A274742 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's that begin with 1 and have exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.

1, 1, 1, 2, 2, 1, 2, 4, 3, 1, 3, 6, 6, 4, 1, 3, 9, 12, 8, 5, 1, 4, 12, 18, 20, 10, 6, 1, 4, 16, 30, 30, 30, 12, 7, 1, 5, 20, 40, 60, 45, 42, 14, 8, 1, 5, 25, 60, 80, 105, 63, 56, 16, 9, 1, 6, 30, 75, 140, 140, 168, 84, 72, 18, 10, 1, 6, 36, 105, 175, 280, 224, 252, 108, 90, 20, 11, 1, 7, 42, 126, 280, 350, 504, 336, 360, 135, 110, 22, 12, 1

Offset

3,4

Comments

It appears that the row sums give the positive integers of A001629. - Omar E. Pol, Jul 09 2016

Links

Table of n, a(n) for n=3..93.

Formula

T(n,k) = binomial(floor((n+k-2)/2),k)*floor((n-k-1)/2).

Example

n=3 => 100 -> T(3,0) = 1.
n=4 => 1001 -> T(4,0) = 1; 1100 -> T(4,1) = 1.
n=5 => 10010, 10100 -> T(5,0) = 1; 10011, 11001 -> T(5,1) = 2;
       11100 -> T(5,2) = 1.
Triangle starts:
1
1, 1
2, 2, 1
2, 4, 3, 1
3, 6, 6, 4, 1
3, 9, 12, 8, 5, 1
4, 12, 18, 20, 10, 6, 1
4, 16, 30, 30, 30, 12, 7, 1
5, 20, 40, 60, 45, 42, 14, 8, 1
5, 25, 60, 80, 105, 63, 56, 16, 9, 1
6, 30, 75, 140, 140, 168, 84, 72, 18, 10, 1
6, 36, 105, 175, 280, 224, 252, 108, 90, 20, 11, 1
7, 42, 126, 280, 350, 504, 336, 360, 135, 110, 22, 12, 1

Mathematica

Table[Binomial[Floor[(n + k - 2)/2], k] Floor[(n - k - 1)/2], {n, 3, 15}, {k, 0, n - 3}] // Flatten (* Michael De Vlieger, Jul 05 2016 *)

Prog

(PARI) t(n, k) = binomial(floor((n+k-2)/2), k) * floor((n-k-1)/2)
trianglerows(n) = for(x=3, n+2, for(y=0, x-3, print1(t(x, y), ", ")); print(""))
trianglerows(13) \\ Felix Fröhlich, Jul 05 2016

Crossrefs

Cf. A046854, A274228.

Columns: A008619, A087811.

Sequence in context: A353368 A300667 A129687 * A128176 A368415 A370075

Adjacent sequences: A274739 A274740 A274741 * A274743 A274744 A274745

Keyword

nonn,tabl

Author

Jeremy Dover, Jul 04 2016

Status

approved