%I #20 Feb 10 2019 05:18:59
%S 286330897,286330943,388098901,955201943,1776186851,1854778853,
%T 2559495863,2647782901,3517793911,3628857863,3866728909,3974453911,
%U 4167637819,4269837799,5083007887,5362197829,5642510933,6034811933,8180784851,8214319903
%N Primes such that iteration of A062028 (n + its digit sum) yields 6 primes in a row.
%C In contrast to A048523, ..., A048527, this definition uses "at least" for the number of successive primes. This allows easier computation of subsequences of terms which yield even more primes in a row.
%C One can nonetheless compute the terms of this sequence by considering possible pre-images under A062028 of terms of A048527. This gives the terms which yield exactly 6 primes in a row (i.e., A320878 \ A320879), and one has to take the union with further iterates of this procedure (which successively yields A320879 \ A320880, etc).
%H Lars Blomberg, <a href="/A320878/b320878.txt">Table of n, a(n) for n = 1..7626</a> (Terms < 10^14. The first 200 from M. F. Hasler)
%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_163.htm">Puzzle 163. P+SOD(P)</a>
%F Numbers n in A048519 for which A062028(n) is in A048527, form the subset A320878 \ A320879.
%o (PARI) is_A320878(n,p=n)={for(i=1,6, isprime(p=A062028(p))||return);isprime(n)}
%o forprime(p=286e6,,is_A320878(p)&& print1(p","))
%o /* much faster, using the precomputed array A048527, as follows: */
%o PP(n)=select(p->p+sumdigits(p)==n,primes([n-9*#digits(n),n-2])) \\ Returns list of prime predecessors for A062028. (PP(n) nonempty <=> n in A320881.)
%o A320878=[]; my(S=A048527); while(#S=Set(concat(apply(PP,S))), A320878=setunion(A320878,S)) \\ Yields 211 terms from A048527[1..3000]
%Y Cf. A062028 (n + digit sum of n), A047791 (A062028(n) is prime), A048519 (primes among these).
%Y Cf. A048523 .. A048527, A320879.
%Y a(1) = A090009(7) = start of first chain of 7 primes under iteration of A062028.
%Y Cf. A320881, A048520.
%Y Cf. A230093 (number of m s.th. m + (sum of digits of m) = n) and references there.
%K nonn,base
%O 1,1
%A _Zak Seidov_ and _M. F. Hasler_, Nov 08 2018