A294172 Maximum value of the cyclic convolution of the first n positive integers with themselves.

1, 5, 13, 28, 50, 83, 126, 184, 255, 345, 451, 580, 728, 903, 1100, 1328, 1581, 1869, 2185, 2540, 2926, 3355, 3818, 4328, 4875, 5473, 6111, 6804, 7540, 8335, 9176, 10080, 11033, 12053, 13125, 14268, 15466, 16739, 18070, 19480, 20951, 22505, 24123, 25828, 27600

Offset: 1

Author

Andres Cicuttin, Feb 10 2018

Keywords

• nonn

Comments

Conjecture: a(n) = (n*(13 + 3*(-1)^n + 24*n + 14*n^2))/48, and then lim_{n -> infinity} a(n)/n^3 = 7/24.

The conjectured formula is true. See links. - Sela Fried, Aug 13 2024.

Examples

			For n = 4, the four possible cyclic convolutions of the first four positive integers with themselves are:
(1,2,3,4).(4,3,2,1) = 1*4 + 2*3 + 3*2 + 4*1 = 4 + 6 + 6 + 4 = 20,
(1,2,3,4).(3,2,1,4) = 1*3 + 2*2 + 3*1 + 4*4 = 3 + 4 + 3 + 16 = 26,
(1,2,3,4).(2,1,4,3) = 1*2 + 2*1 + 3*4 + 4*3 = 2 + 2 + 12 + 12 = 28,
(1,2,3,4).(1,4,3,2) = 1*1 + 2*4 + 3*3 + 4*2 = 1 + 8 + 9 + 8 = 26,
then a(4)=28 because 28 is the maximum among the four values.
		

Links

• Sela Fried, On the maximum value of the cyclic convolution of the first n positive integers with themselves (A294172), 2024.
• Sela Fried, Proofs of some Conjectures from the OEIS, arXiv:2410.07237 [math.NT], 2024. See p. 8.

Crossrefs

Cf. A000292, A299111.

Programs

Mathematica

a[n_] := Max[Table[Range[n].RotateRight[Reverse[Range[n]], k], {k, 0, n - 1}]];
Table[a[n], {n, 1, 45}]

PARI

a(n) = vecmax(vector(n, k, sum(i=1, n, (n-i+1)*(1+(i+k) % n)))); \\ Michel Marcus, Feb 11 2018

Formula

a(n) = Max {x; x=Sum_{i=1..n}(n-i+1)*(1+(i+k) mod n); for k=1..n}.

Conjectures from Colin Barker, Feb 11 2018: (Start)

G.f.: x*(1 + 3*x + 2*x^2 + x^3) / ((1 - x)^4*(1 + x)^2).

a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.

(End)