A307498 -id:A307498 - 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

A308493 Numbers k such that k in base 10 contains the same digits as k in some other base.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 20, 21, 23, 31, 41, 42, 43, 46, 51, 53, 61, 62, 63, 71, 73, 81, 82, 83, 84, 86, 91, 93, 100, 101, 102, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 121, 122, 123, 131, 133, 141, 144, 151, 155, 158, 161, 166, 171, 177, 181

Offset: 1

Jinyuan Wang, Aug 05 2019

Supersequence of A034294 and A307498.
This sequence is infinite because 2*10^k is a term for any k >= 0.
Also 10^k is a term when k >= 0 and so too 10^k*(10^m - 1)/9 for any k > 0 and m >= 0. - Bruno Berselli, Aug 26 2019

			k = 113 is in the sequence because the set of digits of k {1, 3} equals the set of digits of (k in base 110) = 13.

Cf. A034294, A037415, A037422, A037428, A037433, A037437, A037440, A037442, A037443, A249899, A307498.

isok(k) = {my(j=Set(digits(k))); for(b=2, k+1, if((b!=10) && (Set(digits(k, b)) == j), return(1))); return(0);} \\ Michel Marcus, Aug 05 2019

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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