A308493 -id:A308493 - cp's OEIS frontend

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

Erich Friedman

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)

Cf. A307498, A308493.

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

More terms from Jinyuan Wang, Aug 05 2019

Further terms from Giovanni Resta, Aug 06 2019

A307498 Numbers k such that the digits of k in base 10 are a permutation of those of k in some other base.

Original entry on oeis.org

0, 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, 158, 191, 196, 227, 261, 265, 283, 316, 370, 371, 441, 445, 511, 518, 551, 774, 782, 825, 834, 882, 910, 911, 912, 913, 914, 915, 916, 917, 918

Offset: 1

Jinyuan Wang, Aug 05 2019

Supersequence of A034294 and subsequence of A307498.
If the digits of k in base 10 is a permutation of m = (k in base b), 10^j < k < 10^(j+1), then 10^(j/(j+1)) < b < 10^((j+1)/j).
If k > 10, other base can only be 4, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 25, 26, 28, 37, 46, 55, 64, 73, 82.
The digits of k in base 10 is a permutation of k in base 82 iff k = 91.
The largest term is less than 10^25. See proof in A034294.

			13 in base 4 is 31, 227 in base 9 is 272.

Cf. A034294, A115920, A308493.

isok(k) = {my(v = vecsort(digits(k))); k < 10 || sum(j = 3, 82, vecsort(digits(k, j)) == v) > 1;}

cp's OEIS Frontend

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).

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