A261020 - OEIS

%I #44 Feb 22 2018 16:33:38

%S 11,12,13,14,15,16,17,18,19,21,31,41,51,61,71,81,91,111,124,139,142,

%T 193,214,241,319,391,412,421,913,931,1111,1115,1133,1151,1155,1177,

%U 1199,1248,1284,1313,1331,1379,1397,1428,1482,1511,1515,1551,1717,1739,1771,1793

%N 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.

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

%C 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.

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

%C 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

%C 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

%C Each decimal digit must be relatively prime to the decimal digit sum. - _Tom Edgar_, Aug 17 2015

%H Michel Lagneau and David A. Corneth, <a href="/A261020/b261020.txt">Table of n, a(n) for n = 1..10083</a> (all elements < 10^14, first 268 terms from Michel Lagneau)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FiniteGroup.html">Finite Group</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Finite_group">Finite group</a>

%e 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.

%e 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}.

%p nn:=2000:

%p for n from 1 to nn do:

%p x:=convert(n,base,10):nn0:=length(n):

%p lst1:={op(x),x[nn0]}:n0:=nops(lst1):

%p s:=sum('x[i]', 'i'=1..nn0):lst:={}:

%p    if  lst1[1]=1 then

%p     for j from 1 to n0 do:

%p      for l from j to n0 do:

%p       p:=irem(lst1[j]*lst1[l],s):lst:=lst union {p}:

%p      od:

%p     od:

%p     if lst=lst1

%p      then

%p        n3:=nops(lst1):lst2:={}:

%p         for c from 1 to n3 do:

%p           for d from 1 to n3 do:

%p            if irem(lst1[c]*lst1[d], s)=1

%p             then

%p             lst2:=lst2 union {lst1[c]}:

%p             else

%p            fi:

%p           od:

%p          od:

%p            if lst2=lst

%p            then

%p            printf(`%d, `, n):

%p            else

%p            fi:

%p           fi:

%p          fi:

%p      od:

%o (Sage)

%o def is_group(n):

%o     DD=n.digits()

%o     digsum=sum(DD)

%o     D=Set(DD)

%o     if not(1 in D) or 0 in D:

%o         return false

%o     for x in D:

%o         for y in D:

%o             if not(gcd(y,digsum)==1):

%o                 return false

%o             if not((x*inverse_mod(y,digsum))%digsum in D):

%o                 return false

%o     return true

%o [n for n in [1..2000] if is_group(n)] # _Tom Edgar_, Aug 17 2015

%Y Cf. A011531, A261021, A261322.

%K nonn,base

%O 1,1

%A _Michel Lagneau_, Aug 07 2015