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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)^d_(i+1)}+d_(k)^d_(1).
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%I #22 Feb 18 2021 00:46:36

%S 1,2,3,4,5,6,7,8,9,63,448,1899,1942,4155,4355,8503,28375,44569,73687,

%T 1953504,1954329,70860598,522169982

%N 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)^d_(i+1)}+d_(k)^d_(1).

%C 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 m=10^8).

%e For 1899 we have: 9^9 + 9^8 + 8^1 + 1^9 = 430467219.

%e Finally 430467219/1899 = 226681.

%e For 1954329 we have: 9^2 + 2^3 +3^4 + 4^5 + 5^9 + 9^1 + 1^9 = 1954329.

%p with(numtheory): P:=proc(q) local a,b,k,ok,n; for n from 10 to q do a:=[]; b:=n;

%p 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 do if a[k-1]=0 and a[k]=0 then ok:=0; break; else b:=b+a[k-1]^a[k]; fi; od;

%p if ok=1 then if type((b+a[nops(a)]^a[nops(1)])/n,integer) then print(n);

%p fi; fi; od; end: P(10^10);

%o (PARI) isok(n) = d = digits(n); k = #d; (sum(i=1, k-1, j=k-i+1; d[j]^d[(j-1)])+ d[1]^d[k]) % n == 0; \\ _Michel Marcus_, Sep 29 2014

%Y Cf. A243023, A243025.

%K nonn,base,fini,full

%O 1,2

%A _Paolo P. Lava_, May 29 2014

%E a(1)-a(9) prepended and a(23) from _Hiroaki Yamanouchi_, Sep 29 2014