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%I #36 Sep 08 2022 08:46:06
%S 1,1,1,4,2,1,16,7,3,1,65,27,11,4,1,267,108,43,16,5,1,1105,440,173,65,
%T 22,6,1,4597,1812,707,267,94,29,7,1,19196,7514,2917,1105,398,131,37,8,
%U 1,80380,31307,12111,4597,1680,575,177,46,9,1,337284,130883,50503,19196,7085,2488,808,233,56,10,1
%N Riordan array (1/(1-x*C(x)^3), x*C(x)), C(x) the g.f. of A000108.
%C T(n+3,n) = A011826(n+5).
%H G. C. Greubel, <a href="/A236830/b236830.txt">Rows n= 0..100 of triangle, flattened</a>
%F Sum_{k=0..n} T(n,k) = A026726(n).
%F G.f.: 1/((x^2*C(x)^4-x*C(x))*y-x*C(x)^3+1), where C(x) the g.f. of A000108. - _Vladimir Kruchinin_, Apr 22 2015
%F From _Peter Bala_, Feb 18 2018: (Start)
%F T(n,k) = Sum_{i = 0..n-k} Fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i).
%F The n-th row polynomial of row reverse triangle is the n-th degree Taylor polynomial of the rational function (1 - 3*x + 2*x^2)/(1 - 3*x + x^2) * 1/(1 - x)^n about 0. For example, for n = 4, (1 - 3*x + 2*x^2)/(1 - 3*x + x^2) * 1/(1 - x)^4 = 1 + 4*x + 11*x^2 + 27*x^3 + 65*x^4 + O(x^5), giving row 4 as (65, 27, 11, 4, 1). (End)
%e Triangle begins:
%e 1;
%e 1, 1;
%e 4, 2, 1;
%e 16, 7, 3, 1;
%e 65, 27, 11, 4, 1;
%e 267, 108, 43, 16, 5, 1;
%e 1105, 440, 173, 65, 22, 6, 1;
%e 4597, 1812, 707, 267, 94, 29, 7, 1;
%e 19196, 7514, 2917, 1105, 398, 131, 37, 8, 1;
%e Production matrix is:
%e 1 1
%e 3 1 1
%e 6 1 1 1
%e 10 1 1 1 1
%e 15 1 1 1 1 1
%e 21 1 1 1 1 1 1
%e 28 1 1 1 1 1 1 1
%e 36 1 1 1 1 1 1 1 1
%e 45 1 1 1 1 1 1 1 1 1
%e 55 1 1 1 1 1 1 1 1 1 1
%e 66 1 1 1 1 1 1 1 1 1 1 1
%e 78 1 1 1 1 1 1 1 1 1 1 1 1
%e 91 1 1 1 1 1 1 1 1 1 1 1 1 1
%e ...
%p A236830 := (n,k) -> add(combinat:-fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i), i = 0..n-k): seq(seq(A236830(n, k), k = 0..n), n = 0..10); # _Peter Bala_, Feb 18 2018
%t (* The function RiordanArray is defined in A256893. *)
%t c[x_] := (1 - Sqrt[1 - 4 x])/(2 x);
%t RiordanArray[1/(1 - # c[#]^3)&, # c[#]&, 11] // Flatten (* _Jean-François Alcover_, Jul 16 2019 *)
%t Table[Sum[Binomial[2*n-k-j-2, n-k-j]*Fibonacci[2*j-1], {j,0,n-k}], {n, 0, 12}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Jul 18 2019 *)
%o (PARI) T(n,k) = sum(j=0,n-k, binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1));
%o for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Jul 18 2019
%o (Magma) [(&+[Binomial(2*n-k-j-2, n-k-j)*Fibonacci(2*j-1): j in [0..n-k]]): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jul 18 2019
%o (Sage) [[sum( binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1) for j in (0..n-k) ) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Jul 18 2019
%o (GAP) Flat(List([0..12], n-> List([0..n], k-> Sum([0..n-k], j-> Binomial(2*n-k-j-2, n-k-j)*Fibonacci(2*j-1) )))); # _G. C. Greubel_, Jul 18 2019
%Y Cf. (columns): A165201, A026726, A026671, A026674, A026672, A026675, A026673, A026843, A026842 or A026846 or A026849, A026844, A026841 or A026848.
%K nonn,tabl
%O 0,4
%A _Philippe Deléham_, Feb 01 2014
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