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A105809
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Riordan array (1/(1-x-x^2), x/(1-x)).
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17
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1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 5, 7, 7, 4, 1, 8, 12, 14, 11, 5, 1, 13, 20, 26, 25, 16, 6, 1, 21, 33, 46, 51, 41, 22, 7, 1, 34, 54, 79, 97, 92, 63, 29, 8, 1, 55, 88, 133, 176, 189, 155, 92, 37, 9, 1, 89, 143, 221, 309, 365, 344, 247, 129, 46, 10, 1, 144, 232, 364, 530, 674, 709, 591
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OFFSET
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0,4
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COMMENTS
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Previous name was: A Fibonacci-Pascal matrix.
In the column k of this triangle (without leading zeros) is the k-fold iterated partial sums of the Fibonacci numbers, starting with 1. A000045(n+1), A000071(n+3), A001924(n+1), A014162(n+1), A014166(n+1), ..., n >= 0. See the Riordan property. - Wolfdieter Lang, Oct 03 2014
For a combinatorial interpretation of these iterated partial sums see the H. Belbachir and A. Belkhir link. There table 1 shows in the rows these columns. In their notation (with r=k) f^(k)(n) = T(k,n+k).
The A-sequence of this Riordan triangle is [1, 1] (see the recurrence for T(n,k), k>=1, given in the formula section). The Z-sequence is A165326 = [1, repeat(1,-1)]. See the W. Lang link under A006232 for Riordan A- and Z-sequences.
The alternating row sums are A212804. (End)
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LINKS
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FORMULA
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Riordan array (1/(1-x-x^2), x/(1-x)).
Number triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0)=A000045(n);
T(n, m) = T(n-1, m-1)+T(n-1, m).
T(n,k) = Sum_{j=0..n} binomial(j,n+k-j). - Paul Barry, Oct 23 2006
G.f. of row polynomials Sum_{k=0..n} T(n,k)*x^k is (1-z)/((1-z-z^2)*(1-(1+x)*z)) (Riordan property). - Wolfdieter Lang, Oct 04 2014
T(n, k) = binomial(n, k)*hypergeom([1, k/2-n/2, k/2-n/2+1/2],[k+1, -n],-4) for n>0. - Peter Luschny, Oct 10 2014
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EXAMPLE
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The triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
0: 1
1: 1 1
2: 2 2 1
3: 3 4 3 1
4: 5 7 7 4 1
5: 8 12 14 11 5 1
6: 13 20 26 25 16 6 1
7: 21 33 46 51 41 22 7 1
8: 34 54 79 97 92 63 29 8 1
9: 55 88 133 176 189 155 92 37 9 1
10: 89 143 221 309 365 344 247 129 46 10 1
11: 144 232 364 530 674 709 591 376 175 56 11 1
12: 233 376 596 894 1204 1383 1300 967 551 231 67 12 1
13: 377 609 972 1490 2098 2587 2683 2267 1518 782 298 79 13 1
------------------------------------------------------------------
Recurrence from Z-sequence (see a comment above): 8 = T(0,5) = (+1)*5 + (+1)*7 + (-1)*7 + (+1)*4 + (-1)*1 = 8. - Wolfdieter Lang, Oct 04 2014
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MAPLE
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T := (n, k) -> `if`(n=0, 1, binomial(n, k)*hypergeom([1, k/2-n/2, k/2-n/2+1/2], [k+1, -n], -4)); for n from 0 to 13 do seq(simplify(T(n, k)), k=0..n) od; # Peter Luschny, Oct 10 2014
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MATHEMATICA
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T[n_, k_] := Sum[Binomial[n-j, k+j], {j, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)
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PROG
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(Haskell)
a105809 n k = a105809_tabl !! n !! k
a105809_row n = a105809_tabl !! n
a105809_tabl = map fst $ iterate
(\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [0]))) ([1], [1, 1])
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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Use first formula as a more descriptive name, Joerg Arndt, Jun 08 2021
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STATUS
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approved
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