

A274742


Triangle read by rows: T(n,k) (n>=3, 0<=k<=n3) = number of nsequences 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, 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
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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+k2)/2),k)*floor((nk1)/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+k2)/2), k) * floor((nk1)/2)
trianglerows(n) = for(x=3, n+2, for(y=0, x3, print1(t(x, y), ", ")); print(""))
trianglerows(13) \\ Felix Fröhlich, Jul 05 2016


CROSSREFS

Cf. A046854, A274228.
Columns: A008619, A087811.
Sequence in context: A275297 A300667 A129687 * A128176 A144963 A305632
Adjacent sequences: A274739 A274740 A274741 * A274743 A274744 A274745


KEYWORD

nonn,tabl


AUTHOR

Jeremy Dover, Jul 04 2016


STATUS

approved



