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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.

%I A319070 #35 Apr 29 2023 07:01:44
%S A319070 0,1,4,4,16,7,36,12,24,19,100,17,144,39,44,32,256,33,324,41,72,103,
%T A319070 484,40,160,147,108,65,784,57,900,80,152,259,228,66,1296,327,204,93,
%U A319070 1600,99,1764,137,160,487,2116,92,504,165,332,185,2704,135,388
%N 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.
%H A319070 Rémy Sigrist, <a href="/A319070/b319070.txt">Table of n, a(n) for n = 1..10000</a>
%H A319070 Luc Rousseau, <a href="/A319070/a319070.png">Diagram illustrating a(10)=19.</a>
%F A319070 a(1) = 0.
%F A319070 a(p) = (p-1)^2 for p a prime number.
%F A319070 a(p^k) = (p-1)^2*k*p^(k-1) for p^k a prime power.
%F A319070 a(p*q) = 2*(p-1)^2*q + (q-p)^2 for p and q primes (p < q).
%F A319070 a(n) = (n/2 - 1)^2 + 3 if n=2*p with p a prime greater than 2.
%F A319070 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).
%F A319070 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.:
%F A319070 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,
%F A319070 but
%F A319070 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.
%F A319070 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
%e A319070 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.
%t A319070 a[n_] := Module[{x = Differences[Divisors[n]]}, Plus @@ (x*Reverse[x])];
%t A319070 Table[a[n], {n, 1, 55}]
%o A319070 (PARI) arect(n, d, D) = (D-d)*(n/d - n/D);
%o A319070 a(n) = my(vd = divisors(n)); sum(k=1, #vd-1, arect(n, vd[k], vd[k+1])); \\ _Michel Marcus_, Oct 28 2018
%Y A319070 Cf. A191743, A290110 (introducing factorization patterns of sequences of divisors).
%Y A319070 Cf. A165900 (the Fibonacci polynomial).
%K A319070 nonn,look
%O A319070 1,3
%A A319070 _Luc Rousseau_, Sep 09 2018