Philippe Gibone - cp's OEIS frontend

User: Philippe Gibone

Philippe Gibone has authored 3 sequences.

A350513 Number of 2 X n rectangular Celtic knots (up to rotations and symmetries).

3, 21, 648, 15552, 404595, 10812528, 290993472, 7849029276, 211855234563, 5719493462076, 154420985682648, 4169319089851512, 112571188876032435, 3039418274650387848, 82064259021919140432, 2215734684425617523796, 59824833697843168341123, 1615270484817096465311316

Offset: 1

Philippe Gibone, Jan 02 2022

nonn

Jessica Connor and Nick Ward, Celtic Knot Theory

Cf. A032120 (1 X n Celtic knots).

a(n) = if(n==2, 21, (3^(3*n-2) + 3^((3*n-1)\2) + 3^(2*n-1) + 3^(3*n\2))/4) \\ Andrew Howroyd, Jan 03 2022

a(n) = (3^(3*n-2) + 3^((3*n-2)/2) + 3^(2*n-1) + 3^(3*n/2))/4 for even n > 2;
a(n) = (3^(3*n-2) + 3^((3*n-1)/2) + 3^(2*n-1) + 3^((3*n-1)/2))/4 for odd n.
a(n) = (3^(3*n-2) + 3^floor((3*n-1)/2) + 3^(2*n-1) + 3^floor(3*n/2))/4 for n <> 2.

A274230 Number of holes in a sheet of paper when you fold it n times and cut off the four corners.

Original entry on oeis.org

0, 0, 1, 3, 9, 21, 49, 105, 225, 465, 961, 1953, 3969, 8001, 16129, 32385, 65025, 130305, 261121, 522753, 1046529, 2094081, 4190209, 8382465, 16769025, 33542145, 67092481, 134193153, 268402689, 536821761, 1073676289, 2147385345

Offset: 0

Philippe Gibone, Jun 15 2016

The folds are always made so the longer side becomes the shorter side.
We could have counted not only the holes but also all the notches: 4, 6, 9, 15, 25, 45, 81, 153, 289, ... which has the formula a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) and appears to match the sequence A183978. - Philippe Gibone, Jul 06 2016
The same sequence (0,0,1,3,9,21,49,...) turns up when you start with an isosceles right triangular piece of paper and repeatedly fold it in half, snipping corners as you go. Is there an easy way to see why the two questions have the same answer? - James Propp, Jul 05 2016
Reply from Tom Karzes, Jul 05 2016: (Start)
This case seems a little more complicated than the rectangular case, since with the triangle you alternate between horizontal/vertical folds vs. diagonal folds, and the resulting fold pattern is more complex, but I think the basic argument is essentially the same.
Note that with the triangle, the first hole doesn't appear until after you've made 3 folds, so if you start counting at zero folds, you have three leading zeros in the sequence: 0,0,0,1,3,9,21,... (End)
Also the number of subsets of {1,2,...,n} that contain both even and odd numbers. For example, a(3)=3 and the 3 subsets are {1,2}, {2,3}, {1,2,3}; a(4)=9 and the 9 subsets are {1,2}, {1,4}, {2,3}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}. (See comments in A052551 for the number of subsets of {1,2,...,n} that contain only odd and even numbers.) - Enrique Navarrete, Mar 26 2018
Also the number of integer compositions of n + 1 with an odd part other than the first or last. The complementary compositions are counted by A052955(n>0) = A027383(n) + 1. - Gus Wiseman, Feb 05 2022
Also the number of unit squares in the (n+1)-st iteration in the version of the dragon curve where the rotation directions alternate, so that any clockwise rotation is followed by a counterclockwise rotation, and vice versa (see image link below). - Talmon Silver, May 09 2023

Paolo P. Lava, Table of n, a(n) for n = 0..1000
Philippe Gibone, Illustration of a(0)-a(4) (idealized).
Talmon Silver, Illustration of alternating dragon.
N. J. A. Sloane, Illustration for a(4) = 9 (scan of an actual cut-up piece of paper)
N. J. A. Sloane, Illustration for a(5) = 21 (scan of an actual cut-up piece of paper)
Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).

Cf. A000225, A079667, A183978.
See A274626, A274627 for the three- and higher-dimensional analogs.
This is the main diagonal of A274635.
Counting fold lines instead of holes gives A027383.
Bisections are A060867 (even) and A134057 (odd).
Cf. A000712, A001405, A008549, A011782, A026010, A052955, A097613, A126869, A138364, A232580, A351003.

[(2^Ceiling(n/2)-1)*(2^Floor(n/2)-1 ): n in [0..35]]; // Vincenzo Librandi, Jul 02 2016

A274230:=n->(1+2^n-2^((n-3)/2)*(3-3*(-1)^n+2*sqrt(2)+2*(-1)^n*sqrt(2))): seq(A274230(n), n=0..50); # Wesley Ivan Hurt, Jul 07 2016

Table[(2^Ceiling@ # - 1) (2^Floor@ # - 1) &[n/2], {n, 0, 31}] (* Michael De Vlieger, Jun 30 2016 *)

a(n)=2^n+1-(n%2+2)<<(n\2) \\ Charles R Greathouse IV, Jul 05 2016

u(0) = 0; v(0) = 0; u(n+1) = v(n); v(n+1) = 2u(n) + 1; a(n) = u(n)*v(n).
a(n) = (2^ceiling(n/2) - 1)*(2^floor(n/2) - 1).
Proof from Tom Karzes, Jul 05 2016: (Start)
Let r be the number of times you fold along one axis and s be the number of times you fold along the other axis. So r is ceiling(n/2) and s is floor(n/2), where n is the total number of folds.
When unfolded, the resulting paper has been divided into a grid of (2^r) by (2^s) rectangles. The interior grid lines will have diamond-shaped holes where they intersect (assuming diagonal cuts).
There are (2^r-1) internal grid lines along one axis and (2^s-1) along the other. The total number of internal grid line intersections is therefore (2^r-1)*(2^s-1), or (2^ceiling(n/2)-1)*(2^floor(n/2)-1) as claimed. (End)
From Colin Barker, Jun 22 2016, revised by N. J. A. Sloane, Jul 05 2016: (Start)
It follows that:
a(n) = (2^(n/2)-1)^2 for n even, a(n) = 2^n+1-3*2^((n-1)/2) for n odd.
a(n) = 3*a(n-1)-6*a(n-3)+4*a(n-4) for n>3.
G.f.: x^2 / ((1-x)*(1-2*x)*(1-2*x^2)).
a(n) = (1+2^n-2^((n-3)/2)*(3-3*(-1)^n+2*sqrt(2)+2*(-1)^n*sqrt(2))). (End)
a(n) = A000225(n) - 2*A052955(n-2) for n > 1. - Yuchun Ji, Nov 19 2018
a(n) = A079667(2^(n-1)) for n >= 1. - J. M. Bergot, Jan 18 2021
a(n) = 2^(n-1) - A052955(n) = 2^(n-1) - A027383(n) - 1. - Gus Wiseman, Jan 29 2022
E.g.f.: cosh(x) + cosh(2*x) - 2*cosh(sqrt(2)*x) + sinh(x) + sinh(2*x) - 3*sinh(sqrt(2)*x)/sqrt(2). - Stefano Spezia, Apr 06 2022

A209036 Number of permutations of the multiset {1,1,2,2,....,n,n} with exactly two consecutive equal terms.

Original entry on oeis.org

1, 2, 36, 984, 43800, 2868480, 259554960, 31012490880, 4728875800320, 896042510496000, 206523228759724800, 56893926736333209600, 18461230471787348044800, 6968851610446509386803200

Offset: 1

Philippe Gibone, Mar 04 2012

nonn

This is a particular case (p = 1) of the more general: a(p,n) = number of permutations of the multiset {1,1,2,2,....,n,n} with exactly p times two consecutive equal terms. The sequence a(0,p) is A114938.

			a(1,2) = 2, because 1221 and 2112 are the only permutations of {1,1,2,2} where exactly two consecutive terms are equal.

Cf. A114938 (a(0,n)).

for (p = 0; p < 20; p++)
    a[p][0] = 0;
for (n = 0; n < 20; n++)
    a[0][n] = 0;
a[1][0] = 1;
for (n = 0; n < 18; n++)
   for (p = 0; p < 18; p++)
        a[p+1][n + 1] = (2*n - p + 2)*a[p][n] + (2*n - p + 1)*(2*n - p)*a[p+1][n]/2 + p*a[p+1][n] + (p + 1)*(2*n - p)*a[p + 2][n] + (p + 2)*(p + 1)*a[p + 3][n]/2 ;
   for(n = 0; n < 10; n++)
   {
    printf("%d, %ld     ", n, a[2][n]);
    if (n % 5 == 0)
     printf("\n\n");
   }

a(1,1) = 1; a(p,n+1) = a[p, n + 1] = (2*n - p + 2)*a[p-1, n] + (2*n - p + 1)*(2*n - p)*a[p, n]/2 + p*a[p, n] + (p + 1)*(2*n - p)*a[p + 1, n + (p + 2)*(p + 1)*a[p + 2, n]/2.

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User: Philippe Gibone

A350513 Number of 2 X n rectangular Celtic knots (up to rotations and symmetries).

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A274230 Number of holes in a sheet of paper when you fold it n times and cut off the four corners.

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A209036 Number of permutations of the multiset {1,1,2,2,....,n,n} with exactly two consecutive equal terms.

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