A037955 -id:A037955 - cp's OEIS frontend

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

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.

A382817 a(n) = number of primes among the partial sums of row n of Pascal's triangle (A007318).

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 2, 2, 0, 2, 1, 3, 2, 3, 2, 3, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 3, 0, 2, 7, 2, 0, 0, 1, 1, 0, 0, 0, 2, 0, 1, 1, 1, 0, 1, 3, 1, 0, 1, 1, 1, 1, 1, 1, 5, 3, 3, 2, 3, 2, 3, 3, 10, 0, 1, 0, 1, 0, 2, 2, 2, 0, 0, 1, 1, 0, 2, 1, 1, 1, 2, 1, 2, 0

Offset: 0

Clark Kimberling, Apr 07 2025

nonn

			The numbers in A008949 (partial sums of Pascal's triangle) begin thus:
  1
  1    2
  1    3     4
  1    4     7     8
  1    5    11    15    16
  1    6    16    26    31    32
  1    7    22    42    57    63    64
Row n=4 includes exactly 2 primes, so a(4) = 2.

Cf. A007318, A008949, A258483, A382816.

a:= n-> nops(select(isprime, ListTools[PartialSums]
            ([seq(binomial(n, k), k=0..n)]))):
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 07 2025

t = Accumulate /@ Table[Binomial[n, i], {n, 0, 100}, {i, 0, n}]; (* A037955 *)
Map[PrimeQ, t]; Table[Count[m[[n]], True], {n, 1, 100}]

a(n) = my(v=vector(n+1, k, binomial(n,k-1))); #select(isprime, vector(#v, k, sum(i=1, k, v[i]))); \\ Michel Marcus, Apr 13 2025

a(n) = 0 <=> n in { A258483 }.

cp's OEIS Frontend

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

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