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Non-repunit elements of A261020 in nonincreasing order.
2

%I #14 Aug 18 2015 12:41:27

%S 21,31,41,51,61,71,81,91,421,931,3311,5111,5511,7711,8421,9731,9911,

%T 311111,444111,711111,777111,993311,8811111,51111111,55551111,

%U 91111111,93333311,99311111,99991111,441111111,6666611111,7111111111,9333311111,411111111111,555111111111,771111111111,777777111111,911111111111

%N Non-repunit elements of A261020 in nonincreasing order.

%C Permutations of digits of all terms in this sequence are in A261020. There are 2403274 such permutations. About 38% (binomial(32,6) = 906192) of these permutations come from a(61) = 99999911111111111111111111111111.

%C On average, for every number of digits from 1 to 72, there's exactly one element.

%H David A. Corneth, <a href="/A261322/b261322.txt">Table of n, a(n) for n = 1..72</a>

%e {1, 3, 9} forms a group under multiplication in Z/mZ for m = 13 and m = 26 (and no other values of m). m is the sum of digits of a term, so we can solve 9*x + 3*y + 1*z in {13, 26} for (x, y, z) >= (1, 1, 1). Solutions are (x, y, z) in {(1, 1, 1), (2, 2, 2), ..., (1, 1, 14)}. A solution (x, y, z) denotes a term starting with x nines, then followed by y threes, and then by z ones.

%Y Cf. A261020.

%K nonn,fini,full,base

%O 1,1

%A _David A. Corneth_, Aug 14 2015