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 A034294 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). 4
 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, 83, 84, 86, 91, 93, 371, 441, 445, 511, 551, 774, 834, 882, 912, 961, 2116, 5141, 7721, 9471, 15226, 99481, 313725, 315231, 1527465, 3454446, 454003312, 956111321, 2426472326, 3066511287, 5217957101 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS From Jinyuan Wang, Aug 06 2019: (Start) Define j by 10^j < k < 10^(j+1). Let m denote the reversal of k_10. 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). 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. 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) LINKS PROG (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 CROSSREFS Cf. A307498, A308493. Sequence in context: A261556 A033088 A307498 * A304246 A271837 A290950 Adjacent sequences: A034291 A034292 A034293 * A034295 A034296 A034297 KEYWORD base,nice,nonn,fini AUTHOR EXTENSIONS More terms from Jinyuan Wang, Aug 05 2019 Further terms from Giovanni Resta, Aug 06 2019 STATUS approved

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Last modified January 30 17:11 EST 2023. Contains 359945 sequences. (Running on oeis4.)