cp's OEIS Frontend

A243023 Consider a k-digit number m = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + ... + d_(2)*10 + d_(1). Sequence lists the numbers m that divide Sum_{i=1..k-1}{d_(i+1)^d(i)}+d(1)^d(k) (see example below).

%I A243023 #25 Feb 18 2021 00:46:28
%S A243023 1,2,3,4,5,6,7,8,9,63,448,1547,1693,6189068,20443796,67526389
%N A243023 Consider a k-digit number m = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + ... + d_(2)*10 + d_(1). Sequence lists the numbers m that divide Sum_{i=1..k-1}{d_(i+1)^d(i)}+d(1)^d(k) (see example below).
%C A243023 Numbers with two consecutive zeros are not considered, to avoid the case 0^0. Nevertheless, even if we consider 0^0=1 the results do not change (at least up to the last number I tested, that is m=10^8).
%e A243023 For 63 we have 6^3 + 3^6 = 945 and 945/63 = 15.
%e A243023 Obviously also with 36 we have 3^6 + 6^3 = 945 but 945/36 = 105/4.
%e A243023 For 6189068 we have: 6^8 + 0^6 + 9^0 + 8^9 + 1^8 + 6^1 + 8^6 = 136159496.
%e A243023 Finally 136159496/6189068 = 22.
%p A243023 with(numtheory): P:=proc(q) local a,b,k,ok,n; for n from 10 to q do a:=[]; b:=n;
%p A243023 while b>0 do a:=[op(a),b mod 10]; b:=trunc(b/10); od; b:=0; ok:=1; for k from 2 to nops(a)
%p A243023 do if a[k-1]=0 and a[k]=0 then ok:=0; break; else b:=b+a[k]^a[k-1]; fi; od;
%p A243023 if ok=1 then if type((b+a[1]^a[nops(a)])/n,integer) then print(n);
%p A243023 fi; fi; od; end: P(10^10);
%t A243023 fQ[n_] := Block[{id = IntegerDigits@ n}, IntegerQ[ Total[ (id^RotateLeft@ id)]/n]]; k = 1; lst = {}; While[k < 1000000001, If[fQ@k, AppendTo[lst, k]; Print@ k]; k++]; lst (* _Robert G. Wilson v_, Jun 01 2014 *)
%Y A243023 Cf. A243024, A243025.
%K A243023 nonn,base,fini,full
%O A243023 1,2
%A A243023 _Paolo P. Lava_, May 29 2014