A325982 -id:A325982 - cp's OEIS frontend

A325983 Row sums of the triangle A325982.

1, 1, 2, 2, 5, 5, 18, 21, 77, 102, 337, 480, 1449, 2155, 6107, 9348, 25355, 39639, 104188, 165596, 425156, 684926, 1726737, 2813582, 6990175, 11501905, 28232753, 46854161, 113841632, 190362483, 458480128, 771855377, 1844765161, 3124639626, 7417428613, 12633074088

Offset: 1

Stefano Spezia, May 29 2019

nonn

Stefano Spezia, Table of n, a(n) for n = 1..2000

Cf. A004526, A007318, A325982.

List([1..40], n->Sum([0..Int((n-1)/2)], k->Binomial(n-1, k-1)-Binomial(n-k-1, k-1)+1));

[(&+[Binomial(n-1, k-1)-Binomial(n-k-1, k-1)+1: k in [0..Floor((n-1)/2)]]): n in [1..40]];

a := n -> add(binomial(n-1, k-1)-binomial(n-k-1, k-1)+1, k = 0 .. floor((n-1)/2)): seq(a(n), n = 1 .. 40);

T[n_,k_]:=Binomial[n-1,k-1]-Binomial[n-k-1,k-1]+1; a[n_]:=Sum[T[n,k],{k,0,Floor[(n-1)/2]}]; Array[a,40]
Table[If[EvenQ[n], 2^(n - 2) + n/2 + 1 - Binomial[n, n/2]/2 + Fibonacci[n]/2 - LucasL[n]/2, 2^(n - 2) + (n + 1)/2 - Binomial[n - 1, (n - 1)/2]/2 - Fibonacci[n - 3] - 3*Fibonacci[n]/2 + LucasL[n]/2], {n, 1, 40}] (* Vaclav Kotesovec, Jun 20 2021 *)

a(n) = sum(k=0, floor((n-1)/2), binomial(n - 1, k - 1) - binomial(n - k - 1, k - 1) + 1);

a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n - 1, k - 1) - binomial(n - k - 1, k - 1) + 1.
a(n) = Sum_{k=0..A004526(n-1)} A007318(n - 1, k - 1) - A007318(n - k - 1, k - 1) + 1.
Conjecture: a(n) ~ 2^(n-2). - Stefano Spezia, Jun 16 2021
This conjecture is true. Its proof follows from Vaclav Kotesovec's formula used in the 2nd Mathematica code. - Stefano Spezia, Jun 27 2021
Recurrence: n*(15*n^6 - 600*n^5 + 9874*n^4 - 85078*n^3 + 403791*n^2 - 1000762*n + 1013720)*a(n) = (45*n^7 - 1815*n^6 + 30507*n^5 - 273677*n^4 + 1393992*n^3 - 3930412*n^2 + 5362320*n - 2240000)*a(n-1) + 2*(30*n^7 - 1230*n^6 + 20813*n^5 - 186314*n^4 + 944533*n^3 - 2692476*n^2 + 3985604*n - 2395680)*a(n-2) - (255*n^7 - 10470*n^6 + 180608*n^5 - 1685204*n^4 + 9136801*n^3 - 28646510*n^2 + 47833000*n - 32569600)*a(n-3) + (15*n^7 - 585*n^6 + 11089*n^5 - 129179*n^4 + 931040*n^3 - 3910812*n^2 + 8540192*n - 7271040)*a(n-4) + 2*(165*n^7 - 6810*n^6 + 118529*n^5 - 1121692*n^4 + 6211866*n^3 - 20094734*n^2 + 35150196*n - 25693920)*a(n-5) - 4*(15*n^7 - 600*n^6 + 10309*n^5 - 98168*n^4 + 555848*n^3 - 1857428*n^2 + 3364504*n - 2524480)*a(n-6) - 8*(n-7)*(15*n^6 - 510*n^5 + 7099*n^4 - 51282*n^3 + 202026*n^2 - 411828*n + 340960)*a(n-7). - Vaclav Kotesovec, Jun 20 2021

A335322 Triangle read by rows: T(n, k) = binomial(n, floor((n+k+1)/2)) with k <= n.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 10, 5, 5, 1, 1, 15, 15, 6, 6, 1, 1, 35, 21, 21, 7, 7, 1, 1, 56, 56, 28, 28, 8, 8, 1, 1, 126, 84, 84, 36, 36, 9, 9, 1, 1, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1, 792, 792, 495, 495, 220, 220, 66, 66, 12, 12, 1, 1

Offset: 1

Stefano Spezia, May 31 2020

T(n, k) is a tight upper bound of the cardinality of an intersecting Sperner family or antichain of the set {1, 2,..., n}, where every collection of pairwise independent subsets is characterized by an intersection of cardinality at least k (see Theorem 1.3 in Wong and Tay).

Equals A061554 with the first row of the array (resp. the first column of the triangle) removed. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins
n\k|  1   2   3   4   5   6   7   8
---+-------------------------------
1  |  1
2  |  1   1
3  |  3   1   1
4  |  4   4   1   1
5  | 10   5   5   1   1
6  | 15  15   6   6   1   1
7  | 35  21  21   7   7   1   1
8  | 56  56  28  28   8   8   1   1
...

Eric Charles Milner, A Combinatorial Theorem On Systems of Sets, Journal of the London Mathematical Society, 43, (1968), 204-206.
W. H. W. Wong and E. G. Tay, On Cross-intersecting Sperner Families, arXiv:2001.01910 [math.CO], 2020.
Index entries for sequences related to antichains.

Cf. A000372, A001405, A004526, A007318, A007695, A061554, A266696, A325982, A325983.

Cf. A037951 (k=3), A037952 (k=1), A037953 (k=5), A037954 (k=7), A037955 (k=2), A037956 (k=4), A037957 (k=6), A037958 (k=8), A045621 (row sums).

T[n_,k_]:=Binomial[n,Floor[(n+k+1)/2]]; Table[T[n,k],{n,12},{k,n}]//Flatten

T(n, k) = binomial(n, (n+k+1)\2);
vector(10, n, vector(n, k, T(n, k))) \\ Michel Marcus, Jun 01 2020

T(n, k) = A007318(n, A004526(n+k+1)) with k <= n.

cp's OEIS Frontend

A325983 Row sums of the triangle A325982.

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Formula