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%I #34 Dec 08 2020 10:24:01
%S 0,2,3,4,5,7,7,8,9,13,11,17,13,15,14,16,17,25,19,21,16,27,23,25,25,29,
%T 27,23,29,68,31,32,38,49,32,59,37,51,40,57,41,66,43,43,52,55,47,59,49,
%U 57,44,45,53,59,36,81,46,61,59,84,61,63,76,64,38,102,67,81
%N Let n be a positive integer. For each prime divisor p of n, consider the highest power of p which does not exceed n. The sum a(n) of these powers is defined as the power-sum of n.
%C The idea of this sequence comes from the 85th Eötvös-Kürschák Competition of 1985, Class 9 - 12, Category 1, Round 1, Problem 2. In the problem it was asked to prove that there exist infinitely many positive integers n such as n < a(n) (see link).
%C The Kürschák Mathematical Competition, founded in 1894 in Hungary, but known as Eötvös Mathematical Competition until 1938, is the oldest modern mathematical competition for students in the world.
%C For each prime divisor p of n, the highest power of p which does not exceed n is equal to p^(floor(log_p(n))) [for p=2, see A000523 and A053644].
%C There exist two infinite families of numbers (see De Koninck & Mercier reference) that satisfy n < a(n):
%C --> n = 2^k + 2, k>=2 or n is in A052548 \ {3,4}, and,
%C --> n = 2*p with odd prime p or n is in A100484 \ {4}.
%C Other integers like 12, 20, 30, 33, 36, 39, 42, 45, 48, 50 satisfy also n < a(n).
%D J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 683, pp. 89 and 294.
%H Eötvös-Kürschák Competitions, <a href="http://www.batmath.it/matematica/raccolte_es/ek_competitions/ek_competitions.pdf">Problem 2</a>, 85th Eötvös-Kürschák Competition 1985.
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical competitions</a>.
%F a(n) = Sum_{p | n} p^(floor(log_p(n))).
%F a(n) = n iff n = p^k , p prime, k >= 1 (A246655).
%e 12 = 2^2 * 3^1; the highest power of 2 which does not exceed 12 is 2^3 and the highest power of 3 which does not exceed 12 is 3^2, hence a(12) = 2^3 + 3^2 = 2^(floor(log_2(12)) + 3^(floor(log_3(12)) = 17.
%p pf := n -> NumberTheory:-PrimeFactors(n): a := n -> add(p^ilog[p](n), p in pf(n)):
%p seq(a(n), n=1..68); # _Peter Luschny_, Dec 07 2020
%t f[n_, p_] := p^Floor[Log[p, n]]; a[1] = 0; a[n_] := Plus @@ (f[n, #] & /@ FactorInteger[n][[;; , 1]]); Array[a, 100] (* _Amiram Eldar_, Dec 06 2020 *)
%o (PARI) a(n) = my(f=factor(n)); sum(k=1, #f~, my(p=f[k,1]); p^logint(n, p)); \\ _Michel Marcus_, Dec 06 2020
%Y Cf. A000523, A052548, A053644, A100484, A246655.
%K nonn
%O 1,2
%A _Bernard Schott_, Dec 06 2020
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