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Numbers n such that sum of decimal digits of n equals number of prime divisors of n counted with multiplicity and sum of distinct decimal digits of n equals number of distinct primes dividing n.
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%I #7 Apr 23 2017 18:45:39

%S 30,102,1002,1012,1210,2001,2120,3010,10002,10030,20001,20112,20120,

%T 100012,100030,101020,102010,110020,110120,120001,121120,200001,

%U 200120,211100,221120,230010,300010,320320,400010,400140,1000002,1000012,1000140,1000230,1001020,1003002,1004010,1010120,1011300,1013310,1021100

%N Numbers n such that sum of decimal digits of n equals number of prime divisors of n counted with multiplicity and sum of distinct decimal digits of n equals number of distinct primes dividing n.

%C Numbers n such that A007953(n) = A001222(n) and A217928(n) = A001221(n).

%H Giovanni Resta, <a href="/A280911/b280911.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DigitSum.html">Digit Sum</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeFactor.html">Prime Factor</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DistinctPrimeFactors.html">Distinct Prime Factors</a>

%e 20112 is in the sequence because 20112 = 2^4*3*419 (6 prime factors, 3 distinct), 2 + 0 + 1 + 1 + 2 = 6 and 2 + 0 + 1 = 3.

%t Select[Range[1100000], Total[IntegerDigits[#1]] == PrimeOmega[#1] && Total[Union[IntegerDigits[#1]]] == PrimeNu[#1] &]

%Y Cf. A001221, A001222, A007953, A050689, A050690, A057531, A217928.

%K nonn,base,easy

%O 1,1

%A _Ilya Gutkovskiy_, Jan 10 2017