A339378 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.

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, 27, 23, 29, 68, 31, 32, 38, 49, 32, 59, 37, 51, 40, 57, 41, 66, 43, 43, 52, 55, 47, 59, 49, 57, 44, 45, 53, 59, 36, 81, 46, 61, 59, 84, 61, 63, 76, 64, 38, 102, 67, 81

Offset: 1

Author

Bernard Schott, Dec 06 2020

Keywords

• nonn

Comments

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).

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.

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].

There exist two infinite families of numbers (see De Koninck & Mercier reference) that satisfy n < a(n):

--> n = 2^k + 2, k>=2 or n is in A052548 \ {3,4}, and,

--> n = 2*p with odd prime p or n is in A100484 \ {4}.

Other integers like 12, 20, 30, 33, 36, 39, 42, 45, 48, 50 satisfy also n < a(n).

Examples

			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.
		

References

• J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, 2004, Problème 683, pp. 89 and 294.

Links

• Eötvös-Kürschák Competitions, Problem 2, 85th Eötvös-Kürschák Competition 1985.
• Index to sequences related to Olympiads and other Mathematical competitions.

Crossrefs

Cf. A000523, A052548, A053644, A100484, A246655.

Programs

Maple

pf := n -> NumberTheory:-PrimeFactors(n): a := n -> add(p^ilog[p](n), p in pf(n)):
seq(a(n), n=1..68); # Peter Luschny, Dec 07 2020

Mathematica

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 *)

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

Formula

a(n) = Sum_{p | n} p^(floor(log_p(n))).

a(n) = n iff n = p^k , p prime, k >= 1 (A246655).