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A281384 Least integer with more than 1 digit in base n, such that the set of its base-n digits equals the set of its binary digits. 2
2, 9, 4, 5, 6, 49, 8, 9, 10, 11, 12, 13, 14, 225, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 961, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 3969, 64, 65, 66, 67, 68, 69, 70 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,1

COMMENTS

"More than 1 digit in base n" is equivalent to ">= n", which also implies "more than 1 digit in base 2", since n >= 2 by definition. If such a number written in binary is not a string of 1's, then its set of binary digits is {0, 1}, and the smallest number to have the same digits in base n is n = 10[n] itself. If n has all binary digits equal to 1, i.e., n = 2^k-1, then the smallest solution is the next larger number having the digits {0, 1} in base n (and also in base 2), which is, since 11[n] = n+1 is excluded, 100[n] = n^2 = (2^k-1)^2 = 2^{2k} + 2^{k+1} + 1 = 1...01[2]. - M. F. Hasler, Jan 22 2017

LINKS

Table of n, a(n) for n=2..70.

FORMULA

For any n>1: if n belongs to A000225, then a(n)=n^2, otherwise a(n)=n. - Rémy Sigrist, Jan 22 2017

PROG

(PARI) a(n) = my(m=n); while (Set(digits(m, n)) != Set(digits(m, 2)), m++); m;

(PARI) A281384(n) = if(n+1==1<<logint(n), n^2, n) \\ M. F. Hasler, Jan 22 2017

CROSSREFS

First column of triangular array A281383.

Sequence in context: A070700 A248682 A222239 * A203648 A300889 A275807

Adjacent sequences:  A281381 A281382 A281383 * A281385 A281386 A281387

KEYWORD

nonn

AUTHOR

Michel Marcus, Jan 21 2017

STATUS

approved

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Last modified March 25 12:11 EDT 2019. Contains 321470 sequences. (Running on oeis4.)