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.
The conjectured formula is true. See links. - Sela Fried, Aug 13 2024.
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.
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
KEYWORD
nonn
AUTHOR
Andres Cicuttin, Feb 10 2018
STATUS
approved