OFFSET
1,1
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
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
EXAMPLE
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}.
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:
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Aug 07 2015
STATUS
approved