A056939 -id:A056939 - cp's OEIS frontend

A120257 Triangle of Hankel transforms of certain binomial sums.

1, 2, -1, 3, -6, -1, 4, -20, -20, 1, 5, -50, -175, 70, 1, 6, -105, -980, 1764, 252, -1, 7, -196, -4116, 24696, 19404, -924, -1, 8, -336, -14112, 232848, 731808, -226512, -3432, 1, 9, -540, -41580, 1646568, 16818516, -24293412, -2760615, 12870, 1, 10, -825, -108900, 9343620, 267227532, -1447482465

Offset: 0

Paul Barry, Jun 13 2006

Row k is the Hankel transform of Sum_{j=0..n} binomial(k+j, j). Absolute value is reversal of A103905. Diagonal and subdiagonals are essentially signed versions of the central coefficients of certain generalized Pascal-Narayana triangles (A007318, A001263, A056939, A056940, A056941).

			Triangle begins
  1;
  2,   -1;
  3,   -6,     -1;
  4,  -20,    -20,      1;
  5,  -50,   -175,     70,      1;
  6, -105,   -980,   1764,    252,      -1;
  7, -196,  -4116,  24696,  19404,    -924,    -1;
  8, -336, -14112, 232848, 731808, -226512, -3432, 1;

Cf. A120258.

T(n, k) = (-1)^((k+1)\2) * prod(j=0, n-k-1, binomial(2*k+2+j, k+1)/binomial(k+1+j, j)); \\ Michel Marcus, Jan 13 2022

T(n, k) = (cos(Pi*k/2) - sin(Pi*k/2)) * Product_{j=0..n-k-1} C(2k+2+j, k+1)/C(k+1+j, j).

A142470 Triangle T(n, k) = ( (k+2)/(2binomial(k+2, 2)^2) )binomial(n, k)^2binomial(n+1, k)binomial(n+2, k), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 30, 30, 1, 1, 80, 300, 80, 1, 1, 175, 1750, 1750, 175, 1, 1, 336, 7350, 19600, 7350, 336, 1, 1, 588, 24696, 144060, 144060, 24696, 588, 1, 1, 960, 70560, 790272, 1728720, 790272, 70560, 960, 1, 1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1

Offset: 0

Roger L. Bagula, Sep 20 2008

Row sums are 1, 2, 10, 62, 462, 3852, 34974, 338690, 3452306, 36683660, 403472368, ...
From Peter Bala, May 08 2012: (Start)
Define the action of the operator L on a sequence { a(i) }{0<=i<=n} by L{ a(i) }{0<=i<=n} = { a(i)^2 - a(i-1)*a(i+1) }_{0<=i<=n} with the conventions a(-1) = a(n+1) = 0. Extend the action of L to a lower triangular array T by letting L act on the rows of T. Then L acting on Pascal's triangle A007318 produces the triangle of Narayana numbers A001263 and L applied to A001263 produces the present triangle.
Since the Narayana polynomials are real-rooted it follows by a theorem of Branden that the row polynomials of this array are also real-rooted.
(End)

			The triangle begins as:
  1;
  1,    1;
  1,    8,      1;
  1,   30,     30,       1;
  1,   80,    300,      80,        1;
  1,  175,   1750,    1750,      175,        1;
  1,  336,   7350,   19600,     7350,      336,       1;
  1,  588,  24696,  144060,   144060,    24696,     588,      1;
  1,  960,  70560,  790272,  1728720,   790272,   70560,    960,    1;
  1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Petter Brändén, Iterated sequences and the geometry of zeros, arXiv:0909.1927 [math.CO], 2009-2010; J. Reine Angew. Math. 658 (2011), 115-131.

Cf. A001263, A008459, A056939.

A142470:= func< n,k | ( (k+2)/(2*Binomial(k+2, 2)^2) )*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k) >;
[A142470(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 03 2021

f[n_, k_]:= f[n, k]= Binomial[n, k]*Product[j!*(n+j)!/((k+j)!*(n-k+j)!), {j,1,2}];
T[n_, k_]:= Binomial[n, k]*f[n, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Apr 03 2021 *)

def A142470(n, k): return (2/((k+1)^2*(k+2)))*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k)
flatten([[A142470(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 03 2021

Let f(n, k) = binomial(n, k)*Product_{j=1.2} ( j!*(n+j)!/((k+j)!*(n-k+j)!) ), then T(n, k) = 2^(k-n)*f(n, k)*Sum_{j=k..n} binomial(n, j)*binomial(j, k) = binomial(n, k)*f(n, k).
From Peter Bala, May 08 2012: (Start)
T(n, k) = C(n, k)^2 * Product {i=1..2} i!*(n+i)!/((k+i)!*(n-k+i)!) = C(n, k)*C(n+2, k)*C(n+2, k+1)*C(n+2, k+2)/(C(n+2, 1)*C(n+2, 2)).
T(n, k) = 2/((n+1)*(n+2)*(n+3))*C(n, k)*C(n+1, k)*C(n+2, k+2)*C(n+3, k+1) = C(n, k)*A056939(n, k).
(End)
T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k). - G. C. Greubel, Apr 03 2021

Edited by G. C. Greubel, Apr 03 2021

A173882 Triangle T(n, k) = A090443(n-1)/(A090443(k-1)*A090443(n-k-1)) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 24, 24, 1, 1, 60, 240, 60, 1, 1, 120, 1200, 1200, 120, 1, 1, 210, 4200, 10500, 4200, 210, 1, 1, 336, 11760, 58800, 58800, 11760, 336, 1, 1, 504, 28224, 246960, 493920, 246960, 28224, 504, 1, 1, 720, 60480, 846720, 2963520, 2963520, 846720, 60480, 720, 1

Offset: 0

Roger L. Bagula, Mar 01 2010

A090443 is defined as +1 at negative indices here, which keeps the definition valid in the range 0 <= k <= n.

Row sums are 1, 2, 8, 50, 362, 2642, 19322, 141794, 1045298, 7742882, ....

			Triangle begins as:
  1;
  1,   1;
  1,   6,      1;
  1,  24,     24,       1;
  1,  60,    240,      60,        1;
  1, 120,   1200,    1200,      120,        1;
  1, 210,   4200,   10500,     4200,      210,        1;
  1, 336,  11760,   58800,    58800,    11760,      336,       1;
  1, 504,  28224,  246960,   493920,   246960,    28224,     504,      1;
  1, 720,  60480,  846720,  2963520,  2963520,   846720,   60480,    720,   1;
  1, 990, 118800, 2494800, 13970880, 24449040, 13970880, 2494800, 118800, 990, 1;
  ...

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

Cf. A056939.

T:= func< n,k | k eq 0 or k eq n select 1 else 2*Binomial(n-1,k-1)*Binomial(n,k)*Binomial(n+1,k+1)*(n-k)/(n-k+1) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 17 2021

A090443 := proc(n) (n+2)!*(n+1)!*n!/2 ; end proc:
A173882 := proc(n,m) if m=0 or m= n then 1; else A090443(n-1)/A090443(m-1)/A090443(n-m-1) ; end if; end proc:
seq(seq(A173882(n,m),m=0..n),n=0..5) ; # R. J. Mathar, Mar 19 2011

T[n_,k_]:= If[k==0||k==n, 1, 2*Binomial[n-1,k-1]*Binomial[n,k]*Binomial[n+1,k+1]*(n-k)/(n-k+1)];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 17 2021 *)

def T(n,k): return 1 if (k==0 or k==n) else 2*binomial(n-1,k-1)*binomial(n,k)*binomial(n+1,k+1)*(n-k)/(n-k+1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 17 2021

T(n, k) = 2*binomial(n-1,k-1)*binomial(n,k)*binomial(n+1,k+1)*(n-k)/(n-k+1) with T(n, 0) = T(n, n) = 1.

T(n, k) = c(n)/(c(k)*c(n-k)) where c(n) = Product_{j=2..n} (j-1)*j*(j+1) = (n-1)!*n!*(n+1)!/2 and c(0) = c(1) = 1.

A095265 A sequence generated from a 4th degree Pascal's Triangle polynomial.

Original entry on oeis.org

1, 22, 103, 284, 605, 1106, 1827, 2808, 4089, 5710, 7711, 10132, 13013, 16394, 20315, 24816, 29937, 35718, 42199, 49420, 57421, 66242, 75923, 86504, 98025, 110526, 124047, 138628, 154309, 171130, 189131, 208352, 228833, 250614, 273735, 298236

Offset: 1

Gary W. Adamson, May 31 2004

The characteristic polynomial of M = x^4 - 4x^3 + 6x^2 - 4x + 1. (the recursive multipliers are seen in the polynomial with changed signs: (4), (-6), (4), (-1).

			a(13) = 13013 = 4*a(12) - 6*a(11) + 4*a(10) - a(9) = 4*10132 - 6*7711 + 4*5710 - 4089.
a(6) = 1106 since M^6 * [1 0 0 0] = [ 1 6 66 1106].
a(6) = 1106 = f(n) = (20/3)(6)^3 -10*(6^2) +(13/3)*6 = 1440 - 360 + 26.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A056939, A000384.

a:= n-> (20*n^2-30*n+13)*n/3:
seq(a(n), n=1..50);  # Alois P. Heinz, May 25 2013

a[n_] := (MatrixPower[{{1, 0, 0, 0}, {1, 1, 0, 0}, {1, 4, 1, 0}, {1, 10, 10, 1}}, n].{{1}, {0}, {0}, {0}})[[4, 1]]; Table[ a[n], {n, 36}] (* Robert G. Wilson v, Jun 05 2004 *)

a(n+4) = 4*a(n+3) - 6*a(n+2) + 4*a(n+1) - a(n), (multipliers which are present with changed signs in the characteristic polynomial, x^4 - 4x^3 + 6x^2 - 4x + 1. Given the 4 X 4 matrix derived from an A056939 triangle (fill in with zeros): M = [1 0 0 0 / 1 1 0 0 / 1 4 1 0 / 1 10 10 1], then M^n * [1 0 0 0] = [1 n A000384(n) a(n)] where A000384 is the hexagonal series 1, 6, 15, 28... 3. a(n) = (20/3)n^3 - 10n^2 + (13/3)n.

G.f.: x*(21*x^2+18*x+1) / (x-1)^4. - Colin Barker, May 25 2013

Edited and corrected by Robert G. Wilson v, Jun 05 2004

A155834 A triangle sequence of general recursive Sierpinski-Pascal minus general Narayana with adjusted n,m levels and zeros out:k=2; t(n,m)=Pascal(n,m,k-1)-Narayana(n-1,m-1,2*(k-1)).

Original entry on oeis.org

1, 1, 6, 16, 6, 22, 127, 127, 22, 64, 701, 1436, 701, 64, 163, 3117, 11503, 11503, 3117, 163, 382, 12088, 74122, 131494, 74122, 12088, 382, 848, 42890, 413612, 1193930, 1193930, 413612, 42890, 848, 1816, 143562, 2094588, 9280734, 14992440, 9280734

Offset: 4

Roger L. Bagula, Jan 28 2009

Row sums are;
2, 28, 298, 2966, 29566, 304678, 3302560, 38033840, 467861040, 6159690808,
86763791762,...
This level is the Eulerian number level:
only the odd Narayana levels correspond to the recursive Sierpinski-Pascal levels.

			{1, 1},
{6, 16, 6},
{22, 127, 127, 22},
{64, 701, 1436, 701, 64},
{163, 3117, 11503, 11503, 3117, 163},
{382, 12088, 74122, 131494, 74122, 12088, 382},
{848, 42890, 413612, 1193930, 1193930, 413612, 42890, 848},
{1816, 143562, 2094588, 9280734, 14992440, 9280734, 2094588, 143562, 1816},
{3797, 462541, 9928140, 64761204, 158774838, 158774838, 64761204, 9928140, 462541, 3797},
{7814, 1453700, 44960878, 418557816, 1489425900, 2250878592, 1489425900, 418557816, 44960878, 1453700, 7814},
{15914, 4495909, 197226603, 2558716162, 12781854516, 27839586777, 27839586777, 12781854516, 2558716162, 197226603, 4495909, 15914}

A001263, A056939.A056941, A142465, A142467

Clear[A, a0, b0, n, k, m, t, i];
A[n_, 1, m_] := 1; A[n_, n_, m_] := 1;
A[n_, k_, m_] := (m*n - m*k + 1)*A[n - 1, k - 1, m] + (m*k - (m - 1))*A[n - 1, k, m];
t[n_, m_, i_] = Product[Binomial[n + k, m + k]/Binomial[n - m + k, k], {k, 0, i}];
m = 2; a = Table[A[n, k, m - 1] - t[n - 1, k - 1, (2*m - 2)], {n, 4, 14}, { k, 2, n - 1}];
Flatten[a]

Pascal(n,m,k):
a(n,k,m)=(m*n - m*k + 1)*a(n - 1, k - 1, m) + (m*k - (m - 1))*a(n - 1, k, m);
Narayana(n,m,k):
y(n,m,k)=Product[Binomial[n + k, m + k]/Binomial[n - m + k, k], {k, 0, i}];
k=2;
t(n,m)=Pascal(n,m,k-1)-Narayana(n-1,m-1,2*(k-1)).

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A120257 Triangle of Hankel transforms of certain binomial sums.

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A142470 Triangle T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k), read by rows.

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A173882 Triangle T(n, k) = A090443(n-1)/(A090443(k-1)*A090443(n-k-1)) read by rows.

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A095265 A sequence generated from a 4th degree Pascal's Triangle polynomial.

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Maple

Mathematica

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A155834 A triangle sequence of general recursive Sierpinski-Pascal minus general Narayana with adjusted n,m levels and zeros out:k=2; t(n,m)=Pascal(n,m,k-1)-Narayana(n-1,m-1,2*(k-1)).

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A142470 Triangle T(n, k) = ( (k+2)/(2binomial(k+2, 2)^2) )binomial(n, k)^2binomial(n+1, k)binomial(n+2, k), read by rows.