|
Search: id:A060556
|
|
|
| A060556 |
|
Bisection of triangle A060098: odd indexed members of column sequences of A060098 (not counting leading zeros). |
|
+0
6
|
|
| 1, 1, 2, 1, 6, 3, 1, 12, 16, 4, 1, 20, 50, 32, 5, 1, 30, 120, 140, 55, 6, 1, 42, 245, 448, 316, 86, 7, 1, 56, 448, 1176, 1284, 622, 126, 8, 1, 72, 756, 2688, 4170, 3102, 1113, 176, 9, 1, 90, 1200, 5544, 11550, 12122
(list; table; graph; listen)
|
|
|
OFFSET
|
0,3
|
|
|
COMMENT
|
Row sums give A060557. Column sequences without leading zeros give for m=0..5: A000012 (powers of 1), A002378 = 2*A000217, A004320, 4*A040977, A060558, 2*A060559.
Companion triangle (even indexed members) A060102.
With offset 1 for n and k, T(n,k) is the number of (1-2-3)-avoiding trapezoidal words of length n that contain n+1-k 1s. A trapezoidal word (following Riordan) is a sequence (a_1,a_2,...,a_n) of integers with 1<=a_i<=2i-1. For example, T(3,3)=3 counts 122, 132, 133 and T(4,2)=12 counts 1112, 1113, 1114, 1115, 1116, 1117, 1121, 1131, 1141, 1151, 1211, 1311. [From David Callan (callan(AT)stat.wisc.edu), Aug 25 2009]
|
|
FORMULA
|
a(n, m)= A060098(2*n+1-m, m).
G.f. for column m: (x^m)*Po(m+1, x)/(1-x)^(2*m+1), with Po(n, x)=sum(binomial (n, 2*j+1)*x^j, j=0..floor(n/2)) (odd members of row n of Pascal triangle A007318).
a(n, m)= sum(binomial(n-j+m, 2*m)*binomial(m+1, 2*j+1), j=0..floor((m+1)/2)), n >= m >= 0, else zero.
|
|
EXAMPLE
|
{1}; {1,2}; {1,6,3}; {1,12,16,4}; ...; Po(3,x)=3+x.
|
|
CROSSREFS
|
Sequence in context: A016545 A142977 A120108 this_sequence A132813 A034898 A059300
Adjacent sequences: A060553 A060554 A060555 this_sequence A060557 A060558 A060559
|
|
KEYWORD
|
nonn,easy,tabl
|
|
AUTHOR
|
Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Apr 06 2001
|
|
|
Search completed in 0.002 seconds
|
