A340901 Additive with a(p^e) = (-p)^e.

0, -2, -3, 4, -5, -5, -7, -8, 9, -7, -11, 1, -13, -9, -8, 16, -17, 7, -19, -1, -10, -13, -23, -11, 25, -15, -27, -3, -29, -10, -31, -32, -14, -19, -12, 13, -37, -21, -16, -13, -41, -12, -43, -7, 4, -25, -47, 13, 49, 23, -20, -9, -53, -29, -16, -15, -22, -31

Offset: 1

Author

Sebastian Karlsson, Jan 26 2021

Keywords

• sign

Comments

The sequence contains every integer infinitely many times.

Proof (outline):

1. Every integer m > 9 is the sum of distinct odd primes [R. E. Dressler].

2. Any integer k (positive as negative) can be written as k = 4^e - m, for sufficiently large and infinitely many e > 0 and m > 9.

3. Pick an arbitrary integer k and write it like k = 4^e - m. Let p_1, p_2, ..., p_i be distinct odd primes such that p_1 + p_2 + ... + p_i = m. Then a(p_1*p_2*...*p_i*4^e) = 4^e - m = k. Since there are infinitely many representations of any k of the form 4^e - m, this means that there are infinitely many n such that a(n) = k.

Q.E.D.

Examples

			a(20) = a(2^2*5) = (-2)^2 + (-5) = -1.
		

Links

• Sebastian Karlsson, Table of n, a(n) for n = 1..10000
• R. E. Dressler, A stronger Bertrand's postulate with an application to partitions, Proc. Amer. Math. Soc., 33 (1972), 226-228.

Crossrefs

Cf. A001414, A002035, A008472, A008475, A008476, A316523.

Programs

Mathematica

a[n_] := Total@ (((-First[#])^Last[#]) & /@ FactorInteger[n]); a[1] = 0; Array[a, 100] (* Amiram Eldar, May 15 2023 *)

PARI

a(n) = my(f=factor(n)); sum(k=1, #f~, (-f[k,1])^f[k,2]); \\ Michel Marcus, Jan 26 2021
(APL, Dyalog dialect) A340901 ← {1=⍵:0 ⋄ +/{(-⍺)*≢⍵}⌸factors(⍵)} ⍝ Needs also factors function from https://dfns.dyalog.com/c_factors.htm - Antti Karttunen, Feb 16 2024

Python

from sympy import primefactors as pf, multiplicity as mult
def a(n):
    return sum([(-p)**mult(p, n) for p in pf(n)])
for n in range(1, 59):
    print(a(n), end=', ')

Formula

a(A002035(n)) = - A008475(A002035(n)).

a(n^2) = A008475(n^2).