A008474 If n = Product (p_j^k_j) then a(n) = Sum (p_j + k_j).

0, 3, 4, 4, 6, 7, 8, 5, 5, 9, 12, 8, 14, 11, 10, 6, 18, 8, 20, 10, 12, 15, 24, 9, 7, 17, 6, 12, 30, 13, 32, 7, 16, 21, 14, 9, 38, 23, 18, 11, 42, 15, 44, 16, 11, 27, 48, 10, 9, 10, 22, 18, 54, 9, 18, 13, 24, 33, 60, 14, 62, 35, 13, 8, 20, 19, 68, 22, 28, 17, 72, 10, 74, 41, 11, 24, 20, 21

Offset: 1

Author

Olivier Gérard

Keywords

• nonn

Comments

a(1) = 0 by convention, but could equally be taken to be 1 or 2.

Since only the primes p_j with nonzero exponents k_j in the factorization of n are considered in Sum (p_j + k_j), to the empty product (1) should correspond the empty sum (0). a(1) = 0 is thus the most natural choice. - Daniel Forgues, Apr 06 2010

Conjecture: for m > 4, by iterating the map m -> A008474(m) one always reaches 5 [tested up to m = 320000]. - Ivan N. Ianakiev, Nov 10 2014

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Daniel Tsai, A recurring pattern in natural numbers of a certain property, Integers (2021) Vol. 21, Article #A32.
• Eric Weisstein's World of Mathematics, Prime Factorization

Crossrefs

Cf. A107737, A107738, A000026, A027748, A124010, A250030.

Programs

Haskell

a008474 n = sum $ zipWith (+) (a027748_row n) (a124010_row n)
-- Reinhard Zumkeller, Feb 11 2012, Aug 27 2011

Maple

A008474 := proc(n) local e,j; e := ifactors(n)[2]:
add(e[j][1]+e[j][2],j=1..nops(e)) end:
seq (A008474(n), n=1..60);
# Peter Luschny, Jan 17 2011

Mathematica

A008474[n_]:=Plus@@Flatten[FactorInteger[n]]; Table[A008474[n], {n, 200}] (* Zak Seidov, May 23 2005 *)

PARI

{for(k=1, 79,
M=factor(k); smt =0;
for(i=1, matsize(M)[1], for(j=1, matsize(M)[2], smt=smt+M[i,j]));
print1(smt, ", "))} \\\ Douglas Latimer, Apr 27 2012

PARI

a(n)=my(f=factor(n)); sum(i=1,#f~,f[i,1]+f[i,2]) \\ Charles R Greathouse IV, Jun 03 2015

Python

from sympy import factorint
def a(n): return 0 if n == 1 else sum(p+k for p, k in factorint(n).items())
print([a(n) for n in range(1, 79)]) # Michael S. Branicky, Mar 28 2022

Formula

Additive with a(p^e) = p + e.

Extensions

More terms from Zak Seidov, May 23 2005