A319070 a(n) is the area of the surface made of the rectangles with vertices (d, n/d), (D, n/d), (D, n/D), (d, n/D) for all (d, D), pair of consecutive divisors of n.

0, 1, 4, 4, 16, 7, 36, 12, 24, 19, 100, 17, 144, 39, 44, 32, 256, 33, 324, 41, 72, 103, 484, 40, 160, 147, 108, 65, 784, 57, 900, 80, 152, 259, 228, 66, 1296, 327, 204, 93, 1600, 99, 1764, 137, 160, 487, 2116, 92, 504, 165, 332, 185, 2704, 135, 388

Offset: 1

Author

Luc Rousseau, Sep 09 2018

Keywords

• nonn
• look

Examples

			The divisors of n=12 are {1, 2, 3, 4, 6, 12}. The widths of the rectangles from the definition are obtained by difference: {1, 1, 1, 2, 6}. By symmetry, their heights are the same, but in reverse order: {6, 2, 1, 1, 1}. The sought total area is the sum of products width*height of each rectangle, in other words it is the dot product 1*6 + 1*2 + 1*1 + 2*1 + 6*1. Result: 17. So, a(12)=17.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Luc Rousseau, Diagram illustrating a(10)=19.

Crossrefs

Cf. A191743, A290110 (introducing factorization patterns of sequences of divisors).

Cf. A165900 (the Fibonacci polynomial).

Programs

Mathematica

a[n_] := Module[{x = Differences[Divisors[n]]}, Plus @@ (x*Reverse[x])];
Table[a[n], {n, 1, 55}]

PARI

arect(n, d, D) = (D-d)*(n/d - n/D);
a(n) = my(vd = divisors(n)); sum(k=1, #vd-1, arect(n, vd[k], vd[k+1])); \\ Michel Marcus, Oct 28 2018

Formula

a(1) = 0.

a(p) = (p-1)^2 for p a prime number.

a(p^k) = (p-1)^2*k*p^(k-1) for p^k a prime power.

a(p*q) = 2*(p-1)^2*q + (q-p)^2 for p and q primes (p < q).

a(n) = (n/2 - 1)^2 + 3 if n=2*p with p a prime greater than 2.

a(n) = (n/p + F(p-1))^2 + p^2 - F(p-1)^2 if n = p*q, p < q primes; where F denotes the Fibonacci polynomial, F(x) = x^2 - x - 1 (see A165900).

For more complex factorization patterns of n, the formula depends on the factorization pattern of the sequence of divisors of n (see A191743 or A290110), e.g.:

a(p^2*q) = 4*p*q*(p-1)^2 + (q-p^2)^2 if 1 < p < p^2 < q < p*q < p^2*q,

but

a(p^2*q) = 2*p*q*(p-1)^2 + 2*p*(q-p)^2 + (p^2-q)^2 if 1 < p < q < p^2 < p*q < p^2*q.

a(n) = Sum_{i=1..tau(n)-1} (d_[tau(n)-i+1] - d_[tau(n)-i])*(d_[i+1] - d_[i]), where {d_i}, i=1..tau(n) is the increasing sequence of divisors of n. - Ridouane Oudra, Oct 17 2021