OFFSET
1,1
COMMENTS
If any of the values a(9), a(13), a(14), a(17) and a(18) exist, they exceed 10^10.
31 is not a term since reverse(31)^i = 13^i <= reverse(31^i) for all i >= 1.
Proof: inspection shows it holds for i <= 2. For i > 2, 13^i has less digits than 31^i. As 31 is no multiple of 10, 31^i and reverse(31^i) have the same number of digits. So reverse(31)^i = 13^i <= reverse(31^i) and 31 is no term. - David A. Corneth, May 21 2026
From Michael S. Branicky and David A. Corneth, May 20 2026: (Start)
No power of 10 is a term since reverse(10^j)^i = reverse((10^j)^i) = 1 for all i, j >= 0.
No multiple of 10 can be a term since reverse(m*10) = reverse(m) so reverse(m*10)^i = reverse(m)^i and reverse((m*10)^i) = reverse(m^i * 10^i) = reverse(m^i). Thus, a(n) = m*10 is not possible since 0 < m < 10*m would have satisfied the same condition. (End)
Thus, the sequence contains neither the number 1 nor any number whose last digit is 0.
Some other terms: a(10) = 51894, a(11) = 360133, a(12) = 68186, a(15) = 63136, a(16) = 31813, a(19) = 33913, a(20) = 31713, a(25) = 640136, a(39) = 334713, a(85) = 63269036. - Michael S. Branicky, May 21 2026
EXAMPLE
Let R(k) = reverse(k) = A004086(k).
a(1) = 8 since R(8)^1 = 8 <= 8 = R(8^1), but R(8)^2 = 64 > 46 = R(8^2);
a(2) = 4 since R(4)^1 = 4 <= 4 = R(4^1), R(4)^2 = 16 <= 61 = R(4^2), but R(4)^2 = 64 > 46 = R(4^3);
a(3) = 3 since R(3)^1 = 3 <= 3 = R(3^1), R(3)^2 = 9 <= 9 = R(3^2), R(3)^3 = 27 <= 72 = R(3^3), but R(4)^4 = 81 > 18 = R(4^4);
and no lesser numbers have this property.
PROG
(PARI) a(n) = {for(i = 1, oo, if(iscan(i, n), return(i)))}
iscan(c, n) = {my(rc = rev(c)); for(i = 1, n, if(rc^i > rev(c^i), return(0))); rc^(n+1) > rev(c^(n+1))}
rev(k) = fromdigits(Vecrev(digits(k))) \\ David A. Corneth, May 20 2026
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Jean-Marc Rebert, May 20 2026
STATUS
approved