The base 4 trajectory of 270798 = A075421(370) provably does not contain a palindrome. A proof along the lines of Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2, can be based on the formula given below. - The generating function given describes the sequence from a(11) onward; the g.f. for the complete sequence is known but nearly twice as big.
a(0), ..., a(10) as above; for n > 10 and n = 5 (mod 6): a(n) = 5*4^(2*k+10)+15341035*4^k-15 where k = (n+1)/6; n = 0 (mod 6): a(n) = 10*4^(2*k+10)+9792150*4^k-10 where k = n/6; n = 1 (mod 6): a(n) = 20*4^(2*k+10)-1305620*4^k where k = (n-1)/6; n = 2 (mod 6): a(n) = 20*4^(2*k+10)+14361820*4^k-15 where k = (n-2)/6; n = 3 (mod 6): a(n) = 40*4^(2*k+10)+7833720*4^k-10 where k = (n-3)/6; n = 4 (mod 6): a(n) = 80*4^(2*k+10)-1305620*4^k where k = (n-4)/6. G.f.: -15*(1426085120*x^11+749251744*x^10+419191024*x^9-1430263104*x^8-715827880*x^7-369055228*x^6-352343296*x^5-222825800*x^4-155978060*x^3+356521280*x^2+189401930*x+105842255)/((x-1)*(x^2+x+1)*(2*x^3-1)*(2*x^3+1)*(4*x^3-1))