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).

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

Offset: 1

Author

Erich Friedman

Keywords

• base
• nice
• nonn
• fini

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)

Crossrefs

Cf. A307498, A308493.

Programs

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

Extensions

More terms from Jinyuan Wang, Aug 05 2019

Further terms from Giovanni Resta, Aug 06 2019