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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
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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
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
FORMULA
For d >= 3, there are (d - 1) * 17 terms having d digits. - David A. Corneth, Aug 13 2015
EXAMPLE
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, ...}.
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:
PROG
(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
CROSSREFS
Sequence in context: A274612 A261448 A213312 * A271642 A164849 A162671
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Aug 07 2015
STATUS
approved

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Last modified April 16 02:41 EDT 2024. Contains 371696 sequences. (Running on oeis4.)