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n divided by ((sum of digits of n) times (product of digits of n)) is prime.
2

%I #33 Feb 15 2019 10:26:40

%S 12,111,216,432,41112,81216,186624,248832,311472,316224,341712,422144,

%T 714112,1131111,1131732,1191915,1211328,1292112,1418112,2192832,

%U 3112128,4331232,11127424,11311272,18122112,21111192,26726112,28422144,34338816

%N n divided by ((sum of digits of n) times (product of digits of n)) is prime.

%H David A. Corneth, <a href="/A066042/b066042.txt">Table of n, a(n) for n = 1..1744</a> (first 469 terms from Harry J. Smith and Chai Wah Wu)

%H David A. Corneth, <a href="/A066042/a066042.gp.txt">a(n) = [product of digits of a(n)] * [sum of digits of a(n)] * [some prime]</a>

%F Sum digits of n; take product of digits of n; multiply sum by product and divide into n. If prime, add to sequence.

%e a(2) = 111 because 1+1+1 = 3 and 1*1*1 = 1 and 3*1 = 3 and 111/3 = 37 and 37 is prime. [corrected by _Harry J. Smith_, Nov 08 2009]

%t ndspQ[n_]:=Module[{idn=IntegerDigits[n]},FreeQ[idn,0]&&PrimeQ[n/(Total[ idn]Times@@idn)]]; Select[Range[35*10^6],ndspQ] (* _Harvey P. Dale_, Feb 09 2015 *)

%o (PARI) ProdD(x)= { local(p=1); while (x>9 && p>0, p*=x%10; x\=10); return(p*x) } SumD(x)= { local(s=0); while (x>9, s+=x%10; x\=10); return(s + x) } { n=0; for (m=1, 10^12, p=ProdD(m); if (p == 0, next); f=m/(SumD(m)*p); if (frac(f)==0 && isprime(f), write("b066042.txt", n++, " ", m); if (n==100, return)) ) } \\ _Harry J. Smith_, Nov 08 2009

%Y Cf. A038369, A049102, A066146.

%K easy,nonn,base

%O 1,1

%A _Enoch Haga_, Dec 13 2001

%E Checked to over 10^8 (110508539) without finding another example.

%E Offset 1 from _Harry J. Smith_, Nov 08 2009

%E Should have found 34338816, 37121112, and 41174112 < 10^8. Term a(29) from _Harry J. Smith_, Nov 08 2009