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Search: id:A050143
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| A050143 |
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Array T by antidiagonals: T(i,j)=Sum{T(h,k): 0<=h<=i-1, 0<=k<=j}, T(i,0)=1 for i >= 0, T(0,j)=0 for j >= 1. |
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+0
10
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| 1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 4, 7, 1, 0, 1, 5, 12, 15, 1, 0, 1, 6, 18, 32, 31, 1, 0, 1, 7, 25, 56, 80, 63, 1, 0, 1, 8, 33, 88, 160, 192, 127, 1, 0, 1, 9, 42, 129, 280, 432, 448, 255, 1, 0, 1, 10, 52, 180, 450, 832, 1120, 1024, 511, 1, 0, 1
(list; table; graph; listen)
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OFFSET
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1,9
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COMMENT
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Sum[T(i,j),j=0..1] = 2^i, i>=0. For i>=1 and j>=2, T[i,j] = number of paths of i-1 upsteps and j-1 downsteps with some (possibly all or none) of its good vertices marked. A good vertex is one incident with an upstep but not with a downstep. For example, asterisks indicate the good vertices in the path *U*U*UDU*UDDU*, and with i=j=2, T[2,2] = 4 counts UD, *UD, DU, DU*. - David Callan (callan(AT)stat.wisc.edu), Oct 19 2005
Formatted as a triangular array with offset (0,8), it is [0, 1, 0, -1, 1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 1, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 05 2006
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EXAMPLE
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Antidiagonals, each starting on top row: {1}; {0,1}; {0,1,1}; {0,1,3,1}; ...
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CROSSREFS
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Antidiagonal sums are odd-indexed Fibonacci numbers (A001519).
Signed alternating antidiagonal sums are F(n)-2, as in A001911.
T(n, 1)=-1+2^n=A000225(n). T(n+2, 2)=4*A001792(n). Cf. A050147, A050148.
Cf. A055807 (mirror array).
Adjacent sequences: A050140 A050141 A050142 this_sequence A050144 A050145 A050146
Sequence in context: A120323 A099905 A085391 this_sequence A103495 A081719 A121314
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KEYWORD
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nonn,tabl
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu)
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