A273133 a(n) = n minus the bottom entry of the difference table of the divisors of n.

0, 1, 1, 3, 1, 4, 1, 7, 5, 10, 1, 11, 1, 16, 7, 15, 1, 6, 1, 31, 13, 28, 1, 36, 9, 34, 19, 31, 1, -20, 1, 31, 25, 46, 7, 47, 1, 52, 31, 106, 1, -62, 1, 31, 21, 64, 1, 151, 13, 66, 43, 31, 1, -34, 19, 8, 49, 82, 1, 727, 1, 88, 71, 63, 25, -6, 1, 31, 61, 148, 1, 12, 1, 106, 11, 31, 13, 22, 1, 439, 65, 118, 1, 1541

Offset: 1

Author

Omar E. Pol, May 17 2016

Keywords

• sign

Comments

From David A. Corneth, May 20 2016: (Start)

The bottom of the difference table of the divisors of n can be expressed in terms of the divisors of n and use of Pascal's triangle. Suppose a, b, c, d and e are the divisors of n. Then the difference table is as follows (rotated for ease of reading):

a

. . b-a

b . . . . c-2b+a

. . c-b . . . . . d-3c+3b-a

c . . . . d-2c+b . . . . . . e-4d+6c-4b+a

. . d-c . . . . . e-3d+3c-b

d . . . . e-2d+c

. . e-d

e

From here we can see Pascal's triangle occurring. Induction can be used to show that it's the case in general.

(End)

Examples

			For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18, and the difference triangle of the divisors is:
1 . 2 . 3 . 6 . 9 . 18
. 1 . 1 . 3 . 3 . 9
. . 0 . 2 . 0 . 6
. . . 2 .-2 . 6
. . . .-4 . 8
. . . . . 12
The bottom entry is 12, so a(18) = 18 - 12 = 6.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000040, A000225, A007318, A187202, A273102, A273103, A273109.

Programs

Mathematica

Array[# - First@ NestWhile[Differences, Divisors@ #, Length@ # > 1 &] &, 84] (* Michael De Vlieger, May 20 2016 *)

PARI

a(n) = my(d=divisors(n));n-sum(i=1,#d,binomial(#d-1,i-1)*(-1)^(#d-i)*d[i]) \\ David A. Corneth, May 20 2016

Sage

def A273133(n):
    D = divisors(n)
    T = matrix(ZZ, len(D))
    for (m, d) in enumerate(D):
        T[0, m] = d
        for k in range(m-1, -1, -1) :
            T[m-k, k] = T[m-k-1, k+1] - T[m-k-1, k]
    return n - T[len(D)-1, 0]
print([A273133(n) for n in range(1, 85)]) # Peter Luschny, May 18 2016

Formula

a(n) = n - A187202(n).

a(n) = 1, if n is prime.

a(2^k) = 2^k - 1 = A000225(k), k >= 0.