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%I #42 Jan 12 2020 10:50:34
%S 1,2,3,4,5,6,7,8,9,13,21,23,31,41,42,43,46,51,53,61,62,63,71,73,81,82,
%T 83,84,86,91,93,371,441,445,511,551,774,834,882,912,961,2116,5141,
%U 7721,9471,15226,99481,313725,315231,1527465,3454446,454003312,956111321,2426472326,3066511287,5217957101
%N Numbers k not ending in 0 such that for some base b, k_b is the reverse of k_10 (where k_b denotes k written in base b).
%C From _Jinyuan Wang_, Aug 06 2019: (Start)
%C Define j by 10^j < k < 10^(j+1). Let m denote the reversal of k_10.
%C Then 10^(j/(j+1)) < b < 10^((j+1)/j). Proof: for any j > 0, (10^(j+1) in base b) > m > 10^j = (b^j in base b) and (10^j in base b) < m < 10^(j+1) = (b^(j+1) in base b), therefore 10^(j+1) > b^j and 10^j < b^(j+1).
%C k in base 10 is reversed in base 82 iff k = 91. Otherwise, k in base 10 is reversed in another base less than 82. Because for k > 100, j >= 2 so that b < 10^(3/2) < 32; for k < 100, 82 is the largest b.
%C For j >= 25, 10^(25/26) < b < 10^(26/25), but b can't be 10. Therefore the largest term is less than 10^25. (End)
%o (PARI) is(k) = {r = digits(eval(concat(Vecrev(Str(k))))); sum(j = 2, 9, digits(k, j) == r) + sum(j = 11, 82, digits(k, j) == r) > 0 && k%10 > 0; } \\ _Jinyuan Wang_, Aug 05 2019
%Y Cf. A307498, A308493.
%K base,nice,nonn,fini
%O 1,2
%A _Erich Friedman_
%E More terms from _Jinyuan Wang_, Aug 05 2019
%E Further terms from _Giovanni Resta_, Aug 06 2019
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