A261021 a(1)=0; for n > 1, a(n) is the number k such that the set of the decimal digits is an additive group Z/mZ where m is the sum of the decimal digits.

0, 101, 102, 110, 120, 201, 202, 204, 210, 220, 240, 303, 306, 330, 360, 402, 404, 408, 420, 440, 480, 505, 550, 603, 606, 630, 660, 707, 770, 804, 808, 840, 880, 909, 990, 1001, 1002, 1010, 1020, 1100, 1200, 2001, 2002, 2004, 2010, 2020, 2040, 2100, 2200, 2400

Offset: 1

Author

Michel Lagneau, Aug 07 2015

Keywords

• nonn
• base

Comments

By convention, a(1)=0 because the trivial additive group is usually denoted by 0 where 0 is the identity element.

Let d(1)d(2)..d(q) be the q decimal digits of a number k. The principle of the algorithm is to compute all the sums (d(i)+ d(j))/mZ for 1 <= i,j <= q, and also the additive inverse of each element such that if x is in the group, then there exists x' in the group where x+x' = 0.

The sequence is infinite because the numbers 101, 1001, 10001, ... are in the sequence and generate the group {0,1}.

Only terms of A009996 containing at least one 0 have to be checked. Terms that match the criterion and numbers containing at least one 0 formed by permutations of their digits form all terms of this sequence due to commutativity of addition. - David A. Corneth, Aug 13 2015

Examples

			408 is in the sequence because 4+0+8 = 12 and the elements {0, 4, 8} is an additive group, subgroup of (Z/12Z,+) with 6 elements {0, 2, 4, 6, 8, 10}. Each element has an inverse: 2+10 == 0 (mod 12), 4+8 == 0 (mod 12), 6+6 == 0 (mod 12), 8+4 == 0 (mod 12) and 10+2 == 0 (mod 12).
The subsequence having the same property with Z/12Z is {408, 480, 606, 660, 804, 840, 4008, 4080, 4800, 6006, 6060, 6600, 8004, 8040, 8400, 40008, 40080, 40800, 48000, 60006, 60060, 60600, 66000, 80004, 80040, 80400, 84000, ...}.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 281 terms from Michel Lagneau)
• Eric Weisstein's World of Mathematics, Finite Group
• Wikipedia, Finite group

Crossrefs

Cf. A009996, A261020.

Programs

Maple

nn:=3000:
for n from 1 to nn do:
x:=convert(n,base,10):nn0:=length(n):
lst1:={op(x),x[nn0]}:n0:=nops(lst1):
s:=sum('x[i]', 'i'=1..nn0):lst:={}:
   if  lst1[1]=0 then
    for j from 1 to n0 do:
     for l from j to n0 do:
      p:=irem(lst1[j]+lst1[l],s):lst:=lst union {p}:
     od:
    od:
    if lst=lst1
     then
       n3:=nops(lst1):lst2:={}:
        for c from 1 to n3 do:
          for d from 1 to n3 do:
           if irem(lst1[c]+lst1[d], s)=0
            then
            lst2:=lst2 union {lst1[c]}:
            else
           fi:
          od:
         od:
           if lst2=lst
           then
           printf(`%d, `, n):
           else
           fi:
          fi:
         fi:
     od:

PARI

is(n) = {my(d = digits(n),s = Set(digits(n))); if(n==0,return(1));
if(#s==2 || #s==3,return(s[1]==0 && (s[#s] / s[2] == 2^(#s-2)) && hammingweight(d)==2),return(0))}
\\a(n) works for n > 1.
a(n) = {my(qd = ((-1 + sqrt(1 + 8*(n + 15+1/2) / 17)) / 2)\1 + 2, v = vector(qd),i=1,h=2); n -= (binomial(qd-1,2)*17 -16); while(n-(qd-1)*h>0,
n-=(qd-1)*h;i++; h=1 + (i%2 == 0) + (i < 5)); n--; v[1]=i;
v[qd-n\h] = i*2^(n%h-(i%2==0)); sum(i=1,#v,10^(#v-i)*v[i])} \\ David A. Corneth, Aug 13 2015

Formula

For d >= 3, there are (d - 1) * 17 terms having d digits. - David A. Corneth, Aug 13 2015