Displaying 1-10 of 11 results found.
0, 1, 6, 26, 96, 321, 1002, 2972, 8472, 23392, 62912, 165504, 427264, 1085184, 2717184, 6718464, 16427008, 39763968, 95387648, 226951168, 535953408, 1257046016, 2929852416, 6789267456, 15648423936, 35888562176, 81927340032
FORMULA
G.f.: x*((1-x)^4)/(1-2*x)^5.
Triangle of partial row sums (prs) of triangle A055249.
+10
11
1, 4, 1, 13, 5, 1, 38, 18, 6, 1, 104, 56, 24, 7, 1, 272, 160, 80, 31, 8, 1, 688, 432, 240, 111, 39, 9, 1, 1696, 1120, 672, 351, 150, 48, 10, 1, 4096, 2816, 1792, 1023, 501, 198, 58, 11, 1, 9728, 6912, 4608, 2815, 1524, 699, 256, 69, 12, 1, 22784, 16640, 11520
COMMENTS
In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is (((1-z)^2)/(1-2*z)^3)/(1-x*z/(1-z)). This is the third member of the family of Riordan-type matrices obtained from A007318(n,m) (Pascal's triangle read as lower triangular matrix) by repeated application of the prs-procedure.
FORMULA
a(n, m)=sum( A055249(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n<m, (sequence of partial row sums in column m). Column m recursion: a(n, m)= sum(a(j, m), j=m..n-1)+ A055249(n, m), n >= m >= 0, a(n, m) := 0 if n<m. G.f. for column m: (((1-x)^2)/(1-2*x)^3)*(x/(1-x))^m, m >= 0.
T(n, k) = binomial(n, k)*hypergeom([3, k - n], [k + 1], -1). - Peter Luschny, Sep 23 2024
EXAMPLE
[0] 1
[1] 4, 1
[2] 13, 5, 1
[3] 38, 18, 6, 1
[4] 104, 56, 24, 7, 1
[5] 272, 160, 80, 31, 8, 1
[6] 688, 432, 240, 111, 39, 9, 1
[7] 1696, 1120, 672, 351, 150, 48, 10, 1
Fourth row polynomial (n = 3): p(3, x) = 38 + 18*x + 6*x^2 + x^3.
MAPLE
T := (n, k) -> binomial(n, k)*hypergeom([3, k - n], [k + 1], -1):
for n from 0 to 7 do seq(simplify(T(n, k)), k = 0..n) od; # Peter Luschny, Sep 23 2024
Summatory Pascal triangle T(n,k) (0 <= k <= n) read by rows. Top entry is 1. Each entry is the sum of the parallelogram above it.
+10
11
1, 1, 1, 2, 3, 2, 4, 8, 8, 4, 8, 20, 26, 20, 8, 16, 48, 76, 76, 48, 16, 32, 112, 208, 252, 208, 112, 32, 64, 256, 544, 768, 768, 544, 256, 64, 128, 576, 1376, 2208, 2568, 2208, 1376, 576, 128, 256, 1280, 3392, 6080, 8016, 8016, 6080, 3392, 1280, 256
COMMENTS
We may also relabel the entries as U(0,0), U(1,0), U(0,1), U(2,0), U(1,1), U(0,2), U(3,0), ... [That is, T(n,k) = U(n-k, k) for 0 <= k <= n and U(m,s) = T(m+s, s) for m,s >= 0.]
We explain the parallelogram definition of T(n,k).
T(0,0) *
|\
| \
| * T(k,k)
T(n-k,0) * |
\ |
\|
* T(n,k)
The definition implies that T(n,k) is the sum of all T(i,j) such that (i,j) has integer coordinates over the set
{(i,j): a(1,0) + b(1,1), 0 <= a <= n-k, 0 <= b <= k} - {(n,k)}.
The parallelogram can sometimes be degenerate; e.g., when k = 0 or n = k. (End)
T(n,k) is the number of 2-compositions of n having sum of the entries of the first row equal to k (0 <= k <= n). A 2-composition of n is a nonnegative matrix with two rows, such that each column has at least one nonzero entry and whose entries sum up to n. - Emeric Deutsch, Oct 12 2010 Robeva and Sun (2020) let A(m,n) = U(m-1, n-1) be the number of subdivisions of a 2-row grid with m points on the top and n points at the bottom (and such that the lower left point is the origin).
The authors proved that A(m,n) = 2*(A(m,n-1) + A(m-1,n) - A(m-1,n-1)) for m, n >= 2 (with (m,n) <> (2,2)), which is equivalent to a similar recurrence for U(n,k) given in the Formula section below. (They did not explicitly specify the value of A(1,1) = U(0,0) because they did not care about the number of subdivisions of a degenerate polygon with only one side.)
They also proved that, for (m,n) <> (1,1), A(m,n) = (2^(m-2)/(n-1)!) * Q_n(m) =
= (2^(m-2)/(n-1)!) * Sum_{k=1..n} A336244(n,k) * m^(n-k), where Q_n(m) is a polynomial in m of degree n-1. (End) With the square array notation of Petros Hadjicostas, Jul 16 2020 below, U(i,j) is the number of lattice paths from (0,0) to (i,j) whose steps move north or east or have positive slope. For example, representing a path by its successive lattice points rather than its steps, U(1,2) = 8 counts {(0,0),(1,2)}, {(0,0),(0,1),(1,2)}, {(0,0),(0,2),(1,2)}, {(0,0),(1,0),(1,2)}, {(0,0),(1,1),(1,2)}, {(0,0),(0,1),(0,2),(1,2)}, {(0,0),(0,1),(1,1),(1,2)}, {(0,0),(1,0),(1,1),(1,2)}. If north (vertical) steps are excluded, the resulting paths are counted by A049600. - David Callan, Nov 25 2021
LINKS
Fang, E., Jenkins, J., Lee, Z., Li, D., Lu, E., Miller, S. J., ... & Siktar, J. (2019). Central Limit Theorems for Compound Paths on the 2-Dimensional Lattice, arXiv preprint arXiv:1906.10645. Also Fib. Q., 58:1 (2020), 208-225.
FORMULA
G.f.: U(z, w) = Sum_{n >= 0, k >= 0} U(n, k)*z^n*w^k = Sum{n >= 0, k >= 0} T(n, k)*z^(n-k)*w^k = (1-z)*(1-w)/(1 - 2*w - 2*z + 2*z*w).
Maple code gives another explicit formula for U(n, k).
From Jon Stadler (jstadler(AT)capital.edu), Apr 30 2003: (Start)
U(n,k) is the number of ways of writing the vector (n,k) as an ordered sum of vectors, equivalently, the number of paths from (0,0) to (n,k) in which steps may be taken from (i,j) to (p,q) provided (p,q) is to the right or above (i,j).
2*U(n,k) = Sum_{i <= n, j <= k} U(i,j).
U(n,k) = 2*U(n-1,k) + Sum_{i < k} U(n,i).
U(n,k) = Sum_{j=0..n+k} C(n,j-k+1)*C(k,j-n+1)*2^j. (End)
T(n, k) = 2*(T(n-1, k-1) + T(n-1, k)) - (2 - 0^(n-2))*T(n-2, k-1) for n > 1 and 1 < k < n; T(n, 0) = T(n, n) = 2*T(n-1, 0) for n > 0; and T(0, 0) = 1. - Reinhard Zumkeller, Dec 03 2004 Sum_{k=0..n} k*T(n,k) = A181292(n). T(n,k) = Sum_{j=0..min(k, n-k)} (-1)^j*2^(n-j-1)*binomial(k, j)*binomial(n-j, k) for (n,k) != (0,0).
G.f.: G(t,z) = (1-z)*(1-t*z)/(1 - 2*z - 2*t*z + 2*t*z^2). (End)
U(n,k) = 0 if k < 0; else U(k,n) if k > n; else 1 if n <= 1; else 3 if n = 2 and k = 1; else 2*U(n,k-1) + 2*U(n-1,k) - 2*U(n-1,k-1). - David W. Wilson; corrected in the case k > n by Robert Israel, Jun 15 2011 [Corrected by Petros Hadjicostas, Jul 16 2020] U(n,k) = binomial(n,k) * 2^(n-1) * hypergeom([-k,-k], [n+1-k], 2) if n >= k >= 0 with (n,k) <> (0,0). - Robert Israel, Jun 15 2011 [Corrected by Petros Hadjicostas, Jul 16 2020] U(n,k) = Sum_{0 <= i+j <= n+k-1} (-1)^j*C(i+j+1, j)*C(n+i, n)*C(k+i, k). - Masato Maruoka, Dec 10 2019 T(n, k) = 2^(n - 1)*binomial(n, k)*hypergeom([-k, k - n], [-n], 1/2) = A059474(n, k)/2 for n >= 1. - Peter Luschny, Nov 26 2021 T(n, n-k) = T(n, k).
T(n, n-2) = 2* A049611(n-1), n >= 2. T(n, n-3) = 4* A049612(n-2), n >= 3. T(n, n-4) = 8* A055589(n-3), n >= 4. T(n, n-5) = 16* A055852(n-4), n >= 5. T(n, n-6) = 32* A055853(n-5), n >= 6. Sum_{k=0..floor(n/2)} T(n, k) = A181306(n). (End)
EXAMPLE
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins
[0] 1;
[1] 1, 1;
[2] 2, 3, 2;
[3] 4, 8, 8, 4;
[4] 8, 20, 26, 20, 8;
[5] 16, 48, 76, 76, 48, 16;
[6] 32, 112, 208, 252, 208, 112, 32;
...
T(5,2) = 76 is the sum of the elements above it in the parallelogram bordered by T(0,0), T(5-2,0) = T(3,0), T(2,2) and T(5,2). We of course exclude T(5,2) from the summation. Thus
T(5,2) = Sum_{a=0..5-2, b=0..2, (a,b) <> (5-2,2)} T(a(1,0) + b(1,1)) =
= (1 + 1 + 2) + (1 + 3 + 8) + (2 + 8 + 26) + (4 + 20) = 76. [Edited by Petros Hadjicostas, Jul 16 2020] Square array U(n,k) (with rows n >= 0 and columns k >= 0) begins
1, 1, 2, 4, 8, ...
1, 3, 8, 20, 48, ...
2, 8, 26, 76, 208, ...
4, 20, 76, 252, 768, ...
8, 48, 208, 768, 2568, ...
16, 112, 544, 2208, 8016, ...
...
Consider the following 2-row grid with n = 3 points at the top and k = 2 points at the bottom:
A B C
*--*--*
| /
| /
*--*
D E
The sets of the dividing internal lines of the A(3,2) = U(3-1, 2-1) = 8 subdivisions of the above 2-row grid are as follows: { }, {DC}, {DB}, {EB}, {EA}, {DB, DC}, {DB, EB}, and {EA, EB}. See Robeva and Sun (2020).
These are the 2-compositions of n = 3 with sum of first row entries equal to k = 1:
[1; 2], [0,1; 2,0], [0,1; 1,1], [1,0; 0,2], [1,0; 1,1], [0,0,1; 1,1,0], [0,1,0; 1,0,1], and [1,0,0; 0,1,1]. We have T(3,2) = 8 such matrices. See Emeric Deutsch's contribution above. See also Section 2 in Castiglione et al. (2007). (End)
MAPLE
A059576 := proc(n, k) local b, t1; t1 := min(n+k-2, n, k); add( (-1)^b * 2^(n+k-b-2) * (n+k-b-2)! * (1/(b! * (n-b)! * (k-b)!)) * (-2 * n-2 * k+2 * k^2+b^2-3 * k * b+2 * n^2+5 * n * k-3 * n * b), b=0..t1); end;
T := proc (n, k) if k <= n then add((-1)^j*2^(n-j-1)*binomial(k, j)*binomial(n-j, k), j = 0 .. min(k, n-k)) fi end proc: 1; for n to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form # Emeric Deutsch, Oct 12 2010 T := (n, k) -> `if`(n=0, 1, 2^(n-1)*binomial(n, k)*hypergeom([-k, k - n], [-n], 1/2)): seq(seq(simplify(T(n, k)), k=0..n), n=0..10); # Peter Luschny, Nov 26 2021
MATHEMATICA
T[0, 0] = 1; T[n_, k_] := 2^(n-k-1)*n!*Hypergeometric2F1[ -k, -k, -n, -1 ] / (k!*(n-k)!); Flatten[ Table[ T[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Feb 01 2012, after Robert Israel *)
PROG
(Haskell)
a059576 n k = a059576_tabl !! n !! k
a059576_row n = a059576_tabl !! n
a059576_tabl = [1] : map fst (iterate f ([1, 1], [2, 3, 2])) where
f (us, vs) = (vs, map (* 2) ws) where
ws = zipWith (-) (zipWith (+) ([0] ++ vs) (vs ++ [0]))
([0] ++ us ++ [0])
(Magma)
A011782:= func< n | n eq 0 select 1 else 2^(n-1) >;
if k eq 0 or k eq n then return A011782(n); else return 2*T(n-1, k-1) + 2*T(n-1, k) - (2 - 0^(n-2))*T(n-2, k-1);
end if; return T;
end function;
[T(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 02 2022 (SageMath)
if (k==0 or k==n): return 1 if (n==0) else 2^(n-1) # A011782 else: return 2*T(n-1, k-1) + 2*T(n-1, k) - (2 - 0^(n-2))*T(n-2, k-1)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 02 2022
Triangle with columns built from row sums of the partial row sums triangles obtained from Pascal's triangle A007318. Essentially A049600 formatted differently.
+10
9
1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 8, 4, 1, 1, 16, 20, 13, 5, 1, 1, 32, 48, 38, 19, 6, 1, 1, 64, 112, 104, 63, 26, 7, 1, 1, 128, 256, 272, 192, 96, 34, 8, 1, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1, 1024, 2816, 4096
COMMENTS
In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is 1/((1-z)*(1-x*z*(1-z)/(1-2*z))). Column m (without leading zeros) is obtained from convolution of A000012 (powers of 1) with m-fold convoluted A011782.
FORMULA
a(n, m)= Am(n, 0) if n >= m >= 0 and a(n, m) := 0 if n<m; i.e. first column of the lower triangular matrix Am := prs^(m)( A007318) with the lower triangular matrix A007318 (Pascal triangle) and prs^(m) is the partial row sums (prs) mapping for triangular matrices applied m times. See e.g. A055584 for m=4. G.f. for column m: (1/(1-x))*(x*(1-x)/(1-2*x))^m, m >= 0.
T(n, k) = sum_{j=0..n-k} C(n-k, j)*C(k+j-1, k-1). - Paul D. Hanna, Jan 14 2004
EXAMPLE
{1}; {1, 1}; {1, 2, 1}; {1, 4, 3, 1}; {1, 8, 8, 4, 1}; ...
Fourth row polynomial (n=3): p(3,x)= 1+4*x+3*x^2+x^3
MATHEMATICA
t[n_, k_] := Hypergeometric2F1[k, k-n, 1, -1]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after Paul D. Hanna *)
PROG
(PARI) {T(n, k) = if( n<0 || k<0, 0, polcoeff( polcoeff( 1 / ((1 - z) * (1 - x*z * (1 - z) / (1 - 2*z) + z * O(z^n) + x * O(x^k))), k), n))}; /* Michael Somos, Sep 30 2003 */ (PARI) {T(n, k)=if(k>n||n<0||k<0, 0, if(k==0||k==n, 1, sum(j=0, n-k, binomial(n-k, j)*binomial(k+j-1, k-1)); ); )} (Hanna)
Triangle read by rows, T(n,k) = hypergeometric_2F1([n-k+1, -k], [1], -1) for n>=0 and k>=0.
+10
9
1, 1, 2, 1, 3, 4, 1, 4, 8, 8, 1, 5, 13, 20, 16, 1, 6, 19, 38, 48, 32, 1, 7, 26, 63, 104, 112, 64, 1, 8, 34, 96, 192, 272, 256, 128, 1, 9, 43, 138, 321, 552, 688, 576, 256, 1, 10, 53, 190, 501, 1002, 1520, 1696, 1280, 512, 1, 11, 64, 253, 743, 1683, 2972, 4048
COMMENTS
Previous name was: Triangle of coefficients of polynomials v(n,x) jointly generated with A160232; see the Formula section. Row sums: (1,3,8,...), even-indexed Fibonacci numbers.
Alt. row sums: (1,-1,2,-3,...), signed Fibonacci numbers.
Subtriangle of the triangle T(n,k) given by (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 12 2012
FORMULA
u(n,x) = u(n-1,x) + x*v(n-1,x), v(n,x) = u(n-1,x) + 2x*v(n-1,x), where u(1,x) = 1, v(1,x) = 1.
As DELTA-triangle with 0 <= k <= n: T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1), T(0,0) = T(1,0) = T(2,0) = 1, T(1,1) = T(2,2) = 0, T(2,1) = 2 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 12 2012 The following remarks assume the row and column indexing start at 0.
T(n,k) = Sum_{i = 0..k} 2^(k-i)*binomial(n-k,i)*binomial(k,i) = Sum_{i = 0..k} binomial(n-k+i,i)*binomial(k,i).
Riordan array (1/(1 - x), x*(2 - x)/(1 - x)).
O.g.f. 1/(1 - (2*t + 1)*x + t*x^2) = 1 + (1 + 2*t)*x + (1 + 3*t + 4*t^2)*x^2 + ....
Read as a square array, this equals P * transpose(P^2), where P denotes Pascal's triangle A007318. (End) For k<n, T(n,k) = T(n-1,k) + Sum_{i=1..k} T(n-i,k-i). - Glen Whitney, Aug 17 2021
EXAMPLE
First five rows:
1;
1, 2;
1, 3, 4;
1, 4, 8, 8;
1, 5, 13, 20, 16;
First five polynomials v(n,x):
1
1 + 2x
1 + 3x + 4x^2
1 + 4x + 8x^2 + 8x^3
1 + 5x + 13x^2 + 20x^3 + 16x^4
(1, 0, -1/2, 1/2, 0, 0, ...) DELTA (0, 2, 0, 0, 0, ...) begins:
1;
1, 0;
1, 2, 0;
1, 3, 4, 0;
1, 4, 8, 8, 0;
1, 5, 13, 20, 16, 0;
1, 6, 19, 38, 48, 32, 0;
0;
0, 1;
0, 1, 2;
0, 1, 3, 4;
0, 1, 4, 8, 8;
0, 1, 5, 13, 20, 16;
0, 1, 6, 19, 38, 48, 32;
MAPLE
T := (n, k) -> hypergeom([n-k+1, -k], [1], -1):
seq(lprint(seq(simplify(T(n, k)), k=0..n)), n=0..7); # Peter Luschny, May 20 2015
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 13;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + 2*x*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
PROG
(Haskell)
a208341 n k = a208341_tabl !! (n-1) !! (k-1)
a208341_row n = a208341_tabl !! (n-1)
a208341_tabl = map reverse a106195_tabl
(PARI) T(n, k) = sum(i = 0, k, 2^(k-i)*binomial(n-k, i)*binomial(k, i));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print(); ); \\ Michel Marcus, Aug 14 2015
Riordan array (1/(1-2*x), x*(1-x)/(1-2*x)).
+10
7
1, 2, 1, 4, 3, 1, 8, 8, 4, 1, 16, 20, 13, 5, 1, 32, 48, 38, 19, 6, 1, 64, 112, 104, 63, 26, 7, 1, 128, 256, 272, 192, 96, 34, 8, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1024, 2816, 4096, 4048, 2972, 1683, 743, 253, 64, 11
COMMENTS
Extract antidiagonals from the product P * A, where P = the infinite lower triangular Pascal's triangle matrix; and A = the Pascal's triangle array:
1, 1, 1, 1, ...
1, 2, 3, 4, ...
1, 3, 6, 10, ...
1, 4, 10, 20, ...
...
Row sums are Fibonacci(2n+2). Diagonal sums are A006054(n+2). Row sums of inverse are A105523. Product of Pascal triangle A007318 and A046854. A106195 with an appended column of href="/A055587" title="Triangle with columns built from row sums of the partial row sums triangles obtained from Pascal's triangle A007318. Essenti...">A055587. Alternatively, k-th column (k=0, 1, 2) is the binomial transform of bin(n, k).
T(n,k) is the number of ideals in the fence Z(2n) having k elements of rank 1. - Emanuele Munarini, Mar 22 2011 Subtriangle of the triangle given by (0, 2, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012
FORMULA
T(n,k) = Sum_{j=0..n} C(n-k,n-j)*C(j,k).
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-k,i)*2^(n-k-i).
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-i,k)*(-1)^i*2^(n-k-i).
Recurrence: T(n+2,k+1) = 2*T(n+1,k+1)+T(n+1,k)-T(n,k) (End)
u(n,x) = u(n-1,x)+v(n-1,x), v(n,x) = u(n-1,x)+(x+1)v(n-1,x),
where u(1,x)=1, v(1,x)=1. Then v matches A106195 and u matches A207605. (End) T(n+k,k) is the coefficient of x^n y^k in 1/(1-2x-y+xy). - Ira M. Gessel, Oct 30 2012 T(n, k) = hypergeometric_2F1(-n+k, k+1, 1 , -1). - Peter Luschny, May 20 2015
EXAMPLE
Triangle begins
1;
2, 1;
4, 3, 1;
8, 8, 4, 1;
16, 20, 13, 5, 1;
32, 48, 38, 19, 6, 1;
64, 112, 104, 63, 26, 7, 1;
(0, 2, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, ...) begins :
1;
0, 1;
0, 2, 1;
0, 4, 3, 1;
0, 8, 8, 4, 1;
0, 16, 20, 13, 5, 1;
0, 32, 48, 38, 19, 6, 1;
MAPLE
T := (n, k) -> hypergeom([-n+k, k+1], [1], -1):
seq(lprint(seq(simplify(T(n, k)), k=0..n)), n=0..7); # Peter Luschny, May 20 2015
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + (x + 1) v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Table[Hypergeometric2F1[-n+k, k+1, 1, -1], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 15 2020 *)
PROG
(Maxima) create_list(sum(binomial(i, k)*binomial(n-k, n-i), i, 0, n), n, 0, 8, k, 0, n); [Emanuele Munarini, Mar 22 2011]
(Haskell)
a106195 n k = a106195_tabl !! n !! k
a106195_row n = a106195_tabl !! n
a106195_tabl = [1] : [2, 1] : f [1] [2, 1] where
f us vs = ws : f vs ws where
ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
([0] ++ us ++ [0])
(Python)
from sympy import Poly, symbols
x = symbols('x')
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
(Python)
from mpmath import hyp2f1, nprint
def T(n, k): return hyp2f1(k - n, k + 1, 1, -1)
(Magma) [ (&+[Binomial(n-k, n-j)*Binomial(j, k): j in [0..n]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Mar 15 2020 (Sage) [[sum(binomial(n-k, n-j)*binomial(j, k) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 15 2020
CROSSREFS
Column 0 = 1, 2, 4...; (binomial transform of 1, 1, 1...); column 1 = 1, 3, 8, 20...(binomial transform of 1, 2, 3...); column 2: 1, 4, 13, 38...= binomial transform of bin(n, 2): 1, 3, 6...
Triangle of partial row sums (prs) of triangle A055252.
+10
6
1, 5, 1, 19, 6, 1, 63, 25, 7, 1, 192, 88, 32, 8, 1, 552, 280, 120, 40, 9, 1, 1520, 832, 400, 160, 49, 10, 1, 4048, 2352, 1232, 560, 209, 59, 11, 1, 10496, 6400, 3584, 1792, 769, 268, 70, 12, 1, 26624, 16896, 9984, 5376, 2561, 1037, 338, 82, 13, 1, 66304, 43520
COMMENTS
In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is (((1-z)^3)/(1-2*z)^4)/(1-x*z/(1-z)). This is the fourth member of the family of Riordan-type matrices obtained from A007318(n,m) (Pascal's triangle read as lower triangular matrix) by repeated application of the prs-procedure.
FORMULA
a(n, m)=sum( A055252(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n<m, (sequence of partial row sums in column m). Column m recursion: a(n, m)= sum(a(j, m), j=m..n-1)+ A055252(n, m), n >= m >= 0, a(n, m) := 0 if n<m. G.f. for column m: (((1-x)^3)/(1-2*x)^4)*(x/(1-x))^m, m >= 0.
T(n, k) = binomial(n, k)*hypergeom([4, k - n], [k + 1], -1). - Peter Luschny, Sep 23 2024
EXAMPLE
[0] 1
[1] 5, 1
[2] 19, 6, 1
[3] 63, 25, 7, 1
[4] 192, 88, 32, 8, 1
[5] 552, 280, 120, 40, 9, 1
[6] 1520, 832, 400, 160, 49, 10, 1
[7] 4048, 2352, 1232, 560, 209, 59, 11, 1
Fourth row polynomial (n=3): p(3, x)= 63 + 25*x + 7*x^2 + x^3.
MAPLE
T := (n, k) -> binomial(n, k)*hypergeom([4, k - n], [k + 1], -1):
for n from 0 to 7 do seq(simplify(T(n, k)), k = 0..n) od; # Peter Luschny, Sep 23 2024
1, 6, 25, 88, 280, 832, 2352, 6400, 16896, 43520, 109824, 272384, 665600, 1605632, 3829760, 9043968, 21168128, 49152000, 113311744, 259522560, 590872576, 1337982976, 3014656000, 6761218048, 15099494400, 33587986432, 74440507392
COMMENTS
Number of 132-avoiding permutations of [n+5] containing exactly three 123 patterns. - Emeric Deutsch, Jul 13 2001 If X_1,X_2,...,X_n are 2-blocks of a (2n+2)-set X then, for n>=1, a(n-1) is the number of (n+3)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 18 2007
FORMULA
G.f.: (1-x)^2/(1-2*x)^4.
a(n) = A055584(n+1, 1). a(n) = sum(a(j), j=0..n-1)+ A001793(n+1), n >= 1. a(n) = 2^(n-3)(n+1)(n+3)(n+8)/3.
Preceded by 0, this is the binomial transform of the tetrahedral numbers A000292. - Carl Najafi, Sep 08 2011 E.g.f.: (1/6)*(2*x^3 + 15*x^2 + 24*x + 6)*exp(2*x). - G. C. Greubel, Aug 22 2015
EXAMPLE
a(1)=6 because 432516,432561,435126,452136,532146 and 632145 are the only 132-avoiding permutations of 123456, containing exactly three increasing subsequences of length 3.
MATHEMATICA
Table[(1/3)*2^(n-3)*(n+1)*(n+3)*(n+8), {n, 0, 50}] (* G. C. Greubel, Aug 22 2015 *) LinearRecurrence[{8, -24, 32, -16}, {1, 6, 25, 88}, 30] (* Harvey P. Dale, Nov 03 2017 *)
PROG
(PARI) Vec(((1-x)^2)/(1-2*x)^4 + O(x^30)) \\ Michel Marcus, Aug 22 2015
Second binomial transform of binomial(n+3, 3).
+10
3
1, 6, 30, 136, 579, 2358, 9288, 35640, 133893, 494262, 1797714, 6456024, 22930695, 80660934, 281309436, 973599912, 3346483977, 11431295910, 38828142342, 131206405608, 441271936971, 1477621745046, 4927988620080, 16373939547096
COMMENTS
2nd binomial transform of binomial(n+3, 3), A000292. 3rd binomial transform of (1,3,3,1,0,0,0,0,...).
FORMULA
a(n) = 3^n*(n^3 + 24*n^2 + 137*n + 162)/162.
G.f.: (1 - 2*x)^3/(1 - 3*x)^4.
E.g.f.: (6 + 18*x + 9*x^2 + x^3)*exp(3*x)/6. - G. C. Greubel, Oct 18 2018
MATHEMATICA
LinearRecurrence[{12, -54, 108, -81}, {1, 6, 30, 136}, 50] (* G. C. Greubel, Oct 18 2018 *)
PROG
(PARI) x='x+O('x^30); Vec((1-2*x)^3/(1-3*x)^4) \\ G. C. Greubel, Oct 18 2018 (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-2*x)^3/(1-3*x)^4)); // G. C. Greubel, Oct 18 2018
Triangle read by rows: T(n,k) = (k+1)*(k+2)*(k+3)*binomial(n,k)/6 (0 <= k <= n).
+10
1
1, 1, 4, 1, 8, 10, 1, 12, 30, 20, 1, 16, 60, 80, 35, 1, 20, 100, 200, 175, 56, 1, 24, 150, 400, 525, 336, 84, 1, 28, 210, 700, 1225, 1176, 588, 120, 1, 32, 280, 1120, 2450, 3136, 2352, 960, 165, 1, 36, 360, 1680, 4410, 7056, 7056, 4320, 1485, 220, 1, 40, 450, 2400, 7350
COMMENTS
Sum of entries in row n = (2^n/48)*(n+4)*(n^2 + 11n + 12) = A049612(n+1).
EXAMPLE
Triangle starts:
1;
1, 4;
1, 8, 10;
1, 12, 30, 20;
1, 16, 60, 80, 35;
1, 20, 100, 200, 175, 56;
1, 24, 150, 400, 525, 336, 84;
MAPLE
T:=(n, k)->(k+1)*(k+2)*(k+3)*binomial(n, k)/6: for n from 0 to 10 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
MATHEMATICA
Flatten[Table[(k+1)(k+2)(k+3) Binomial[n, k]/6, {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, May 14 2012 *)
Search completed in 0.011 seconds
|