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 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. 2

%I #48 Jan 17 2024 09:15:00

%S 0,101,102,110,120,201,202,204,210,220,240,303,306,330,360,402,404,

%T 408,420,440,480,505,550,603,606,630,660,707,770,804,808,840,880,909,

%U 990,1001,1002,1010,1020,1100,1200,2001,2002,2004,2010,2020,2040,2100,2200,2400

%N 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.

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

%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 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.

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

%C 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

%H David A. Corneth, <a href="/A261021/b261021.txt">Table of n, a(n) for n = 1..10000</a> (first 281 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>

%F For d >= 3, there are (d - 1) * 17 terms having d digits. - _David A. Corneth_, Aug 13 2015

%e 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).

%e 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, ...}.

%p nn:=3000:

%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]=0 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)=0

%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 (PARI) is(n) = {my(d = digits(n),s = Set(digits(n))); if(n==0,return(1));

%o if(#s==2 || #s==3,return(s[1]==0 && (s[#s] / s[2] == 2^(#s-2)) && hammingweight(d)==2),return(0))}

%o \\a(n) works for n > 1.

%o 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,

%o n-=(qd-1)*h;i++; h=1 + (i%2 == 0) + (i < 5)); n--; v[1]=i;

%o 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

%Y Cf. A009996, A261020.

%K nonn,base

%O 1,2

%A _Michel Lagneau_, Aug 07 2015

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Last modified May 25 17:00 EDT 2024. Contains 372804 sequences. (Running on oeis4.)