A099575 -id:A099575 - cp's OEIS frontend

A099577 Diagonal sums of triangle A099575.

1, 1, 2, 2, 6, 7, 13, 15, 38, 47, 85, 104, 245, 313, 558, 706, 1594, 2080, 3674, 4753, 10429, 13817, 24246, 31875, 68497, 91804, 160301, 213345, 451166, 610247, 1061413, 1426503, 2978230, 4058629, 7036859, 9533213, 19694622, 27007760

Offset: 0

Paul Barry, Oct 23 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A099574, A099575.

[(&+[Binomial(n-k+Floor(k/2)+1, 1+Floor(k/2))*(1+Floor(k/2))/(n-k+1): k in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022

A099577 := proc(n)
    local a,k ;
    a := 0 ;
    for k from 0 to floor(n/2) do
        a := a+add(binomial(n-k+j,j),j=0..floor(k/2)) ;
    end do:
    a ;
end proc:
seq(A099577(n),n=0..50); # R. J. Mathar, Nov 28 2014

Table[Sum[Binomial[n-k+Floor[k/2]+1, 1+Floor[k/2]]*(1+Floor[k/2])/(n-k+1), {k, 0, Floor[n/2]}], {n, 0, 40}] (* G. C. Greubel, Jul 24 2022 *)

[sum( binomial(n-k+(k//2)+1, 1+(k//2))*(1+(k//2))/(n-k+1) for k in (0..(n//2)) ) for n in (0..40)] # G. C. Greubel, Jul 24 2022

a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..floor(k/2)} binomial(n-k+j, j).

a(n) = Sum_{k=0..floor(n/2)} binomial(n - k + floor(k/2) + 1, 1 + floor(k/2))*(1 + floor(k/2))/(n-k+1). - G. C. Greubel, Jul 24 2022

A099576 Row sums of triangle A099575.

Original entry on oeis.org

1, 2, 6, 12, 35, 72, 210, 440, 1287, 2730, 8008, 17136, 50388, 108528, 319770, 692208, 2042975, 4440150, 13123110, 28614300, 84672315, 185122080, 548354040, 1201610592, 3562467300, 7821594872, 23206929840, 51037462560, 151532656696

Offset: 0

Paul Barry, Oct 23 2004

Andrew Howroyd, Table of n, a(n) for n = 0..200

Cf. A029907, A099575, A099577.

[(&+[Binomial(n+Floor(k/2)+1, Floor(k/2)+1)*(1+Floor(k/2))/(n+1): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022

seq(op([(1+n/(n+1))*binomial(3*n+1,n),2*binomial(3*n+3,n)]),n=0..20);

a[n_] := Sum[Binomial[n + j, j], {k, 0, n}, {j, 0, k/2}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 06 2018 *)
a[n_] := Binomial[2*n+2, n]*Hypergeometric2F1[-n, n+1, -2*n-2, -1]; Flatten[Table[a[n], {n, 0, 28}]] (* Detlef Meya, Dec 25 2023 *)

a(n) = sum(k=0, n, sum(j=0, floor(k/2), binomial(n+j, j))); \\ Andrew Howroyd, Feb 13 2018

[sum( binomial(n+(k//2)+1, (k//2)+1)*(1+(k//2))/(n+1) for k in (0..n) ) for n in (0..40)] # G. C. Greubel, Jul 24 2022

a(n) = Sum_{k=0..n} Sum_{j=0..floor(k/2)} binomial(n+j, j).
Conjecture: 4*n*(n-1)*(3*n+2)*(n+2)*a(n) - 36*(n-1)*(n+1)*a(n-1) - 3*n*(3*n+5)*(3*n-1)*(3*n-2)*a(n-2) = 0. - R. J. Mathar, Nov 28 2014
From Robert Israel, May 08 2018: (Start)
a(2*n) = (1+n/(n+1))*binomial(3*n+1,n).
a(2*n+1) = 2*binomial(3*n+3,n).
The conjecture follows from this. (End)
a(n) = (1/(n+1))*Sum_{k=0..n} binomial(n + floor(k/2) + 1, floor(k/2) + 1)*(1 + floor(k/2)). - G. C. Greubel, Jul 24 2022
a(n) = binomial(2*n+2, n)*hypergeom([-n, n+1], [-2*n-2], -1). - Detlef Meya, Dec 25 2023

A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).

Original entry on oeis.org

1, 1, 4, 5, 21, 28, 120, 165, 715, 1001, 4368, 6188, 27132, 38760, 170544, 245157, 1081575, 1562275, 6906900, 10015005, 44352165, 64512240, 286097760, 417225900, 1852482996, 2707475148, 12033222880, 17620076360, 78378960360, 114955808528

Offset: 0

Paul Barry, Oct 23 2004

Main diagonal of triangle A099575.
With offset 2, this is the number of compositions of n-1 into floor(n/2) parts. - T. D. Noe, Jan 05 2007
From Petros Hadjicostas, Jul 19 2018: (Start)
We clarify the above comment by T. D. Noe. The number of compositions of N into K positive parts is C(N-1, K-1). This was proved by MacMahon in 1893 (and probably by others before him). The number of compositions of N into K nonnegative parts is C(N+K-1, K-1) because for every composition b_1 + ... + b_K = N with b_i >= 0 for all i, we may create another composition c_1 + ... + c_K = N+K with c_i = b_i + 1 >= 1.
The statement of T. D. Noe above means that, for n>=2, a(n-2) is the number of compositions of N = n-1 into K = floor(n/2) nonnegative parts. Thus, a(n-2) = C(N+K-1, K-1) = C(n-1+floor(n/2)-1, floor(n/2)-1) = C(floor((3(n-2)+2)/2), floor((n-2)/2)).
This interpretation is important for T. D. Noe's comments for sequence A030077, whose unknown general formula remains an unsolved problem (as of July 2018).
It should be noted, however, that for most authors "composition" means "composition into positive parts". The phrase "weak composition" is sometimes used for a "composition into nonnegative parts".
(End)

			From _Petros Hadjicostas_, Jul 19 2018: (Start)
With n=2 there are a(2-2) = a(0) = 1 compositions of 2-1 = 1 into floor(2/2) = 1 nonnegative parts, namely 1 (only).
With n=3 there are a(3-2) = a(1) = 1 compositions of 3-1 = 2 into floor(3/2) = 1 nonnegative parts, namely 2 (only).
With n=4 there are a(4-2) = a(2) = 4 compositions of 4-1 = 3 into floor(4/2) = 2 nonnegative parts, namely 0+3, 3+0, 1+2, and 2+1.
With n=5 there are a(5-2) = a(3) = 5 compositions of 5-1 = 4 into floor(5/2) = 2 nonnegative parts, namely 0+4, 4+0, 1+3, 3+1, and 2+2.
With n=6 there are a(6-2) = a(4) = 21 compositions of 6-1 = 5 into floor(6/2) = 3 nonnegative parts, namely the 3 permutations of 1+1+3, the 3 permutations of 1+2+2, the 3 permutations of 0+0+5, the 6 permutations of 0+1+4, and the 6 permutations of 0+2+3.
(End)

T. D. Noe, Table of n, a(n) for n=0..200

Cf. A025174 (bisection), A030077, A045721 (bisection), A099575, A127040.

[(&+[Binomial(n+j, j): j in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022

A099578:=n->binomial(floor((3*n+2)/2), floor(n/2)); seq(A099578(k), k=0..50); # Wesley Ivan Hurt, Nov 01 2013

Table[Binomial[Floor[(3n+2)/2], Floor[n/2]], {n, 0, 50}] (* Wesley Ivan Hurt, Nov 01 2013 *)
CoefficientList[Series[-((Sqrt[4 -27 x^2] -2(Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]] + Sqrt[3] Sin[2/3 ArcSin[(3 Sqrt[3] x)/2]]))/(3 x Sqrt[4 -27 x^2])), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 15 2016 *)
a[n_] := Binomial[2*n+1, n]*Hypergeometric2F1[-n, n+1, -2*n-1, -1]; Flatten[Table[a[n], {n, 0, 29}]] (* Detlef Meya, Dec 25 2023 *)

a(n) = binomial((3*n+2)\2, n\2); \\ Michel Marcus, Nov 02 2013

[binomial((3*n+2)//2, n//2) for n in (0..40)] # G. C. Greubel, Jul 24 2022

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, k).
40*n*(n+1)*a(n) +36*n*(n-2)*a(n-1) -6*(45*n^2-23)*a(n-2) -27*(3*n-4)*(3*n-5)*a(n-3) = 0. - R. J. Mathar, Oct 30 2014
From Benedict W. J. Irwin, Aug 15 2016: (Start)
G.f.: -( (sqrt(4-27*x^2) - 2*(cos(arcsin(3*sqrt(3)*x/2)/3) + sqrt(3)*sin(2*arcsin(3*sqrt(3)*x/2)/3)) )/( 3*x*sqrt(4-27*x^2)) ).
E.g.f.: Hypergeometric2F3(2/3,4/3;1/2,1,3/2;27*x^2/16) + x*Hypergeometric2F3(4/3,5/3;3/2,3/2,2;27*x^2/16).
(End)
Recurrence: 4*n*(n+1)*(6*n-1)*a(n) = 18*n*a(n-1) + 3*(3*n-2)*(3*n-1)*(6*n+5)*a(n-2). - Vaclav Kotesovec, Aug 15 2016
a(n) = binomial(2*n+1, n)*hypergeom([-n, n+1], [-2*n-1], -1). - Detlef Meya, Dec 25 2023

A099573 Reverse of number triangle A054450.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233

Offset: 0

Paul Barry, Oct 23 2004

			First few rows of the array:
  1, 1, 2, 3,  5,  8, ... (A000045)
  1, 1, 3, 4,  8, 12, ... (A052952)
  1, 1, 4, 5, 12, 17, ... (A054451)
  1, 1, 5, 6, 17, 23, ... (A099571)
  1, 1, 6, 7, 23, 30, ... (A099572)
  ...
Triangle begins as:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 3, 3;
  1, 1, 4, 4,  5;
  1, 1, 5, 5,  8,  8;
  1, 1, 6, 6, 12, 12, 13;
  1, 1, 7, 7, 17, 17, 21, 21;
  1, 1, 8, 8, 23, 23, 33, 33, 34;
  1, 1, 9, 9, 30, 30, 50, 50, 55, 55;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000045, A029907 (row sums), A052951, A052952, A054450, A054451, A052952.

Cf. A099571, A099572, A099574 (diagonal sums), A099575.

[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2)))
flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)

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A099577 Diagonal sums of triangle A099575.

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A099576 Row sums of triangle A099575.

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A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).

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A099573 Reverse of number triangle A054450.

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