9)(m).340, f1(m,2) = 65.(9)(m).34.65.(9)(m).340, etc. Let g(s,t,r) = s*(10^((L+t)*(1+r))-1)/(10^(L+t)-1) where L = number of digits of s, in fact g(s,t,r) = (s.(0)(t))(r).s so the function g is the same function that has been defined in the sequence A101704. If s is in the sequence then all numbers of the form g(s,t,r) for nonnegative integers t and r are in the sequence. Next term is greater than 11*10^9. It seems that the eleven next terms are 65340065340, 65934659340, 65999999340, 653400065340, 659340659340 659999999340, 6534000065340, 6534653465340, 6593400659340, 6599346599340 and 6599999999340. Is it true that, all terms of this sequence are of the form g(f1(m,n),r,t)?
LINKS
Ray Chandler, Table of n, a(n) for n = 1..10000
FORMULA
EXAMPLE
g(65340,0,2)= (65340)(3) = 653406534065340 is in the sequence because reversal(653406534065340) = 43560435604356 = (1/15)*653406534065340.
MATHEMATICA
Do[If[n == 15*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 0, 11000000000, 30}]
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Farideh Firoozbakht, Jan 04 2005
EXTENSIONS
More terms from Ray Chandler, Oct 09 2017
STATUS
approved