A247552 Least numbers x such that the ratio of the sum of all the cyclic permutations of x, plus the unpermuted number, and x itself is equal to n.

1, 11, 111, 1111, 11111, 111111, 428571, 11111111, 111111111, 1111111111, 1818, 111111111111, 230769, 428571428571, 111111111111111, 1111111111111111, 4705882352941176, 111111111111111111, 473684210526315789, 11111111111111111111, 142857, 18181818

Offset: 1

Author

Paolo P. Lava, Sep 19 2014

Keywords

• nonn
• base

Comments

Mainly repdigit numbers with n 1's.

If x has m digits with sum s then the sum of the m cyclic permutations of x (including x itself) is s*(10^m-1)/9, since each digit occurs once in each position. My program uses this to test potential (m, s) pairs. - Jens Kruse Andersen, Sep 23 2014

If appears that the number of digits of a(n) is n-1 if and only if n is a full reptend prime (A001913). - Michel Marcus, Sep 24 2014

There are 106 repdigit numbers with n 1's in the first 5000 terms. - Jens Kruse Andersen, Sep 30 2014

Examples

			428571 is the minimum number such that 428571 + 142857 + 714285 + 571428 + 857142 + 285714 = 2999997 and 2999997 / 428571 = 7.
1818 is the minimum number such that 1818 + 8181 + 1818 + 8181 = 19998 and 19998 / 1818 = 11.
		

Links

• Jens Kruse Andersen, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A247315, A247316, A247317.

Programs

Maple

P:=proc(q) local a, b, c, d, j, n, t, v;
v:=array(1..100); for j from 1 to 100 do v[j]:=0; od; t:=0;
for n from 1 to q do a:=n; b:=a; c:=ilog10(a);
for k from 1 to c do a:=(a mod 10)*10^c+trunc(a/10); b:=b+a; od;
if type(b/n,integer) then if b/n=t+1
then t:=t+1; lprint(t,n); while v[t+1]>0 do t:=t+1; lprint(t,v[t]); od;
else if b/n>t+1 then if v[b/n]=0 then v[b/n]:=n; fi; fi;
fi; fi; od; end: P(10^6);

PARI

isok(n, k) = {d = digits(k); nbd = #d; sp = 0; for (i=1, nbd, dpk = vector(nbd-1, j, d[j+1]); dpk = concat(dpk, d[1]); sp += subst(Pol(dpk, x), x, 10); d = dpk;); sp == k*n;}
a(n) = {k = 1; while(! isok(n, k), k++;); k ;} \\ Michel Marcus, Sep 21 2014

PARI

a(n)=my(r=0,m,g,s,x); for(m=1, n, r=10*r+1; g=n/gcd(r, n); forstep(s=g, 9*m, g, x=s*r/n; if(#digits(x)==m && sumdigits(x)==s, return(x))))
vector(30, n, a(n)) \\ Faster program. Jens Kruse Andersen, Sep 23 2014

Extensions

a(12)-a(22) from Jens Kruse Andersen, Sep 23 2014