A261020 Numbers k such that the set of the decimal digits is a subgroup of the multiplicative group (Z/mZ)* where m is the sum of the decimal digits of k.

11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91, 111, 124, 139, 142, 193, 214, 241, 319, 391, 412, 421, 913, 931, 1111, 1115, 1133, 1151, 1155, 1177, 1199, 1248, 1284, 1313, 1331, 1379, 1397, 1428, 1482, 1511, 1515, 1551, 1717, 1739, 1771, 1793

Offset: 1

Author

Michel Lagneau, Aug 07 2015

Keywords

• nonn
• base

Comments

(Z/mZ)* is the multiplicative group of units of Z/mZ.

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 products d(i)*d(j) (mod m) for 1 <= i,j <= q, and also the multiplicative inverse of each element such that if x is in the group, then there exists x' in the group where x*x' = 1.

The sequence is infinite because the numbers 11, 111, 1111, ... are in the sequence and generate the trivial subgroup {1}.

Only zerofree elements of A009996 have to be checked. Terms that match the criterion and permutations of their digits form all terms of this sequence due to commutativity of multiplication. - David A. Corneth, Aug 08 2015

To reduce cases, only check terms from A009995 (containing a 1 but no 0) for values m from digsum(term) to 81. - David A. Corneth, Aug 13 2015

Each decimal digit must be relatively prime to the decimal digit sum. - Tom Edgar, Aug 17 2015

Examples

			139 is a term because 1+3+9 = 13 and the elements {1, 3, 9} form a subgroup of the multiplicative group (Z/13Z)* with 12 elements. Each element is invertible: 1*1 == 1 (mod 13), 3*9 == 1 (mod 13) and 9*3 == 1 (mod 13). The other numbers of the sequence having the same property with (Z/13Z)* are 139, 193, 319, 391, 913, and 931.
1248 is in the sequence because 1+2+4+8 = 15 and the elements {1, 2, 4, 8} form a subgroup of the multiplicative group (Z/15Z)* with 8 elements: {1,2,4,7,8,11,13,14}.
		

Links

• Michel Lagneau and David A. Corneth, Table of n, a(n) for n = 1..10083 (all elements < 10^14, first 268 terms from Michel Lagneau)
• Eric Weisstein's World of Mathematics, Finite Group
• Wikipedia, Finite group

Crossrefs

Cf. A011531, A261021, A261322.

Programs

Maple

nn:=2000:
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]=1 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)=1
            then
            lst2:=lst2 union {lst1[c]}:
            else
           fi:
          od:
         od:
           if lst2=lst
           then
           printf(`%d, `, n):
           else
           fi:
          fi:
         fi:
     od:

Sage

def is_group(n):
    DD=n.digits()
    digsum=sum(DD)
    D=Set(DD)
    if not(1 in D) or 0 in D:
        return false
    for x in D:
        for y in D:
            if not(gcd(y,digsum)==1):
                return false
            if not((x*inverse_mod(y,digsum))%digsum in D):
                return false
    return true
[n for n in [1..2000] if is_group(n)] # Tom Edgar, Aug 17 2015