A236830 Riordan array (1/(1-x*C(x)^3), x*C(x)), C(x) the g.f. of A000108.
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, 22, 6, 1, 4597, 1812, 707, 267, 94, 29, 7, 1, 19196, 7514, 2917, 1105, 398, 131, 37, 8, 1, 80380, 31307, 12111, 4597, 1680, 575, 177, 46, 9, 1, 337284, 130883, 50503, 19196, 7085, 2488, 808, 233, 56, 10, 1
Offset: 0
Examples
Triangle begins:
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, 22, 6, 1;
4597, 1812, 707, 267, 94, 29, 7, 1;
19196, 7514, 2917, 1105, 398, 131, 37, 8, 1;
Production matrix is:
1 1
3 1 1
6 1 1 1
10 1 1 1 1
15 1 1 1 1 1
21 1 1 1 1 1 1
28 1 1 1 1 1 1 1
36 1 1 1 1 1 1 1 1
45 1 1 1 1 1 1 1 1 1
55 1 1 1 1 1 1 1 1 1 1
66 1 1 1 1 1 1 1 1 1 1 1
78 1 1 1 1 1 1 1 1 1 1 1 1
91 1 1 1 1 1 1 1 1 1 1 1 1 1
...
Links
- G. C. Greubel, Rows n= 0..100 of triangle, flattened
Crossrefs
Programs
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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
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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
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Maple
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
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Mathematica
(* The function RiordanArray is defined in A256893. *) c[x_] := (1 - Sqrt[1 - 4 x])/(2 x); RiordanArray[1/(1 - # c[#]^3)&, # c[#]&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *) 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 *)
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PARI
T(n,k) = sum(j=0,n-k, binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1)); for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 18 2019
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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
Formula
Sum_{k=0..n} T(n,k) = A026726(n).
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
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = Sum_{i = 0..n-k} Fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i).
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)
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