A095667 -id:A095667 - cp's OEIS frontend

A095666 Pascal (1,4) triangle.

4, 1, 4, 1, 5, 4, 1, 6, 9, 4, 1, 7, 15, 13, 4, 1, 8, 22, 28, 17, 4, 1, 9, 30, 50, 45, 21, 4, 1, 10, 39, 80, 95, 66, 25, 4, 1, 11, 49, 119, 175, 161, 91, 29, 4, 1, 12, 60, 168, 294, 336, 252, 120, 33, 4, 1, 13, 72, 228, 462, 630, 588, 372, 153, 37, 4, 1, 14, 85, 300, 690, 1092

Offset: 0

Wolfdieter Lang, Jun 11 2004

This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - Reinhard Zumkeller, Apr 08 2012

			Triangle begins:
  [4];
  [1,4];
  [1,5,4];
  [1,6,9,4];
  [1,7,15,13,4];
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A020714(n-1), n >= 1, 4 if n=0.
Alternating row sums are [4, -3, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+4), A055999(n+1), A060488(n+3), A095667-71, A095819.

a095666 n k = a095666_tabl !! n !! k
a095666_row n = a095666_tabl !! n
a095666_tabl = [4] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]
-- Reinhard Zumkeller, Apr 08 2012

a(n,k):=(1+3*k/n)*binomial(n,k) # Mircea Merca, Apr 08 2012

A095666[n_, k_] := If[n == k,  4, (3*k/n + 1)*Binomial[n, k]];
Table[A095666[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1 + 3*k/n)*binomial(n,k). - Mircea Merca, Apr 08 2012

A213550 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 5, 2, 15, 9, 3, 35, 25, 13, 4, 70, 55, 35, 17, 5, 126, 105, 75, 45, 21, 6, 210, 182, 140, 95, 55, 25, 7, 330, 294, 238, 175, 115, 65, 29, 8, 495, 450, 378, 294, 210, 135, 75, 33, 9, 715, 660, 570, 462, 350, 245, 155, 85, 37, 10, 1001, 935, 825, 690, 546

Offset: 1

Clark Kimberling, Jun 16 2012

Principal diagonal:  A002418
Antidiagonal sums:  A005585
row 1,  (1,3,6,...)**(1,2,3,...):  A000332
row 2,  (1,3,6,...)**(2,3,4,...):  A005582
row 3,  (1,3,6,...)**(3,4,5,...):  A095661
row 4,  (1,3,6,...)**(4,5,6,...):  A095667
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....5....15...35....70....126
2....9....25...55....105...182
3....13...35...75....140...238
4....17...45...95....175...294
5....21...55...115...210...350

Cf. A213500, A213548.

b[n_] := n (n + 1)/2; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213550 *)
d = Table[t[n, n], {n, 1, 40}] (* A002418 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A005585 *)

T(n,k) = 5*T(n,k-1) - 10*T(n,k-2) + 10*T(n,k-3) - 5*T(n,k-4) + T(n,k-5).

G.f. for row n: f(x)/g(x), where f(x) = n-(n-1)*x and g(x) = (1 - x)^5.

A374378 Iterated rascal triangle R2: T(n,k) = Sum_{m=0..2} binomial(n-k,m)*binomial(k,m).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 19, 15, 6, 1, 1, 7, 21, 31, 31, 21, 7, 1, 1, 8, 28, 46, 53, 46, 28, 8, 1, 1, 9, 36, 64, 81, 81, 64, 36, 9, 1, 1, 10, 45, 85, 115, 126, 115, 85, 45, 10, 1, 1, 11, 55, 109, 155, 181, 181, 155, 109, 55, 11, 1

Offset: 0

Kolosov Petro, Jul 06 2024

Triangle T(n,k) is the second triangle R2 among the rascal-family triangles; A374452 is triangle R3; A077028 is triangle R1.
Triangle T(n,k) equals Pascal's triangle A007318 through row 2i+1, i=2 (i.e., row 5).
Triangle T(n,k) equals Pascal's triangle A007318 through column i, i=2 (i.e., column 2).

			Triangle begins:
--------------------------------------------------
k=     0   1   2   3    4    5    6   7   8   9 10
--------------------------------------------------
n=0:   1
n=1:   1   1
n=2:   1   2   1
n=3:   1   3   3   1
n=4:   1   4   6   4    1
n=5:   1   5  10  10    5    1
n=6:   1   6  15  19   15    6    1
n=7:   1   7  21  31   31   21    7   1
n=8:   1   8  28  46   53   46   28   8   1
n=9:   1   9  36  64   81   81   64  36   9   1
n=10:  1  10  45  85  115  126  115  85  45  10  1

Kolosov Petro, Table of n, a(n) for n = 0..1325
Amelia Gibbs and Brian K. Miceli, Two Combinatorial Interpretations of Rascal Numbers, arXiv:2405.11045 [math.CO], 2024.
Jena Gregory, Brandt Kronholm, and Jacob White, Iterated rascal triangles, Aequationes mathematicae, 2023.
Jena Gregory, Iterated rascal triangles, Theses and Dissertations. 1050., The University of Texas Rio Grande Valley, 2022.
Philip K. Hotchkiss, Student Inquiry and the Rascal Triangle, arXiv:1907.07749 [math.HO], 2019.
Philip K. Hotchkiss, Generalized Rascal Triangles, Journal of Integer Sequences, Vol. 23, 2020.
Petro Kolosov, Identities in Iterated Rascal Triangles, 2024.

Cf. A077028, A007318, A000027, A000217, A005448, A056108, A212656, A000292, A095667, A095666, A006261.

t[n_, k_]:=Sum[Binomial[n - k, m]*Binomial[k, m], {m, 0, 2}]; Column[Table[t[n, k], {n, 0, 12}, {k, 0, n}], Center]

T(n,k) = 1 + k*(n-k) + (1/4)*(k-1)*k*(n-k-1)*(n-k).
Row sums give A006261(n).
Diagonal T(n+1, n) gives A000027(n).
Diagonal T(n+2, n) gives A000217(n).
Diagonal T(n+3, n) gives A005448(n).
Diagonal T(n+4, n) gives A056108(n).
Diagonal T(n+5, n) gives A212656(n).
Column k=3 difference binomial(n+6, 3) - T(n+6, 3) gives C(n+3,3)=A007318(n+3,3).
Column k=4 difference binomial(n+7, 4) - T(n+7, 4) gives fifth column of (1,4)-Pascal triangle A095667.
G.f.: (1 + 3*x^4*y^2 - (2*x + 3*x^3*y)*(1 + y) + x^2*(1 + 5*y + y^2))/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Jul 09 2024

A185779 Third accumulation array of Pascal's triangle (as a rectangle), by antidiagonals.

Original entry on oeis.org

1, 4, 4, 10, 17, 10, 20, 45, 45, 20, 35, 95, 126, 95, 35, 56, 175, 281, 281, 175, 56, 84, 294, 546, 662, 546, 294, 84, 120, 462, 966, 1358, 1358, 966, 462, 120, 165, 690, 1596, 2534, 2941, 2534, 1596, 690, 165, 220, 990, 2502, 4410, 5790, 5790, 4410, 2502, 990, 220, 286, 1375, 3762, 7272, 10620, 12021, 10620, 7272, 3762, 1375, 286, 364, 1859, 5467, 11484, 18432, 23229, 23229, 18432, 11484, 5467, 1859, 364, 455

Offset: 1

Clark Kimberling, Feb 03 2011

Using "Axxxxxx < Ayyyyyy" to mean that Ayyyyyy is the accumulation array of Axxxxxx, as defined at A144112:
A185779 < A144225 < A007318 < A014430 < A077023 < A185779, where each of these is formatted as a rectangle (e.g., A007318 is Pascal's triangle).  See A185778.
row 1: A000292
row 2: A095667

			Northwest corner:
1....4...10...20...35
4....17..45...95...175
10...45..126..281..546
20...95..281..662..1358

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

Cf. A007318, A014430, A077023, A144112, A144225, A185778.

f[n_, k_] := Binomial[n + k + 4, n + 2] - (k + 3)*(k + 4)/2 - (k + 2)* n*(k*n + n + 3*k + 7)/4; TableForm[Table[f[n, k], {n, 1, 5}, {k, 1, 5}]]
Table[f[n - k + 1, k], {n, 10}, {k, n, 1, -1}] // Flatten (* G. C. Greubel, Jul 12 2017 *)

T(n,k) = C(n+k+4,n+2) - (k+3)*(k+4)/2 - (k+2)*n*(k*n+n+3*k+7)/4, for k>=1, n>=1.

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A095666 Pascal (1,4) triangle.

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A213550 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

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A374378 Iterated rascal triangle R2: T(n,k) = Sum_{m=0..2} binomial(n-k,m)*binomial(k,m).

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A185779 Third accumulation array of Pascal's triangle (as a rectangle), by antidiagonals.

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