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 A105809 A Fibonacci-Pascal matrix. 17

%I

%S 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,

%T 33,46,51,41,22,7,1,34,54,79,97,92,63,29,8,1,55,88,133,176,189,155,92,

%U 37,9,1,89,143,221,309,365,344,247,129,46,10,1,144,232,364,530,674,709,591

%N A Fibonacci-Pascal matrix.

%C Row sums are A027934, antidiagonal sums are A010049(n+1). Inverse is A105810.

%C From _Wolfdieter Lang_, Oct 04 2014: (Start)

%C 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

%C 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).

%C 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.

%C The alternating row sums are A212804. (End)

%H Reinhard Zumkeller, <a href="/A105809/b105809.txt">Rows n = 0..120 of table, flattened</a>

%H H. Belbachir and A. Belkhir, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Belbachir/belb2.html">Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers</a>, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.

%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>

%F Riordan array (1/(1-x-x^2), x/(1-x)).

%F Number triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0)=A000045(n);

%F T(n, m) = T(n-1, m-1)+T(n-1, m).

%F T(n,k) = Sum_{j=0..n} C(j,n+k-j). - _Paul Barry_, Oct 23 2006

%F G.f. of row polynomials sum(T(n,k)*x^k, k=0..n) is (1-z)/((1-z-z^2)*(1-(1+x)*z)) (Riordan property). - _Wolfdieter Lang_, Oct 04 2014

%F 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

%e The triangle T(n,k) begins:

%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...

%e 0: 1

%e 1: 1 1

%e 2: 2 2 1

%e 3: 3 4 3 1

%e 4: 5 7 7 4 1

%e 5: 8 12 14 11 5 1

%e 6: 13 20 26 25 16 6 1

%e 7: 21 33 46 51 41 22 7 1

%e 8: 34 54 79 97 92 63 29 8 1

%e 9: 55 88 133 176 189 155 92 37 9 1

%e 10: 89 143 221 309 365 344 247 129 46 10 1

%e 11: 144 232 364 530 674 709 591 376 175 56 11 1

%e 12: 233 376 596 894 1204 1383 1300 967 551 231 67 12 1

%e 13: 377 609 972 1490 2098 2587 2683 2267 1518 782 298 79 13 1

%e ... reformatted and extended - _Wolfdieter Lang_, Oct 03 2014

%e ------------------------------------------------------------------

%e 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

%p 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

%t 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 *)

%o a105809 n k = a105809_tabl !! n !! k

%o a105809_row n = a105809_tabl !! n

%o a105809_tabl = map fst \$ iterate

%o (\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [0]))) ([1], [1,1])

%o -- _Reinhard Zumkeller_, Aug 15 2013

%Y Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A109906, A111006, A114197, A162741, A228074.

%Y Cf.A165326 (Z-sequence), A027934 (row sums, see comment above), A212804 (alternating row sums). - _Wolfdieter Lang_, Oct 04 2014

%K easy,nonn,tabl

%O 0,4

%A _Paul Barry_, May 04 2005

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Last modified January 21 21:30 EST 2020. Contains 331128 sequences. (Running on oeis4.)