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a(n) is p1^1 + p2^2 + ... + pk^k where {p1,p2,...,pk} are the distinct prime factors in ascending order in the prime factorization of n.
3

%I #30 Sep 19 2024 05:24:57

%S 0,2,3,2,5,11,7,2,3,27,11,11,13,51,28,2,17,11,19,27,52,123,23,11,5,

%T 171,3,51,29,136,31,2,124,291,54,11,37,363,172,27,41,354,43,123,28,

%U 531,47,11,7,27,292,171,53,11,126,51,364,843,59,136,61,963,52,2,174,1342,67,291,532,370,71,11,73

%N a(n) is p1^1 + p2^2 + ... + pk^k where {p1,p2,...,pk} are the distinct prime factors in ascending order in the prime factorization of n.

%C From _Bernard Schott_, May 07 2021: (Start)

%C a(n) depends only on prime factors of n (see formulas).

%C Primes are fixed points of this sequence.

%C Terms are in increasing order in A344023. (End)

%H Michel Marcus, <a href="/A343300/b343300.txt">Table of n, a(n) for n = 1..10000</a>

%F a(p^k) = p for p prime and k>=1.

%F From _Bernard Schott_, May 07 2021: (Start)

%F a(A033845(n)) = 11;

%F a(A033846(n)) = 27;

%F a(A033847(n)) = 51;

%F a(A033848(n)) = 123;

%F a(A033849(n)) = 28;

%F a(A033850(n)) = 52;

%F a(A033851(n)) = 54;

%F a(A288162(n)) = 171. (End)

%e a(60) = 136 because the distinct prime factors of 60 are {2, 3, 5} and 2^1 + 3^2 + 5^3 = 136.

%p a:= n-> (l-> add(l[i]^i, i=1..nops(l)))(sort(map(i-> i[1], ifactors(n)[2]))):

%p seq(a(n), n=1..73); # _Alois P. Heinz_, Sep 19 2024

%t {0}~Join~Table[Total[(a=First/@FactorInteger[k])^Range@Length@a],{k, 2, 100}]

%o (PARI) a(n) = my(f=factor(n)); sum(k=1, #f~, f[k,1]^k); \\ _Michel Marcus_, Apr 11 2021

%Y Cf. A027748, A344023 (terms ordered).

%Y See also A033845-A033851, A288162.

%K nonn

%O 1,2

%A _Giorgos Kalogeropoulos_, Apr 11 2021