A321203 -id:A321203 - cp's OEIS frontend

A321201 Irregular triangle T with the nontrivial solutions of 2e2 + 3e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.

1, 0, 0, 1, 2, 0, 1, 1, 0, 2, 3, 0, 2, 1, 1, 2, 4, 0, 0, 3, 3, 1, 2, 2, 5, 0, 1, 3, 4, 1, 0, 4, 3, 2, 6, 0, 2, 3, 5, 1, 1, 4, 4, 2, 7, 0, 0, 5, 3, 3, 6, 1, 2, 4, 5, 2, 8, 0, 1, 5, 4, 3, 7, 1, 0, 6, 3, 4, 6, 2, 9, 0, 2, 5, 5, 3, 8, 1, 1, 6, 4, 4, 7, 2, 10, 0

Offset: 2

Wolfdieter Lang, Nov 05 2018

The length of row n is 2*A(n), with A(n) = A008615(n+2) for n >= 2: 2*[1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, ...].
The trivial solution for n = 0 is [0, 0]. There is no solution for n = 1.
The row sums are given in A321202.
If a partition of n with parts 2 or 3 (with inclusive or) is written as 2^{e2} 3^{e3}, where e2 and e3 are nonnegative numbers, then in row n, all pairs [e2, e3] are given, for n >= 2, ordered with increasing values of e2.
The corresponding irregular triangle with the multinomial numbers n!/((n - (e2 + e3))!*e2!*e3!) is given in A321203. It gives the coefficients of x^n = x^{2*{e2} + 3*{e3}} of (1 + x^2 + x^3)^n, for n >= 2.

			The triangle T(n, k) begins (pairs are separated by commas):
  n\k  0  1   2  3   4  5   6  7 ...
  2:   1  0
  3:   0  1
  4:   2  0
  5:   1  1
  6:   0  2,  3  0
  7:   2  1
  8:   1  2,  4  0
  9:   0  3,  3  1
  10:  2  2,  5  0
  11:  1  3,  4  1
  12:  0  4,  3  2,  6  0
  13:  2  3,  5  1,
  14:  1  4,  4  2,  7  0
  15:  0  5,  3  3,  6  1
  16:  2  4,  5  2,  8  0
  17:  1  5,  4  3,  7  1
  18:  0  6,  3  4,  6  2,  9  0
  19:  2  5,  5  3,  8  1
  20:  1  6,  4  4,  7  2, 10  0
  ...
n=8: the two solutions of 2*e2 + 3*e3 = 8 are [e2, e3] = [1, 2] and = [4, 0], and 1 < 4, therefore row 8 is 1  2  4  0, with a comma after the first pair.

Cf. A008615, A321202, A321203.

row[n_] := Reap[Do[If[2 e2 + 3 e3 == n, Sow[{e2, e3}]], {e2, 0, n/2}, {e3, 0, n/3}]][[2, 1]];
Table[row[n], {n, 2, 20}] // Flatten (* Jean-François Alcover, Nov 23 2018 *)

T(n, k) gives all pairs [e2, e3] solving 2*e2 + 3*e3 = n, ordered with increasing value of e2, for n >= 2. The trivial solution [0, 0] for n = 0 is not recorded. There is no solution for n = 1.

Missing row 2 inserted by Jean-François Alcover, Nov 23 2018

A176806 Consider asymmetric 1-D random walk with set of possible jumps {-1,+1,+2}. Sequence gives number of paths of length n ending at origin.

Original entry on oeis.org

1, 0, 2, 3, 6, 20, 35, 105, 238, 588, 1512, 3630, 9339, 23166, 58487, 148148, 373230, 949416, 2406248, 6122142, 15591856, 39729000, 101432982, 259049230, 662421643, 1695149220, 4341026900, 11125755615, 28530984915, 73213888650, 187980163110, 482906682675

Offset: 0

Sergey Perepechko, Apr 26 2010

It appears that a(n) is the coefficient of x^n in the expansion (1+x^2+x^3)^n. - Joerg Arndt, Jul 01 2011 [For the proof see the formula section. - Wolfdieter Lang, Nov 05 2018]

			a(3) = 3: (+2-1-1) or (-1+2-1) or (-1-1+2).
From _Wolfdieter Lang_, Nov 05 2018: (Start)
a(8) = (1/8!)*(d/dt)^8 (1 + t^2 + t^3)^8 becomes for t = 0: 238. (See the comment with the conjecture by _Joerg Arndt_, now proved.)
a(8) = 168 + 70 = 238, the row sum of row n = 8 of A321203, arising from the two [e2, e3] pairs [1, 2] and [4, 0], given in row n = 8 of A321201.
(End)

Cf. A001005, A321201, A321203.

a:=n->add(binomial(n,k)*binomial(k,3*k-n),k=floor((n+2)/3)..floor(n/2));

Table[Sum[Binomial[n, k]*Binomial[k, 3*k-n], {k, Floor[(n+2)/3], Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Mar 01 2016 *)

a(n) = Sum_{k=floor((n+2)/3)..floor(n/2)} binomial(n,k)*binomial(k,3*k-n).
G.f. g(x) satisfies (31*x^3 + 18*x^2 - x - 4)*g(x)^3 + (x+3)*g(x) + 1 = 0.
Recurrence: 2*n*(2*n-1)*(52*n-79)*a(n) + (n-1)*(52*n^2-79*n+36)*a(n-1) - 6*(n-1)*(156*n^2-315*n+106)*a(n-2) - 31*(n-1)*(n-2)*(52*n-27)*a(n-3) = 0.
a(n) ~ c * d^n / sqrt(Pi*n), where d = 2.61071861327603934981864900838405862... is the root of the equation -31 - 18*d + d^2 + 4*d^3 = 0 and c = 0.57803237802255683003114674597591616... is the root of the equation -31 + 324*c^2 - 1248*c^4 + 1664*c^6 = 0. - Vaclav Kotesovec, Mar 01 2016
From Wolfdieter Lang, Nov 05 2018: (Start)
G.f. G(x) = x*(d/dx)log(F^{[-1]}) = f(x)/(1 - (x*f(x))^2 - 2*(x*f(x))^3) = f(x)/(3 - 2*f(x) + (x*f(x))^2), where f(x) = F^{[-1]}(x)/x, and F^{[-1]})(x) is the compositional inverse of F(y) = x/(1 + x^2 + x^3); that is, F(F^{[-1]}(x)) = x, identically. This G can be proved to solve the equation given above for the g.f. g, if one applies the identity for f (as done above in the last formula for G): (x*f(x))^3 + (x*f(x))*2 - f(x) + 1 = 0 (following from the equation for F^{[-1]}). The expansion of f is given in A001005.
The g.f. G(x) can be computed from the general Lagrange series for the function h(t) with derivative h(t)" = 1/phi(t), where phi(t) = (1 + t^2 + t^3), and the inversion of x = y/phi(y) = F(y). Then one finds G(x) = (d/dx)h(F^{[-1]}(x)) = (1/phi(F^{[-1]}(x)))*(d/dx)F^{[-1]}(x), which becomes with the above mentioned identity for f(x) = F^{[-1]}(x)/x the result G(x) = f(x)/(3 - 2*f(x) + (x*f(x))^2).
From this special Lagrange series derivation and the proof that the g.f. g from above coincides with G, the conjecture, given by Joerg Arndt as a comment above, has been proved. This uses [t^n]phi(t)^n = (1/n!)*(d/dt)^n phi(t)^n, evaluated at t = 0, which appears in the considered Lagrange series.
a(n) = Sum_{2*e + 3*e3 = n} n!/((n - (e2+e3))!*e2!*e3!), n >= 2, with a(0) = 1 and a(1) = 0. This is the row sum of the irregular table A321203 of these multinomial numbers for the solutions for the pairs (e2, e3). The pairs of solutions are given in A321201.
(End)

A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) given in A321196.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, -1, 2, 0, 1, 2, -2, 3, 0, 1, -5, 5, -3, 4, 0, 1, 8, -12, 9, -4, 5, 0, 1, -21, 21, -21, 14, -5, 6, 0, 1, 42, -56, 40, -32, 20, -6, 7, 0, 1, -96, 114, -108, 66, -45, 27, -7, 8, 0, 1, 222, -270, 225, -180, 100, -60, 35, -8, 9, 0, 1

Offset: 0

Wolfdieter Lang, Nov 12 2018

Riordan triangle T = (f(t), t*f(t)), with f(t) = F^{[-1]}(t)/t, where F^{[-1]}(t) is the compositional inverse of t = F(x) = x/(1 + x^2 - x^3). The expansion of f(t) is given by {(-1)^n*A001005(n)}_{n >= 0}, the sequence of column k = 0.
This gives the inverse matrix (with upper diagonals filled with zeros) of the Riordan matrix from A321196 for any finite dimension.
The inverse of the Riordan matrix (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) is obtained from the triangle t(n, k) = (-1)^(n-k)*T(n, k), with vanishing upper diagonals.
The row sums give A321199. The alternating row sums give A321200.
The finite A- and Z-sequences of this inverse Riordan triangle of (F(x)/x, F(x)) are A = [1, 0, +1, -1] generated by 1/(F(x)/x), and Z = [0, +1, -1] generated by 1/F(x) - 1/x. See the W. Lang link for A- and Z- sequences in A006232 with references.
For the Boas-Buck column recurrences of Riordan triangles see the Aug 10 2017 remark in A046521, also for two references. For this Bell-type Riordan triangle the Boas-Buck sequence b is generated by B(t) = (log(f(t)))' = (1/(1/f(t) - t^2*f(t) + 2*t^3*f(t)^2) - 1)/t, and b(n) = (-1)^(n)*A176806(n+1), for n >= 0, because the parity of e_3 in the rows n of A321201 coincides with the one of n. See A321203 for the multinomials with negative signs for odd row numbers.

			The triangle begins:
n\k     0     1     2     3    4    5   6   7  8  9  10 ...
-----------------------------------------------------------
0:      1
1:      0     1
2:      1     0     1
3:     -1     2     0     1
4:      2    -2     3     0    1
5:     -5     5    -3     4    0    1
6:      8   -12     9    -4    5    0   1
7:    -21    21   -21    14   -5    6   0   1
8:     42   -56    40   -32   20   -6   7   0  1
9:    -96   114  -108    66  -45   27  -7   8  0  1
10:   222  -270   225  -180  100  -60  35  -8  9  0   1
...
------------------------------------------------------------
Recurrence (from A- and Z-sequence): Z: T(5, 0) = T(4, 1) - T(4, 2) = -2 - 3 = -5. A: T(5, 2) = T(4, 1) + T(4, 3) - T(4, 4) = - 2 + 0 - 1 = -3.
Recurrence column k = 2 (Boas-Buck type sequence b = (-1)^(n+1)* = {0, 2, -3, 6, ...}): T(5, 2) = (3/3)*(b(2)*T(2, 2) + b(1)*T(3, 2) + b(0)*T(4, 2)) = -3*1 + 2*0 + 0*4 = -3.

Cf. A001005, A176806, A321196, A321199, A321200, A321201, A321203.

T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k + 1] - T[n - 1, k + 2] /; n >= k >= 0;Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* James C. McMahon, Dec 10 2023 *)

Recurrence (from A-and Z-sequences): T(n, k) = 0 for n < k, T(0, 0) = 1. Z: T(n, 0) = T(n-1, 1) - T(n-1, 2), n >= 1; A: T(n, k) = T(n-1, k-1) + T(n-1, k+1) - T(n-1, k+2), n >= k >= 0.
Recurrence for column k (Boas-Buck type): T(n, n) = 1; T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), n >= m+1 >= 1. For b see the Boas-Buck comment above.
G.f of row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k: G(x, z) = f(z)/(1 - x*z*f(z)), with f(z) = F^{[-1]}(z)/z, where F^{[-1]}(z) is the compositional inverse of z = F(y) = y/(1 + y^2 - y^3).
G.f of column k: Gcol(k, x) = x^k*f(x)^{k+1}.

A319204 Sequence used for the Boas-Buck type recurrence for Riordan triangle A319203.

Original entry on oeis.org

0, -2, -3, 6, 20, -5, -105, -98, 420, 1008, -990, -6501, -2574, 31603, 52052, -107250, -411944, 81328, 2343042, 2413456, -9883800, -25327722, 23371634, 168185131, 77113020, -835281800, -1452148815, 2847865635, 11561517870, -1613666430, -66318892875, -72637680690, 280330495200, 750725215020

Offset: 0

Wolfdieter Lang, Oct 29 2018

sign

See A319203 for the Boas-Buck type recurrence.

			a(5) = (1/6!)*[d^6/dx^6 (1 - x^2 - x^3)^6] for x = 0, which is -5.
a(5) = +15 - 20 = -5; from the sum of the signed row n=6 in A321203, with parity of e2 + e3 from A321201 even and odd.

Cf. A319203, A321201, A321203.

O.g.f.: (log(f(x)))' = (1/(1/f(x) + x^2*f(x) + 2*x^3*f(x)^2) - 1)/x, with the expansion of f given in A319201. f(x) = F^{[-1]}(x)/x, where F(t) = t/(1 - t^2 - t^3).
a(n) = (1/(n+1)!)*[d^(n+1)/dx^(n+1) (1 - x^2 - x^3)^(n+1)] evaluated at x = 0, for n >= 0. (Cf. Joerg Arndt's conjecture for A176806, which is proved there.)
a(n-1) = Sum_{2*e + 3*e3 = n} (-1)^(e2+e3)*n!/((n - (e2+e3))!*e2!*e3!), n >= 2, with a(0) = 0. The pairs (e2, e3) are given in A321201; see also the multinomial coefficient table A321203 and add the sign factors.

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A321201 Irregular triangle T with the nontrivial solutions of 2*e2 + 3*e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.

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A176806 Consider asymmetric 1-D random walk with set of possible jumps {-1,+1,+2}. Sequence gives number of paths of length n ending at origin.

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A321198 Triangular Riordan matrix T = R^(-1) for triangular Riordan matrix R = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)) given in A321196.

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A319204 Sequence used for the Boas-Buck type recurrence for Riordan triangle A319203.

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A321201 Irregular triangle T with the nontrivial solutions of 2e2 + 3e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.