A224220 - OEIS

%I #21 Feb 27 2017 02:58:32

%S 32,75,8,245,12,21,16,1089,15,1859,21,39,28,4335,24,6137,24,57,40,

%T 11109,33,115,39,45,52,22707,35,27869,40,93,64,55,51,47915,57,111,76,

%U 65559,48,75809,56,129,88,99405,69,329,60,119,65,143259,72,265,63,95,112,198417,87,219539

%N a(n) = smallest number k with property that if the base-n expansion of k is reversed, the result is a nontrivial multiple of k.

%C In other words, k divides (reversal of k in base n), and (k-reversed)/k > 1.

%C The numbers are written in base 10.

%C Theorem: The length of k (in base n) is 2 iff n>=5 and n+1 is composite, otherwise 4.

%D N. J. A. Sloane, paper in preparation.

%D See A214927 for further references and links.

%H Michel Marcus, <a href="/A224220/b224220.txt">Table of n, a(n) for n = 3..400</a>

%H T. J. Kaczynski, <a href="http://www.jstor.org/stable/2689056">Note on a Problem of Alan Sutcliffe</a>, Math. Mag., 41 (1968), 84-86.

%H Lara Pudwell, <a href="http://faculty.valpo.edu/lpudwell/papers/mm005281.pdf">Digit Reversal Without Apology</a>, Mathematics Magazine, Vol. 80 (2007), pp. 129-132. Also arXiv:math/0511366 [math.HO], 2005.

%H Alan Sutcliffe, <a href="http://www.jstor.org/stable/2689015">Integers That Are Multiplied When Their Digits Are Reversed</a>, Mathematics Magazine, 39 (1966), 282-287.

%F If n=3 or n>3 and n+1 is prime, a(n) = (n^2-1)(n+1) (cf. A152619).

%e The numbers a(n) for n = 3, ..., 11 written in base n are 1012, 1023, 13, 1045, 15, 25, 17, 1089, 14.

%e For example, 1012 (base 3) = 32 (base 10), and 2101 (base 3) = 64 (base 10) = 2*32.

%t Table[k = 2; While[Nand[IntegerQ@ #, # != 1] &[FromDigits[#, n]/k] &@ Reverse@ IntegerDigits[k, n], k++]; k, {n, 3, 60}] (* _Michael De Vlieger_, Feb 26 2017 *)

%o (PARI) isok(k, n) = {my(rk = fromdigits(Vecrev(digits(k, n)), n)); !(rk % k) && (rk > k);}

%o a(n) = {my(k = 1); while (!isok(k, n), k++); k;} \\ _Michel Marcus_, Feb 26 2017

%Y Cf. A214927, A006093, A072668, A152619.

%K nonn,base

%O 3,1

%A _N. J. A. Sloane_, Apr 01 2013