cp's OEIS Frontend

A254008 Numbers that divide the sum of the reverse of their divisors (A069192).

%I A254008 #17 Sep 08 2022 08:46:11
%S A254008 1,6,15,37,87,397,435,639,1086,2574,8997,10986,26994,101918,161406,
%T A254008 207321,793170,834657,890827,1976355,3224625,7591023,8999997,9970098,
%U A254008 11052598,17930619,18368601,21739023,38014197,89999997,109999986,116141652,177765554,201264120
%N A254008 Numbers that divide the sum of the reverse of their divisors (A069192).
%H A254008 Lars Blomberg, <a href="/A254008/b254008.txt">Table of n, a(n) for n = 1..41</a>
%e A254008 Divisors of 8997 are 1, 3, 2999, 8997 and the sum of their reverse is 1 + 3 + 9992 + 7998 = 17994. Finally, 17994 / 8997 = 2.
%p A254008 with(numtheory):
%p A254008 T:=proc(w) local x,y,z; x:=w; y:=0;
%p A254008 for z from 1 to ilog10(x)+1 do y:=10*y+(x mod 10); x:=trunc(x/10); od; y; end:
%p A254008 P:=proc(q) local a,b,k; global n;
%p A254008 for n from 1 to q do a:=sort([op(divisors(n))]); b:=add(T(a[k]),k=1..nops(a));
%p A254008 if type(b/n,integer) then print(n); fi; od; end: P(10^9);
%o A254008 (Magma) [n: n in [1..10^6] | &+[Seqint(Reverse(Intseq(d))): d in Divisors(n)] mod n eq 0]; // _Bruno Berselli_, Jan 22 2015
%Y A254008 Cf. A069192, A254009.
%K A254008 nonn,base
%O A254008 1,2
%A A254008 _Paolo P. Lava_, Jan 22 2015
%E A254008 a(19)-a(33) from _Lars Blomberg_, Feb 27 2015