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Squarefree numbers whose product of prime indices is twice their sum of prime indices.
4

%I #18 Jul 20 2020 10:55:04

%S 65,154,190

%N Squarefree numbers whose product of prime indices is twice their sum of prime indices.

%C A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

%C This sequence is finite. Proof: k = p_1*p_2*...*p_t is a term iff q_1*q_2*...*q_t = 2*(q_1 + q_2 + ... + q_t), where q_i = pi(p_i) and q_1 < q_2 < ... < q_t. If t = 2, then 1/2 = 1/q_1 + 1/q_2. Thus q_1 <= 3, we have k = prime(3)*prime(6) = 65. If t = 3, then 1/2 = 1/(q_1*q_2) + 1/(q_1*q_3) + 1/(q_2*q_3). Thus q_1*q_2 <= 5, we have k = prime(1)*prime(4)*prime(5) = 154 or k = prime(1)*prime(3)*prime(8) = 190. If t > 3, then 1/2 = Sum_{i=1..t} q_i/(q_1*q_2*...*q_t) < Sum_{i=1..t} i/t! < 1/2, a contradiction. - _Jinyuan Wang_, Jun 27 2020

%F A003963(a(n)) = 2 * A056239(a(n)).

%e The sequence of terms together with their prime indices starts:

%e 65: {3,6}

%e 154: {1,4,5}

%e 190: {1,3,8}

%p q:= n-> (l-> andmap(i-> i[2]=1, l) and (h-> mul(i, i=h)=2*add(i,

%p i=h))(map(i-> numtheory[pi](i[1]), l)))(ifactors(n)[2]):

%p select(q, [$1..1000])[]; # _Alois P. Heinz_, Sep 12 2019

%t primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];

%t Select[Range[10000],SquareFreeQ[#]&&SameQ[Times@@primeMS[#],2*Plus@@primeMS[#]]&]

%Y Intersection of A005117 and A326151.

%Y Product of prime indices is A003963.

%Y Sum of prime indices is A056239.

%Y Cf. A000720, A001222, A069016, A112798, A301987, A325041, A325042, A326152.

%K nonn,bref,fini,full

%O 1,1

%A _Gus Wiseman_, Sep 12 2019