cp's OEIS Frontend

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.

%I A247552 #28 Sep 05 2021 15:18:30
%S A247552 1,11,111,1111,11111,111111,428571,11111111,111111111,1111111111,1818,
%T A247552 111111111111,230769,428571428571,111111111111111,1111111111111111,
%U A247552 4705882352941176,111111111111111111,473684210526315789,11111111111111111111,142857,18181818
%N 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.
%C A247552 Mainly repdigit numbers with n 1's.
%C A247552 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
%C A247552 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
%C A247552 There are 106 repdigit numbers with n 1's in the first 5000 terms. - _Jens Kruse Andersen_, Sep 30 2014
%H A247552 Jens Kruse Andersen, <a href="/A247552/b247552.txt">Table of n, a(n) for n = 1..1000</a>
%e A247552 428571 is the minimum number such that 428571 + 142857 + 714285 + 571428 + 857142 + 285714 = 2999997 and 2999997 / 428571 = 7.
%e A247552 1818 is the minimum number such that 1818 + 8181 + 1818 + 8181 = 19998 and 19998 / 1818 = 11.
%p A247552 P:=proc(q) local a, b, c, d, j, n, t, v;
%p A247552 v:=array(1..100); for j from 1 to 100 do v[j]:=0; od; t:=0;
%p A247552 for n from 1 to q do a:=n; b:=a; c:=ilog10(a);
%p A247552 for k from 1 to c do a:=(a mod 10)*10^c+trunc(a/10); b:=b+a; od;
%p A247552 if type(b/n,integer) then if b/n=t+1
%p A247552 then t:=t+1; lprint(t,n); while v[t+1]>0 do t:=t+1; lprint(t,v[t]); od;
%p A247552 else if b/n>t+1 then if v[b/n]=0 then v[b/n]:=n; fi; fi;
%p A247552 fi; fi; od; end: P(10^6);
%o A247552 (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;}
%o A247552 a(n) = {k = 1; while(! isok(n, k), k++;); k ;} \\ _Michel Marcus_, Sep 21 2014
%o A247552 (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))))
%o A247552 vector(30, n, a(n)) \\ Faster program. _Jens Kruse Andersen_, Sep 23 2014
%Y A247552 Cf. A247315, A247316, A247317.
%K A247552 nonn,base
%O A247552 1,2
%A A247552 _Paolo P. Lava_, Sep 19 2014
%E A247552 a(12)-a(22) from _Jens Kruse Andersen_, Sep 23 2014