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Lucas numbers which are divisible by the sum of their digits.
1

%I #10 Feb 08 2021 02:54:22

%S 1,3,4,7,18,322,5778,505019158607,84722519070079276,

%T 1473646213395791149646646123,

%U 105249261265075663875711417309855979021650214636

%N Lucas numbers which are divisible by the sum of their digits.

%C a(12) has 83 digits and it is too large to include in the data section. - _Amiram Eldar_, Feb 08 2021

%H Amiram Eldar, <a href="/A117789/b117789.txt">Table of n, a(n) for n = 1..17</a>

%F a(n) = A000204(A337449(n+1)). - _Amiram Eldar_, Feb 08 2021

%e 322 is in the sequence because it is a Lucas number and it is divisible by the sum of its digits, 3+2+2 = 7.

%t Select[LinearRecurrence[{1, 1}, {1, 3}, 230], Divisible[#, Plus @@ IntegerDigits[#]] &] (* _Amiram Eldar_, Feb 08 2021 *)

%o (PARI) {m=370;a=1;b=3;print1(a,",",b,",");for(n=3,m,c=b+a;a=b;b=c;s=0;k=b;while(k>0,d=divrem(k,10);k=d[1];s=s+d[2]);if(b%s==0,print1(b,",")))} \\ _Klaus Brockhaus_, Apr 17 2006

%Y Intersection of A000204 and A005349.

%Y Cf. A337449.

%K base,nonn

%O 1,2

%A Luc Stevens (lms022(AT)yahoo.com), Apr 15 2006

%E a(9) corrected, a(10) and a(11) from _Klaus Brockhaus_, Apr 17 2006