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Smallest number between consecutive odd primes whose sum of prime factors is a minimum.
2

%I #17 Mar 28 2014 13:10:41

%S 4,6,8,12,15,18,20,24,30,32,40,42,45,48,54,60,64,70,72,75,81,84,90,

%T 100,102,105,108,112,120,128,135,138,144,150,154,162,165,168,175,180,

%U 189,192,196,198,200,216,225,228,231,234,240,243,256,260,264,270,275

%N Smallest number between consecutive odd primes whose sum of prime factors is a minimum.

%C The prime factors are counted with multiplicity, as in A001414.

%H Lei Zhou, <a href="/A239439/b239439.txt">Table of n, a(n) for n = 1..10000</a>

%e For n = 1, the 1st and 2nd odd prime numbers are 3 and 5. 4 is the only number between them. So a(1) = 4;

%e ...

%e For n = 10, the 10th and 11th odd prime numbers are 31 and 37. Testing from 32 to 36:

%e 32 = 2^5, sum of prime factors = 2*5 = 10;

%e 33 = 3*11, sum of prime factors = 3+11 = 14;

%e 34 = 2*17, sum of prime factors = 2+17 = 19;

%e 35 = 5*7, sum of prime factors = 5+7 = 12;

%e 36 = 2^2*3^2, sum of prime factors = 2*2+3*2 = 10;

%e 32 and 36 have the minimum sum of prime factors, i.e., 10, and 32 is the smaller number of the two. So a(10) = 32.

%t Table[p1 = Prime[n]; p2 = Prime[n + 1]; a = p2; Do[f = FactorInteger[i]; l = Length[f]; sum = 0; Do[sum = sum + f[[j, 1]]*f[[j, 2]], {j, 1, l}]; If[sum < a, a = sum; s = i];, {i, p1 + 1, p2 - 1}]; s, {n, 2, 58}]

%Y Cf. A000040, A001414.

%K nonn,easy

%O 1,1

%A _Lei Zhou_, Mar 18 2014