%I #11 Sep 05 2022 19:32:50
%S 0,2,3,4,5,5,7,6,6,7,11,7,0,9,8,8,0,8,0,9,10,13,0,9,10,2,9,11,0,10,0,
%T 10,14,2,12,10,0,2,3,11,0,12,0,15,11,2,0,11,14,12,3,4,0,11,16,13,3,2,
%U 0,12,0,2,13,12,5,16,0,4,3,14,0,12,0,2,13
%N Sum of the palindromic primes dividing n (with repetition).
%C a(p)=0 for primes p which are not in A002385.
%C The first runs of 2, 3 and 4 identical values start at a(5), a(370) and a(2776191), respectively. - _Giovanni Resta_, Sep 09 2015
%F Let n=product_i p(i)^e(i), then a(n)=sum_{i,p(i) in A002113} e(i)*p(i).
%F a(n) <= A001414(n).
%p A262049 := proc(n)
%p local a,d ;
%p a := 0 ;
%p for d in ifactors(n)[2] do
%p if isA002113(op(1,d)) then
%p a := a+op(1,d)*op(2,d) ;
%p end if;
%p end do:
%p a ;
%p end proc:
%p seq(A262049(n),n=1..120) ;
%t palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; {0}~Join~
%t Table[Total@ Select[Flatten[Table[#1, {#2}] & @@@ FactorInteger@ n], palQ], {n, 2, 75}] (* _Michael De Vlieger_, Sep 09 2015 *)
%t Join[{0},Table[Total[Times@@@Select[FactorInteger[n],PalindromeQ[#[[1]]]&]],{n,2,80}]] (* _Harvey P. Dale_, Sep 05 2022 *)
%Y Cf. A002113, A002385.
%K nonn,base
%O 1,2
%A _R. J. Mathar_, Sep 09 2015