cp's OEIS Frontend

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.

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A385817 Irregular triangle read by rows listing the lengths of maximal runs (sequences of consecutive elements increasing by 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 1, 1, 2, 4, 1, 1, 1, 3, 1, 2, 2, 1, 3, 5, 2, 1, 1, 1, 2, 1, 4, 1, 1, 1, 2, 3, 2, 2, 3, 1, 4, 6, 1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 3, 1, 5, 1, 2, 1, 2, 1, 2, 2, 4, 2, 1, 1, 3, 3, 3, 2, 4, 1, 5, 7, 2, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 1, 1, 2, 1
Offset: 0

Views

Author

Gus Wiseman, Jul 14 2025

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
This is the triangle A245563, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A245563.

Examples

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 2
   3: 1 1
   4: 3
   5: 2 1
   6: 1 2
   7: 4
   8: 1 1 1
   9: 3 1
  10: 2 2
  11: 1 3
  12: 5
  13: 2 1 1
  14: 1 2 1
  15: 4 1
  16: 1 1 2
  17: 3 2
  18: 2 3
  19: 1 4
  20: 6
  21: 1 1 1 1
		

Crossrefs

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A000071 = A000045-1, before A023758.
Positions of firsts appearances appear to be A001629.
Positions of rows of the form (1,1,...) appear to be A055588 = A001906+1.
First term of each row appears to be A083368.
Row sums appear to be A200648, before A000120.
Row lengths after the first row appear to be A200650+1, before A069010 = A037800+1.
Before the removals we had A245563 (except first term), see A245562, A246029, A328592.
For anti-run ranks we have A385816, before A348366, firsts A052499.
Standard composition numbers of rows are A385818, before A385889.
For anti-runs we have A385886, before A384877, firsts A384878.

Programs

  • Mathematica
    DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}]]

A083368 a(n) is the position of the highest one in A003754(n).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 8, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 9, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2
Offset: 1

Views

Author

Gary W. Adamson, Jun 04 2003

Keywords

Comments

Previous name was: A Fibbinary system represents a number as a sum of distinct Fibonacci numbers (instead of distinct powers of two). Using representations without adjacent zeros, a(n) = the highest bit-position which changes going from n-1 to n.
A003754(n), when written in binary, is the representation of n.
Often one uses Fibbinary representations without adjacent ones (the Zeckendorf expansion).
a(A000071(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
From Gus Wiseman, Jul 24 2025: (Start)
Conjecture: To obtain this sequence, start with A245563 (maximal run lengths of binary indices), then remove empty and duplicate rows (giving A385817), then take the first term of each remaining row. Some variations:
- For sum instead of first term we appear to have A200648.
- For length instead of first term we appear to have A200650+1.
- For last instead of first term we have A385892.
(End)

Examples

			27 is represented 110111, 28 is 111010; the fourth position changes, so a(28)=4.
		

References

  • Jay Kappraff, Beyond Measure: A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, page 460.

Crossrefs

A035612 is the analogous sequence for Zeckendorf representations.
A001511 is the analogous sequence for power-of-two representations.
Appears to be the first element of each row of A385817, see A083368, A200648, A200650, A385818, A385892.
A000120 counts ones in binary expansion.
A245563 gives run lengths of binary indices, see A089309, A090996, A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, ranks A385816.

Programs

  • Haskell
    a083368 n = a083368_list !! (n-1)
    a083368_list = concat $ h $ drop 2 a000071_list where
       h (a:fs@(a':_)) = (map (a035612 . (a' -)) [a .. a' - 1]) : h fs
    -- Reinhard Zumkeller, Aug 10 2014

Formula

For n = F(a)-1 to F(a+1)-2, a(n) = A035612(F(a+1)-1-n).
a(n) = a(k)+1 if n = ceiling(phi*k) where phi is the golden ratio; otherwise a(n) = 1. - Tom Edgar, Aug 25 2015

Extensions

Edited by Don Reble, Nov 12 2005
Shorter name from Joerg Arndt, Jul 27 2025

A123585 Triangle T(n,k), 0<=k<=n, given by [1, -1, 1, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 0, 2, 2, -1, 1, 5, 3, -1, -2, 4, 10, 5, 0, -4, -4, 12, 20, 8, 1, -2, -13, -4, 31, 38, 13, 1, 3, -11, -33, 3, 73, 71, 21, 0, 6, 6, -42, -74, 34, 162, 130, 34, -1, 3, 24, 0, -130, -146, 128, 344, 235, 55, -1, -4, 21, 72, -50, -352
Offset: 0

Views

Author

Philippe Deléham, Nov 13 2006

Keywords

Examples

			Triangle begins:
1;
1, 1;
0, 2, 2;
-1, 1, 5, 3;
-1, -2, 4, 10, 5;
0, -4, -4, 12, 20, 8;
1, -2, -13, -4, 31, 38, 13;
1, 3, -11, -33, 3, 73, 71, 21;
0, 6, 6, -42, -74, 34, 162, 130, 34;
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[CoefficientList[Series[1/(1 - (1 + y)*x + (1 - y^2)*x^2), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Oct 16 2017 *)

Formula

Sum_{k,0<=k<=n} T(n,k) = 2^n = A000079(n).
T(n,0) = A010892(n).
T(n,n) = Fibonacci(n+1) = A000045(n+1).
T(n+1,1) = A099254(n).
T(n+1,n) = A001629(n+2).
Sum_{k, 0<=k<=[n/2]} T(n-k,k) = A003269(n).
T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-2) - T(n-2,k), n>0.
Sum_{k, 0<=k<=n} x^k*T(n,k) = (-1)^n*A003683(n+1), (-1)^n*A006130(n), A000007(n), A010892(n), A000079(n), A030195(n+1) for x=-3, -2, -1, 0, 1, 2 respectively . - Philippe Deléham, Dec 01 2006
T(n+2,n) = A129707(n+1).- Philippe Deléham, Dec 18 2011
G.f.: 1/(1-(1+y)*x+(1-y^2)*x^2). - Philippe Deléham, Dec 18 2011

A246177 Triangle read by rows: T(n,k) is the number of weighted lattice paths in B(n) such that the area between the x-axis and the path is k.

Original entry on oeis.org

1, 1, 2, 3, 1, 5, 2, 1, 8, 5, 3, 1, 13, 10, 8, 4, 2, 21, 20, 18, 12, 7, 3, 1, 34, 38, 39, 30, 22, 12, 7, 2, 1, 55, 71, 80, 70, 57, 39, 26, 14, 7, 3, 1, 89, 130, 160, 154, 138, 106, 81, 52, 34, 18, 10, 4, 2, 144, 235, 312, 327, 315, 267, 220, 163, 118, 78, 49, 28, 16, 7, 3, 1
Offset: 0

Views

Author

Emeric Deutsch, Aug 20 2014

Keywords

Comments

The members of B(n) are paths of weight n that start at (0,0), end on but never go below the horizontal axis, and whose steps are of the following four kinds: an (1,0)-step with weight 1, an (1,0)-step with weight 2, a (1,1)-step with weight 2, and a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.
Apparently, number of terms in row n is 1+floor(n^2/8).
Sum of entries in row n is A004148(n+1) (the 2ndary structure numbers).
T(n,0) = A000045(n+1) (the Fibonacci numbers).
T(n,1) = A001629(n-1) (n>=1).

Examples

			Row 3 is 3,1; indeed, B(3) consists of the paths hhh, hH, Hh, UD with areas 0,0,0,1, respectively.
Triangle starts:
   1;
   1;
   2;
   3,   1;
   5,   2,   1;
   8,   5,   3,   1;
  13,  10,   8,   4,   2;
  21,  20,  18,  12,   7,   3,  1;
  34,  38,  39,  30,  22,  12,  7,  2,  1;
  55,  71,  80,  70,  57,  39, 26, 14,  7,  3,  1;
  89, 130, 160, 154, 138, 106, 81, 52, 34, 18, 10, 4, 2;
		

Crossrefs

Programs

  • Maple
    g := 1/(1-z-z^2-t*z^3*A[1]): for j to 15 do A[j] := 1/(1-t^j*z-t^j*z^2-t^(2*j+1)*z^3*A[j+1]) end do: gser := simplify(series(g, z = 0, 20)): for n from 0 to 17 do P[n] := sort(coeff(gser, z, n)) end do: for n from 0 to 17 do seq(coeff(P[n], t, j), j = 0 .. floor((1/8)*n^2)) end do; # yields sequence in triangular form
    # second Maple program:
    b:= proc(n, y) option remember; `if`(y<0 or y>n, 0, `if`(n=0, 1,
          expand(b(n-1, y)*x^y +`if`(n>1, b(n-2, y)*x^y+b(n-2, y+1)*
          x^(y+1/2), 0) +b(n-1, y-1)*x^(y-1/2))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
    seq(T(n), n=0..20);  # Alois P. Heinz, Aug 20 2014
  • Mathematica
    b[n_, y_] := b[n, y] = If[y<0 || y>n, 0, If[n == 0, 1, Expand[b[n-1, y]*x^y + If[n>1, b[n-2, y]*x^y + b[n-2, y+1]*x^(y+1/2), 0] + b[n-1, y-1]*x^(y-1/2)]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 0]]; Table[T[n], {n, 0, 20}] // Flatten (* Jean-François Alcover, May 27 2015, after Alois P. Heinz *)

Formula

The trivariate g.f. G=G(t,s,z), where t marks area, s marks length (=number of steps), and z marks weight, satisfies G = 1+szG+sz^2G+ts^2z^3G(t,ts,z)G. This follows at once from the fact that every nonempty path is of the form hC or HC or UCDC, where h denotes a (1,0)-step of weight 1, H denotes a (1,0)-step of weight 2, U denotes a (1,1)-step, D denotes a (1,-1)-step, and the C's denote paths, not necessarily the same. From the equation one can find G(t,s,z) as a continued fraction (the Maple program makes use of this).

A332724 Number of length n - 1 ordered set partitions of {1..n} where no element of any block is greater than any element of a non-adjacent consecutive block.

Original entry on oeis.org

0, 0, 1, 6, 14, 32, 65, 128, 243, 452, 826, 1490, 2659, 4704, 8261, 14418, 25030, 43252, 74437, 127648, 218199, 371920, 632306, 1072486, 1815239, 3066432, 5170825, 8705118, 14632958, 24562952, 41177801, 68947520, 115313979, 192656924, 321554986, 536191418
Offset: 0

Views

Author

Gus Wiseman, Mar 03 2020

Keywords

Comments

In other words, parts of not-immediately-subsequent blocks are increasing.

Examples

			The a(2) = 1 through a(4) = 14 ordered set partitions:
  {{1,2}}  {{1},{2,3}}  {{1},{2},{3,4}}
           {{1,2},{3}}  {{1},{2,3},{4}}
           {{1,3},{2}}  {{1,2},{3},{4}}
           {{2},{1,3}}  {{1},{2,4},{3}}
           {{2,3},{1}}  {{1,2},{4},{3}}
           {{3},{1,2}}  {{1},{3},{2,4}}
                        {{1,3},{2},{4}}
                        {{1},{3,4},{2}}
                        {{1},{4},{2,3}}
                        {{2},{1},{3,4}}
                        {{2},{1,3},{4}}
                        {{2},{1,4},{3}}
                        {{2,3},{1},{4}}
                        {{3},{1,2},{4}}
		

Crossrefs

Column k = n - 1 of A332673, which has row-sums A332872.
Ordered set-partitions are A000670.
Unimodal compositions are A001523.
Unimodal normal sequences appear to be A007052.
Non-unimodal normal sequences are A328509.

Programs

  • Mathematica
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    Table[Length[Select[Join@@Permutations/@sps[Range[n]],Length[#]==n-1&&!MatchQ[#,{_,{_,a_,_},,{_,b_,_},_}/;a>b]&]],{n,0,8}]
  • PARI
    \\ here b(n) is A001629(n).
    b(n) = {((n+1)*fibonacci(n-1) + (n-1)*fibonacci(n+1))/5}
    a(n) = {if(n==0, 0, b(n) + 4*b(n-1) + b(n-2))} \\ Andrew Howroyd, Apr 17 2021

Formula

From Andrew Howroyd, Apr 17 2021: (Start)
a(n) = A001629(n) + 4*A001629(n+1) + A001629(n+2) for n > 0.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n > 4.
G.f.: x*(1 + 4*x + x^2)/(1 - x - x^2)^2.
(End)

Extensions

Terms a(9) and beyond from Andrew Howroyd, Apr 17 2021

A039913 Triangular "Fibonacci array".

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 2, 3, 3, 2, 3, 5, 4, 5, 3, 5, 8, 7, 7, 8, 5, 8, 13, 11, 12, 11, 13, 8, 13, 21, 18, 19, 19, 18, 21, 13, 21, 34, 29, 31, 30, 31, 29, 34, 21, 34, 55, 47, 50, 49, 49, 50, 47, 55, 34, 55, 89, 76, 81, 79, 80, 79, 81, 76, 89, 55, 89, 144, 123, 131, 128, 129, 129, 128
Offset: 0

Views

Author

Keywords

Comments

Sum of n-th row = 2*A001629(n+1). - Reinhard Zumkeller, Oct 07 2012

Examples

			0;
1 1;
1 2 1;
2 3 3 2;
3 5 4 5 3;
...
		

Crossrefs

Cf. A108035.

Programs

  • Haskell
    a039913 n k = a039913_tabl !! n !! k
    a039913_row n = a039913_tabl !! n
    a039913_tabl = [[0], [1, 1]] ++ f [0] [1, 1] where
       f us@(u:us') vs@(v:vs') = ws : f vs ws where
         ws = [u + v, u + v + v] ++ zipWith (+) us vs'
    -- Reinhard Zumkeller, Oct 07 2012

Formula

a(0, n)=Fib(n), a(1, n)=Fib(n+2), a(r, n)=a(r-1, n)+a(r-2, n), r >= 2.
G.f.: (x+y)/((1-x-x^2)*(1-y-y^2)). [U coordinates]

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Sep 28 2000

A091562 Triangle read by rows, related to Pascal's triangle, starting with 1, 0, 0.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 2, 5, 7, 5, 2, 3, 10, 17, 17, 10, 3, 5, 20, 41, 51, 41, 20, 5, 8, 38, 91, 136, 136, 91, 38, 8, 13, 71, 195, 339, 405, 339, 195, 71, 13, 21, 130, 403, 799, 1107, 1107, 799, 403, 130, 21, 34, 235, 812, 1807, 2845, 3297, 2845, 1807, 812, 235, 34
Offset: 0

Views

Author

Christian G. Bower, Jan 20 2004

Keywords

Examples

			Triangle begins:
  1;
  0,0;
  1,1,1;
  1,2,2,1;
  2,5,7,5,2;
  ...
		

Crossrefs

Row sums: A054878, column 0: A000045(n-1), column 1: A001629.
Cf. A090171, A090172, A090173, A090174, A091533, A205575 (same recurrence).
Cf. A090172.

Formula

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) + T(n-2, k-2) for n >= 2, k >= 0, with initial conditions specified by first two rows.
G.f.: A(x, y) = (1-x-x*y)/(1-x-x*y-x^2-x^2*y-x^2*y^2).

A128502 Convolution array for Chebyshev's S(n,x)=U(n,x/2) polynomials.

Original entry on oeis.org

1, 2, 3, -2, 4, -6, 5, -12, 3, 6, -20, 12, 7, -30, 30, -4, 8, -42, 60, -20, 9, -56, 105, -60, 5, 10, -72, 168, -140, 30, 11, -90, 252, -280, 105, -6, 12, -110, 360, -504, 280, -42, 13, -132, 495, -840, 630, -168, 7, 14, -156, 660, -1320, 1260, -504, 56, 15, -182, 858, -1980, 2310, -1260, 252, -8, 16, -210, 1092
Offset: 0

Views

Author

Wolfdieter Lang Apr 04 2007

Keywords

Comments

S1(n,x):=sum(S(n-k,x)*S(k,x),k=0..n)= sum(a(n,m)*x^(n-2*m),m=0..floor(n/2)).
The unsigned column sequences, m>=0, divided by (m+1) give Pascal triangle column sequences for m+1.
G.f. for column m sequence: ((-1)^m)*(m+1)*(x^(2*m))/(1-x)^(m+2), m>=0.
Row polynomials P1(n,x):= sum(a(n,m)*x^m,m=0..floor(n/2)) (increasing powers of x).
Written as a triangle with increasing powers of x this is A294519. - Wolfdieter Lang, Nov 12 2017

Examples

			[1];[2];[3,-2],[4,-6];[5,-12,3];[6,-20,12];[7,-30,30,-4];[8,-42,60,-20];...
n=4: [5,-12,3] stands for the polynomial S1(4,x) = 5*x^4-12*x^2+3 = 2*(S(4,x)*1+S(3,x)*S(1,x))+S(2,x)*S(2,x).
n=4: [5,-12,3] stands also for the row polynomial P1(4,x) = 5-12*x+3*x^2.
		

Crossrefs

Row sums (signed array) give A099254. Unsigned row sums are A001629(n+2).
Cf. A115139 (with offset n>=0 is S(n, x) array, decreasing powers of x).
Cf. A294519 (as triangle).

Formula

a(n,m)=binomial(n-m,m)*(n+1-m)*(-1)^m, m=0..floor(n/2), n>=0.
a(n,m)=binomial(n+1-m,m+1)*(m+1)*(-1)^m, m=0..floor(n/2), n>=0.
G.f. for S1(n,x): 1/(1-x*z+z^2)^2.
G.f. for P1(n,x): 1/(1-z+x*z^2)^2.

A191830 Expansion of x^2*(2-3*x)/(1-x-x^2)^2.

Original entry on oeis.org

0, 0, 2, 1, 4, 5, 10, 16, 28, 47, 80, 135, 228, 384, 646, 1085, 1820, 3049, 5102, 8528, 14240, 23755, 39592, 65931, 109704, 182400, 303050, 503161, 834868, 1384397, 2294290, 3800080, 6290788, 10408679, 17213696, 28454415, 47014380, 77647104, 128186062
Offset: 0

Views

Author

Paul Curtz, Jun 17 2011

Keywords

Comments

a(2*n) mod 2 = 0;
a(4*n) mod 4 = 0;
a(5*n) mod 5 = 0 and a(5*n+1) mod 5 = 0;
a(n) = 2*A001629(n) - 3*A001629(n-1). - Johannes W. Meijer, Jun 27 2011

Crossrefs

Programs

  • Maple
    A191830:= proc(n) option remember: if n<=1 then 0 else procname(n-1)+procname(n-2)+A000045(n-5) fi: end proc: with(combinat): A000045:=fibonacci: seq(A191830(n),n=0..30); # Johannes W. Meijer, Jun 27 2011
  • Mathematica
    CoefficientList[Series[x^2(2-3x)/(1-x-x^2)^2,{x,0,40}],x] (* or *) LinearRecurrence[{2,1,-2,-1},{0,0,2,1},40] (* Harvey P. Dale, Mar 16 2015 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-2,1,2]^n*[0;0;2;1])[1,1] \\ Charles R Greathouse IV, Jul 06 2017

Formula

G.f.: x^2*(2-3*x)/(1-x-x^2)^2.
a(n) = a(n-1) + a(n-2) + A000045(n-5), a(0) = a(1) = 0.
a(0)=0, a(1)=0, a(2)=2, a(3)=1, a(n)=2*a(n-1)+a(n-2)-2*a(n-3)-a(n-4). - Harvey P. Dale, Mar 16 2015

A245825 Triangle read by rows: T(n,k) is the number of the vertices of the Fibonacci cube G_n that have degree k (0<=k<=n).

Original entry on oeis.org

1, 0, 2, 0, 2, 1, 0, 1, 3, 1, 0, 0, 5, 2, 1, 0, 0, 3, 7, 2, 1, 0, 0, 1, 10, 7, 2, 1, 0, 0, 0, 9, 14, 8, 2, 1, 0, 0, 0, 4, 23, 16, 9, 2, 1, 0, 0, 0, 1, 22, 34, 19, 10, 2, 1, 0, 0, 0, 0, 14, 50, 44, 22, 11, 2, 1, 0, 0, 0, 0, 5, 55, 77, 56, 25, 12, 2, 1, 0, 0, 0, 0, 1, 40, 117, 106, 69, 28, 13, 2, 1, 0, 0, 0, 0, 0, 20, 131, 188, 140, 83, 31, 14, 2, 1
Offset: 0

Views

Author

Emeric Deutsch, Aug 03 2014

Keywords

Comments

The Fibonacci cube G_n is obtained from the n-cube Q_n by removing all the vertices that contain two consecutive 1s.
Sum of entries in row n is the Fibonacci number F_{n+2}.
Sum of entries in column k (k>=1) is the Fibonacci number F_{2k+3}. - Emeric Deutsch, Jun 22 2015
Sum(k*T(n,k), k=0..n) = 2*sum(F(k)*F(n+1-k),k=0..n+1) = 2*A001629(n+1).

Examples

			Row 2 is 0,2,1 because the Fibonacci cube G_2 is the path-tree P_3 having 2 vertices of degree 1 and 1 vertex of degree 2.
Triangle starts:
1;
0,2;
0,2,1;
0,1,3,1;
0,0,5,2,1;
0,0,3,7,2,1;
0,0,1,10,7,2,1;
		

Crossrefs

Programs

  • Maple
    T := proc (n, k) options operator, arrow: sum(binomial(n-2*i, k-i)*binomial(i+1, n-k-i+1), i = 0 .. k) end proc: seq(seq(T(n, k), k = 0 .. n), n = 0 .. 13);

Formula

T(n,k) = sum(binomial(n-2i, k-i)*binomial(i+1,n-k-i+1), i=0..k).
G.f.: (1 + t*z + (1 - t)*t*z^2)/((1 - t*z)*(1 - t*z^2) - t*z^3).
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