A261322
Non-repunit elements of A261020 in nonincreasing order.
Original entry on oeis.org
21, 31, 41, 51, 61, 71, 81, 91, 421, 931, 3311, 5111, 5511, 7711, 8421, 9731, 9911, 311111, 444111, 711111, 777111, 993311, 8811111, 51111111, 55551111, 91111111, 93333311, 99311111, 99991111, 441111111, 6666611111, 7111111111, 9333311111, 411111111111, 555111111111, 771111111111, 777777111111, 911111111111
Offset: 1
{1, 3, 9} forms a group under multiplication in Z/mZ for m = 13 and m = 26 (and no other values of m). m is the sum of digits of a term, so we can solve 9*x + 3*y + 1*z in {13, 26} for (x, y, z) >= (1, 1, 1). Solutions are (x, y, z) in {(1, 1, 1), (2, 2, 2), ..., (1, 1, 14)}. A solution (x, y, z) denotes a term starting with x nines, then followed by y threes, and then by z ones.
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.
Original entry on oeis.org
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
Offset: 1
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, ...}.
-
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:
-
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
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