This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A105809 #62 Mar 08 2025 09:59:03
%S A105809 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 A105809 33,46,51,41,22,7,1,34,54,79,97,92,63,29,8,1,55,88,133,176,189,155,92,
%U A105809 37,9,1,89,143,221,309,365,344,247,129,46,10,1,144,232,364,530,674,709,591
%N A105809 Riordan array (1/(1 - x - x^2), x/(1 - x)).
%C A105809 Previous name was: A Fibonacci-Pascal matrix.
%C A105809 From _Wolfdieter Lang_, Oct 04 2014: (Start)
%C A105809 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.
%C A105809 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 A105809 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. (End)
%H A105809 Reinhard Zumkeller, <a href="/A105809/b105809.txt">Rows n = 0..120 of table, flattened</a>
%H A105809 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 A105809 Hung Viet Chu, <a href="https://arxiv.org/abs/2106.03659">Partial Sums of the Fibonacci Sequence</a>, arXiv:2106.03659 [math.CO], 2021.
%H A105809 Anton Vladimirovich Shutov, <a href="https://www.chebsbornik.ru/jour/article/view/1878">On some analogue of the Gelfond problem for Zeckendorf representations</a>, Chebyshevskii Sbornik (2024) Vol. 25, No. 5, 195-215. See p. 215. (In Russian)
%H A105809 <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F A105809 Riordan array (1/(1-x-x^2), x/(1-x)).
%F A105809 Triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0) = A000045(n+1);
%F A105809 T(n, m) = T(n-1, m-1) + T(n-1, m).
%F A105809 T(n, k) = Sum_{j=0..n} binomial(j, n+k-j). - _Paul Barry_, Oct 23 2006
%F A105809 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
%F A105809 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
%F A105809 From _Wolfdieter Lang_, Feb 13 2025: (Start)
%F A105809 Array A(k, n) = Sum_{j=0..n} F(j+1)*binomial(k-1+n-j, k-1), k >= 0, n >= 0, with F = A000045, (from Riordan triangle k-th convolution in columns without leading 0s).
%F A105809 A(k, n) = F(n+1+2*k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n+1+j, j), (from iteration of partial sums).
%F A105809 Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(n-j+1) * binomial(j-1, k-1), 0 <= k <= n.
%F A105809 T(n, k) = F(n+1+k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n - (k-1-j), j). (End)
%F A105809 T(n, k) = A027926(n, n+k), for 0 <= k <= n. - _Wolfdieter Lang_, Mar 08 2025
%e A105809 The triangle T(n,k) begins:
%e A105809 n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
%e A105809 0: 1
%e A105809 1: 1 1
%e A105809 2: 2 2 1
%e A105809 3: 3 4 3 1
%e A105809 4: 5 7 7 4 1
%e A105809 5: 8 12 14 11 5 1
%e A105809 6: 13 20 26 25 16 6 1
%e A105809 7: 21 33 46 51 41 22 7 1
%e A105809 8: 34 54 79 97 92 63 29 8 1
%e A105809 9: 55 88 133 176 189 155 92 37 9 1
%e A105809 10: 89 143 221 309 365 344 247 129 46 10 1
%e A105809 11: 144 232 364 530 674 709 591 376 175 56 11 1
%e A105809 12: 233 376 596 894 1204 1383 1300 967 551 231 67 12 1
%e A105809 13: 377 609 972 1490 2098 2587 2683 2267 1518 782 298 79 13 1
%e A105809 ... reformatted and extended - _Wolfdieter Lang_, Oct 03 2014
%e A105809 ------------------------------------------------------------------
%e A105809 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 A105809 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 A105809 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 (Haskell)
%o A105809 a105809 n k = a105809_tabl !! n !! k
%o A105809 a105809_row n = a105809_tabl !! n
%o A105809 a105809_tabl = map fst $ iterate
%o A105809 (\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [0]))) ([1], [1,1])
%o A105809 -- _Reinhard Zumkeller_, Aug 15 2013
%Y A105809 Cf. A165326 (Z-sequence), A027934 (row sums), A010049(n+1) (antidiagonal sums), A212804 (alternating row sums), inverse is A105810.
%Y A105809 Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A109906, A111006, A114197, A162741, A228074.
%K A105809 easy,nonn,tabl
%O A105809 0,4
%A A105809 _Paul Barry_, May 04 2005
%E A105809 Use first formula as a more descriptive name, _Joerg Arndt_, Jun 08 2021