OFFSET
1,1
COMMENTS
a(n) is the unique fixed point of the 1089-trick map T among all (n+4)-digit numbers (verified exhaustively for n = 1..4; conjectured for all n >= 1).
The digit pattern of a(n) is: 1, 0, followed by (n-1) nines, then 8, 9, 0. So a(1) = 10890 has 0 middle nines, a(2) = 109890 has 1, a(3) = 1099890 has 2, etc.
Each a(n)/10 is a primitive term of A001232 (i.e., of the form 11*(10^(n+1) - 1), not a concatenation of smaller A001232 terms). Note that 9*a(n) != reverse(a(n)) for any n (since a(n) ends in 0), so this sequence and A001232 are disjoint.
Proof that T(a(n)) = a(n): Write R = 10^(n+1) - 1 (repdigit of nines). Then a(n) = 110*R. reverse(110*R) = 99*R (leading zero drops). diff = 110*R - 99*R = 11*R. reverse(11*R) = 99*R. T(a(n)) = 11*R + 99*R = 110*R = a(n). QED.
REFERENCES
R. Havenaar, "Fixed Points of Digit-Operation Dynamical Systems: Algebraic Structure and Five Infinite Families", SYNTRIAD Research, 2026.
LINKS
FORMULA
a(n) = 110*(10^(n+1) - 1).
a(n) = 990*A002275(n+1).
a(n) = 10*a(n-1) + 990 for n >= 2, with a(1) = 10890.
G.f.: 10890*x/(1 - 10*x) + 990*x^2/((1 - x)*(1 - 10*x)).
EXAMPLE
For n = 1: a(1) = 110*(100 - 1) = 110*99 = 10890. Check: 10890 - 09801 = 1089; 1089 + 9801 = 10890. So T(10890) = 10890.
For n = 2: a(2) = 110*(1000 - 1) = 110*999 = 109890. Check: 109890 - 098901 = 10989; 10989 + 98901 = 109890.
MATHEMATICA
Table[110 (10^(n + 1) - 1), {n, 1, 18}]
PROG
(Python)
def a(n): return 110 * (10**(n+1) - 1)
print([a(n) for n in range(1, 19)])
CROSSREFS
Cf. A023093 (numbers k such that k and 9*k are anagrams).
KEYWORD
nonn,easy,base
AUTHOR
Remco Havenaar, Feb 27 2026
STATUS
approved
