A330492 a(n) = sum of second differences of the sorted divisors of n.

0, 0, 1, 0, 2, 0, 3, 4, 4, 0, 5, 0, 6, 8, 7, 0, 8, 0, 9, 12, 10, 0, 11, 16, 12, 16, 13, 0, 14, 0, 15, 20, 16, 24, 17, 0, 18, 24, 19, 0, 20, 0, 21, 28, 22, 0, 23, 36, 24, 32, 25, 0, 26, 40, 27, 36, 28, 0, 29, 0, 30, 40, 31, 48, 32, 0, 33, 44, 34, 0, 35, 0, 36

Offset: 2

Author

Michel Lagneau, Dec 16 2019

Keywords

• nonn

Comments

The sums of the first differences of the divisors of n are given by the sequence b(n) = n - 1.

Let the set {D(i)} = {d(i + 1) - d(i)} where the d(i) are the divisors of an integer m listed in ascending order with i = 1, 2 , ..., tau(n)-1. The sequence is given by a(n) = Sum_{k = 1..tau(n)-2} (D(k + 1) - D(k)).

Examples

			a(12) = 5 because the divisors of 12 are {1, 2, 3, 4, 6, 12} and {D(i)} = {d(i+1)-d(i)} ={1, 1, 1, 2, 6}, Sum_{D(i), i = 1..4} {D(i+1)-D(i)} = 0 + 0 + 1 + 4 = 5.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 2..8191
• Antti Karttunen, Data supplement: n, a(n) computed for n = 2..65537

Crossrefs

Cf. A000005, A027750, A036263, A193829.

Programs

Maple

with(numtheory):nn:=100:
for n from 2 to nn do:
d:=divisors(n):n0:=nops(d):T:=array(1..n0-1,[0$n0-1]):
  for j from 1 to n0-1 do:
   T[j]:=d[j+1]-d[j]:
  od:
   s:=sum(‘T[i+1]-T[i] ’,‘i’=1..n0-2): printf(`%d, `,s):
od:
*** alternative program using the formula ***
with(numtheory):nn:=100:
for n from 2 to nn do:
d:=divisors(n):t:=tau(n):s:=d[t]-d[t-1]+d[1]-d[2] :
  printf(`%d, `,s):
od:

Mathematica

Array[Total@ Differences[Divisors@ #, 2] &, 73, 2] (* Michael De Vlieger, Dec 16 2019 *)

PARI

a(n) = my(d=divisors(n)); d[#d] - d[#d-1] + d[1] - d[2]; \\ Michel Marcus, Feb 05 2020

Python

from sympy import primefactors
def a(n): p = primefactors(n)[0]; return (n//p - 1) * (p - 1)
print([a(n) for n in range(2, 75)]) # Michael S. Branicky, Apr 04 2021

Formula

a(n) = d(tau(n)) - d(tau(n) - 1) + d(1) - d(2) where d(i) are the divisors of n.

a(prime(n)) = 0 and a(2k) = k-1, k = 1, 2, ...

a(p^2) = (p-1)^2 if p prime, with the generalization a(p^m) = (p-1)(p^(m-1) - 1).

a(n) = (n/p-1)*(p-1), where p is the least prime factor of n. - Nathaniel Gregg, Apr 04 2021