A294172 Maximum value of the cyclic convolution of 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

1,2

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.

Links

Table of n, a(n) for n=1..45.

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)

Example

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.

Mathematica

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

Prog

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

Crossrefs

Cf. A000292, A299111.

Sequence in context: A316537 A175254 A211636 * A055328 A027962 A023537

Adjacent sequences: A294169 A294170 A294171 * A294173 A294174 A294175

Keyword

nonn

Author

Andres Cicuttin, Feb 10 2018

Status

approved