A008476 -id:A008476 - cp's OEIS frontend

A182938 If n = Product (p_j^e_j) then a(n) = Product (binomial(p_j, e_j)).

1, 2, 3, 1, 5, 6, 7, 0, 3, 10, 11, 3, 13, 14, 15, 0, 17, 6, 19, 5, 21, 22, 23, 0, 10, 26, 1, 7, 29, 30, 31, 0, 33, 34, 35, 3, 37, 38, 39, 0, 41, 42, 43, 11, 15, 46, 47, 0, 21, 20, 51, 13, 53, 2, 55, 0, 57, 58, 59, 15, 61, 62, 21, 0, 65, 66

Offset: 1

Peter Luschny, Jan 16 2011

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Plot of Sum_{k=1..n} a(k) / n^2 for n = 1..1000000

Cf. A000026, A001414, A008473, A008474, A008475, A008476, A008477, A028310, A069799.

Cf. A027748, A124010, A007318, A328745.

a182938 n = product $ zipWith a007318'
   (a027748_row n) (map toInteger $ a124010_row n)
-- Reinhard Zumkeller, Feb 18 2012

A182938 := proc(n) local e,j; e := ifactors(n)[2]:
mul (binomial(e[j][1], e[j][2]), j=1..nops(e)) end:
seq (A182938(n), n=1..100);

a[n_] := Times @@ (Map[Binomial @@ # &, FactorInteger[n], 1]);
Table[a[n], {n, 1, 100}] (* Kellen Myers, Jan 16 2011 *)

a(n)=prod(i=1,#n=factor(n)~,binomial(n[1,i],n[2,i])) \\ M. F. Hasler

for(n=1, 100, print1(direuler(p=2, n, (1 + X)^p)[n], ", ")) \\ Vaclav Kotesovec, Mar 28 2025

a(A185359(n)) = 0. - Reinhard Zumkeller, Feb 18 2012
Dirichlet g.f.: Product_{p prime} (1 + p^(-s))^p. - Ilya Gutkovskiy, Oct 26 2019
Conjecture: Sum_{k=1..n} a(k) ~ c * n^2, where c = 0.33754... - Vaclav Kotesovec, Mar 28 2025

Given terms checked with new PARI code by M. F. Hasler, Jan 16 2011

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

Original entry on oeis.org

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

Sebastian Karlsson, Jan 26 2021

sign

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.

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

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.

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

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

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

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=', ')

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

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

cp's OEIS Frontend

A182938 If n = Product (p_j^e_j) then a(n) = Product (binomial(p_j, e_j)).

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

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