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Numbers n such that both numbers n/(d_1*d_2* ...*d_k) and n/(d_1+d_2+ ... +d_k) are prime, where d_1 d_2 ... d_k is the decimal expansion of n.
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%I #26 Aug 12 2017 03:36:10

%S 11133,11331,13131,31113,112116,121116,13111212,111311115,11114112112,

%T 111212112112,1111111711311,1111171111113,11111111112611112,

%U 11111111121161112,11111112111161112,11111119111131111,11111131111119111,11111139111111111,11111193111111111,11111211161111112,11111611111211112,11116111112111112,11116111211111112

%N Numbers n such that both numbers n/(d_1*d_2* ...*d_k) and n/(d_1+d_2+ ... +d_k) are prime, where d_1 d_2 ... d_k is the decimal expansion of n.

%C For n in this sequence, let prime p = n/(d_1*d_2* ...*d_k) so that n = d_1*d_2* ...*d_k * p. Then n/(d_1+d_2+ ... +d_k) equals either p or some prime dividing d_1*d_2* ...*d_k, that is 2, 3, 5, or 7. The latter case never takes place and thus n/(d_1*d_2* ...*d_k) = n/(d_1+d_2+ ... +d_k) is the same prime. So this sequence is a subsequence of both A034710 and A066307. - _Max Alekseyev_, Aug 19 2013

%H Chai Wah Wu, <a href="/A107650/b107650.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1717 from Max Alekseyev).

%e 111311115 is in the sequence because

%e 111311115/(1*1*1*3*1*1*1*1*5) and 111311115/(1+1+1+3+1+1+1+1+5)

%e are prime(since 1*1*1*3*1*1*1*1*5=1+1+1+3+1+1+1+1+5, the primes are equal).

%t Do[h = IntegerDigits[m]; l = Length[h]; If[Min[h] > 0 && PrimeQ[m/Sum[h[[k]], {k, l}]] && PrimeQ[m/Product[ h[[k]], {k, l}]], Print[m]], {m, 265000000}]

%K base,nonn

%O 1,1

%A _Farideh Firoozbakht_, May 21 2005

%E a(9)-a(10) from _Sean A. Irvine_, Nov 28 2010

%E Terms a(11) onward from _Max Alekseyev_, Aug 20 2013