A144399 -id:A144399 - cp's OEIS frontend

A144385 Triangle read by rows: T(n,k) is the number of partitions of [1, 2, ..., k] into exactly n blocks, each of size 1, 2 or 3 (n >= 0, 0 <= k <= 3n).

1, 0, 1, 1, 1, 0, 0, 1, 3, 7, 10, 10, 0, 0, 0, 1, 6, 25, 75, 175, 280, 280, 0, 0, 0, 0, 1, 10, 65, 315, 1225, 3780, 9100, 15400, 15400, 0, 0, 0, 0, 0, 1, 15, 140, 980, 5565, 26145, 102025, 323400, 800800, 1401400, 1401400, 0, 0, 0, 0, 0, 0, 1, 21, 266, 2520, 19425, 125895, 695695, 3273270, 12962950, 42042000, 106506400, 190590400, 190590400

Offset: 0

David Applegate and N. J. A. Sloane, Dec 07 2008, Dec 17 2008

Row n has 3n+1 entries.

			Triangle begins:
[1]
[0, 1, 1, 1]
[0, 0, 1, 3, 7, 10, 10]
[0, 0, 0, 1, 6, 25, 75, 175, 280, 280]
[0, 0, 0, 0, 1, 10, 65, 315, 1225, 3780, 9100, 15400, 15400]
[0, 0, 0, 0, 0, 1, 15, 140, 980, 5565, 26145, 102025, 323400, 800800, 1401400, 1401400]

Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394, 2017.
David Applegate and N. J. A. Sloane, The Gift Exchange Problem (arXiv:0907.0513, 2009)

See A144399, A144402, A144417, A111246 for other versions of this triangle.
Column sums give A001680, row sums give A144416. Taking last nonzero entry in each row gives A025035.
Diagonals include A000217, A001296, A027778, A144516; also A025035.
A generalization of the triangle in A144331 (and in several other entries).
Cf. A144643.

T := proc(n, k)
option remember;
if n = k then 1;
elif k < n then 0;
elif n < 1 then 0;
else T(n - 1, k - 1) + (k - 1)*T(n - 1, k - 2) + 1/2*(k - 1)*(k - 2)*T(n - 1, k - 3);
end if;
end proc;
for n from 0 to 12 do lprint([seq(T(n,k),k=0..3*n)]); od:

t[n_, n_] = 1; t[n_ /; n >= 0, k_] /; 0 <= k <= 3*n := t[n, k] = t[n-1, k-1] + (k-1)*t[n-1, k-2] + (1/2)*(k-1)*(k-2)*t[n-1, k-3]; t[, ] = 0; Table[t[n, k], {n, 0, 12}, {k, 0, 3*n}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)

T(n, k) = T(n - 1, k - 1) + (k - 1)*T(n - 1, k - 2) + (1/2)*(k - 1)*(k - 2)*T(n - 1, k - 3).

E.g.f.: Sum_{ n >= 0, k >= 0 } T(n, k) y^n x^k / k! = exp( y*(x+x^2/2+x^3/6) ). That is, the coefficient of y^n is the e.g.f. for row n. E.g. the e.g.f. for row 2 is (1/2)*(x+x^2/2+x^3/6)^2 = 1*x^2/2! + 3*x^3/3! + 7*x^4/4! + 10*x^5/5! + 10*x^6/6!.

A144000 Rectangular array by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares for which x + y == 0 (mod 3); then R(m,n) is the number of marked squares in the rectangle [0,m]x[0,n].

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 4, 4, 3, 3, 5, 6, 5, 3, 4, 6, 8, 8, 6, 4, 5, 8, 10, 11, 10, 8, 5, 5, 9, 12, 13, 13, 12, 9, 5, 6, 10, 14, 16, 16, 16, 14, 10, 6, 7, 12, 16, 19, 20, 20, 19, 16, 12, 7, 7, 13, 18, 21, 23, 24, 23, 21, 18, 13, 7, 8, 14, 20, 24, 26, 28, 28, 26, 24, 20, 14, 8, 9, 16, 22, 27, 30

Offset: 1

Clark Kimberling, Sep 07 2008

Row 3n is given by 2n*(1,2,3,4,5,6,...).

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A143996, A143997, A143998, A144399, A144001.

A := proc(n,k) ## n = 0 .. infinity and k = 0 .. n
    if  1 = (n-k+1) mod 3 then
        floor((2*(k+1)*(n-k+1)+1) / 3)
    else
        floor((2*(k+1)*(n-k+1)) / 3)
    end if
end proc: # Yu-Sheng Chang, Jan 01 2020

b[n_, m_] := If[Mod[n, 3] == 1, Floor[(2*m*n + 1)/3],  Floor[2*m*n/3]]; a:= Table[a[n, m], {n, 1, 25}, {m, 1, 25}]; Table[a[[k, n - k + 1]], {n, 1, 20}, {k, 1, n}]//Flatten (* G. C. Greubel, Dec 05 2017 *)

R(m,n) = floor((2*m*n + 1)/3) if n == 1 (mod 3) and floor(2*m*n/3) otherwise.

A144001 Rectangular array read by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares for which x + y == 0 (mod 3); then R(m,n) is the number of unmarked squares in the rectangle [0,m] X [0,n].

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 4, 4, 4, 4, 2, 2, 4, 5, 5, 5, 4, 2, 3, 5, 6, 7, 7, 6, 5, 3, 3, 6, 7, 8, 9, 8, 7, 6, 3, 3, 6, 8, 9, 10, 10, 9, 8, 6, 3, 4, 7, 9, 11, 12, 12, 12, 11, 9, 7, 4, 4, 8, 10, 12, 14, 14, 14, 14, 12, 10, 8, 4, 4, 8, 11, 13, 15, 16, 16, 16, 15, 13, 11, 8, 4, 5, 9

Offset: 1

Clark Kimberling, Sep 07 2008

Row 3n is given by n*(1,2,3,4,5,6,...).

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A143996, A143997, A143998, A144399, A144000.

A[oid] := proc(n,k) ## n = 0 .. infinity and k = 0 .. n
    if 1 = (n-k+1) mod 3 then
        (n-k+1)*(k+1) - floor((2*(n-k+1)*(k+1) + 1)/3)
    else
        (n-k+1)*(k+1) - floor(2*(n-k+1)*(k+1)/3)
    end if
end proc: # Yu-Sheng Chang, Jan 07 2020

b[n_, m_]:= If[Mod[n, 3] == 1, m*n - Floor[(2*m*n + 1)/3], m*n - Floor[2*m*n/3]]; TableForm[Table[b[n, m], {n, 1, 6}, {m, 1, 6}]]
a:= Table[a[n, m], {n, 1, 25}, {m, 1, 25}]; Table[a[[k, n - k + 1]], {n, 1, 20}, {k, 1, n}]//Flatten (* G. C. Greubel, Dec 05 2017 *)

R(m,n) = m*n - floor((2*m*n + 1)/3) if n ==1 (mod 3) and m*n - floor(2*m*n/3) otherwise.

cp's OEIS Frontend

A144385 Triangle read by rows: T(n,k) is the number of partitions of [1, 2, ..., k] into exactly n blocks, each of size 1, 2 or 3 (n >= 0, 0 <= k <= 3n).

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A144000 Rectangular array by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares for which x + y == 0 (mod 3); then R(m,n) is the number of marked squares in the rectangle [0,m]x[0,n].

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A144001 Rectangular array read by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares for which x + y == 0 (mod 3); then R(m,n) is the number of unmarked squares in the rectangle [0,m] X [0,n].

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Mathematica

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