A095666 -id:A095666 - cp's OEIS frontend

A096940 Pascal (1,5) triangle.

5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302

Offset: 0

Wolfdieter Lang, Jul 16 2004

This is the fifth member, q=5, 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), A095666 (q=4), 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)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  5;
  1,  5;
  1,  6,  5;
  1,  7, 11,   5;
  1,  8, 18,  16,   5;
  1,  9, 26,  34,  21,   5;
  1, 10, 35,  60,  55,  26,   5;
  1, 11, 45,  95, 115,  81,  31,   5;
  1, 12, 56, 140, 210, 196, 112,  36,   5;
  1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
  1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.

David A. Corneth, Table of n, a(n) for n = 0..9999
Wolfdieter Lang, First 10 rows.

Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.

a(n,k):=piecewise(n=0,5,0Mircea Merca, Apr 08 2012

a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ David A. Corneth, Aug 28 2019

Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; 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): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

A096956 Pascal (1,6) triangle.

Original entry on oeis.org

6, 1, 6, 1, 7, 6, 1, 8, 13, 6, 1, 9, 21, 19, 6, 1, 10, 30, 40, 25, 6, 1, 11, 40, 70, 65, 31, 6, 1, 12, 51, 110, 135, 96, 37, 6, 1, 13, 63, 161, 245, 231, 133, 43, 6, 1, 14, 76, 224, 406, 476, 364, 176, 49, 6, 1, 15, 90, 300, 630, 882, 840, 540, 225, 55, 6, 1, 16, 105, 390, 930

Offset: 0

Wolfdieter Lang, Aug 13 2004

Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.
This is the sixth member, q=6, 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), A095666 (q=4), A096940 (q=5).
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)=(6-5*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(6-5*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k) = A022097(n-2), n >= 2, with n=1 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  [0]  6;
  [1]  1,  6;
  [2]  1,  7,  6;
  [3]  1,  8, 13,  6;
  [4]  1,  9, 21, 19,  6;
  [5]  1, 10, 30, 40, 25,  6;
  ...

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
Wolfdieter Lang, First 10 rows.

Row sums: A005009(n-1), n>=1, 6 if n=0; g.f.: (6-5*x)/(1-2*x). Alternating row sums are [6, -5, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+6), A056115, A096957-9, A097297-A097300.

a(n,k):=piecewise(n=0,6,0Mircea Merca, Apr 08 2012

A096956[n_, k_] := If[n == k, 6, (5*k/n + 1)*Binomial[n, k]];
Table[A096956[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n,m)=0 if m > n, a(0,0) = 6; 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): (6-5*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1+5*k/n)*binomial(n,k), for n > 0. - Mircea Merca, Apr 08 2012

A060488 Number of 4-block ordered tricoverings of an unlabeled n-set.

Original entry on oeis.org

4, 13, 28, 50, 80, 119, 168, 228, 300, 385, 484, 598, 728, 875, 1040, 1224, 1428, 1653, 1900, 2170, 2464, 2783, 3128, 3500, 3900, 4329, 4788, 5278, 5800, 6355, 6944, 7568, 8228, 8925, 9660, 10434, 11248, 12103, 13000, 13940, 14924, 15953, 17028, 18150, 19320

Offset: 3

Vladeta Jovovic, Mar 20 2001

A covering of a set is a tricovering if every element of the set is covered by exactly three blocks of the covering.
If Y is a 4-subset of an n-set X then, for n>=6, a(n-3) is the number of 3-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007
Also the number of balls in a triangular pyramid of which all balls located on the edges have been removed such that the remaining pyramid's edges each consist of two adjacent balls. The layers of pyramids of this form start (from the top) 3, 7, 12, 18, 25, 33,... (A055998) with one smaller additional layer 1, 3, 6, 10, 15, 21,... (A000217) at the bottom. Thus, a(n) = A000217(n) + Sum_{k=1..n} A055998(k). Example: a(4) = (3+7+12+18)+10 = 50. - K. G. Stier, Dec 12 2012

Vincenzo Librandi, Table of n, a(n) for n = 3..1000
Amya Luo, Pattern Avoidance in Nonnesting Permutations, Undergraduate Thesis, Dartmouth College (2024). See p. 11.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Essentially the same as A026054. - Vladeta Jovovic, Jun 15 2006
Column k=4 of A060492.
Cf. A060091, A060491.
Fourth column (m=3) of (1, 4)-Pascal triangle A095666.

[(n-2)*(n-1)*(n+9)/6: n in [3..60]]; // Vincenzo Librandi, Jun 15 2011

Table[ 3 (n - 1) (n - 2)/2! + n (n - 1) (n - 2)/3!, {n, 3, 62}] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011 *)
Table[(n-2)(n-1)(n+9)/6,{n,3,50}] (* or *) LinearRecurrence[{4,-6,4,-1}, {4,13,28,50},50] (* Harvey P. Dale, Jul 21 2012 *)

a(n)=(n-2)*(n-1)*(n+9)/6 \\ Charles R Greathouse IV, Jun 14 2011

a(n) = binomial(n+3, 3) - 6*binomial(n+1, 1) + 8*binomial(n, 0) - 3*binomial(n-1, -1).
G.f.: -y^3*(-4+3*y)/(-1+y)^4.
E.g.f. for ordered k-block tricoverings of an unlabeled n-set is exp(-x+x^2/2+x^3/3*y/(1-y)) * sum(k>=0, 1/(1-y)^binomial(k, 3)*exp(-x^2/2*1/(1-y)^n)*x^k/k! ).
a(n) = (n+9)*binomial(n-1, 2)/3.
a(n) = (n-2)*(n-1)*(n+9)/6. - Zak Seidov, Jun 15 2006
a(3)=4, a(4)=13, a(5)=28, a(6)=50, a(n) = 4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 21 2012

A029609 Central numbers in the (2,3)-Pascal triangle A029600.

Original entry on oeis.org

1, 5, 15, 50, 175, 630, 2310, 8580, 32175, 121550, 461890, 1763580, 6760390, 26001500, 100291500, 387793800, 1502700975, 5834015550, 22687838250, 88363159500, 344616322050, 1345644686100, 5260247409300, 20583576819000, 80619009207750, 316026516094380, 1239796332370260

Offset: 0

Mohammad K. Azarian

For n > 0 also central terms of (1,4)-Pascal triangle A095666. - Reinhard Zumkeller, Apr 08 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A000108, A029600, A095666.

a029609 n = a029600 (2*n) n  -- Reinhard Zumkeller, Apr 08 2012

a[n_]=(5*Binomial[2*n,n]-3KroneckerDelta[n,0])/2; Array[a,27,0] (* Stefano Spezia, Feb 14 2025 *)

From Peter Bala, Aug 16 2011: (Start)
a(n) = (5/2)*binomial(2*n,n) for n >= 1.
O.g.f.: -3/2+5/2*1/sqrt(1-4*x) = 1+5*x+15*x^2+50*x^3+... = 1+5*x*d/dx(log(C(x))), where C(x) is the o.g.f. for the Catalan numbers A000108. (End)
E.g.f.: (5*exp(2*x)*BesselI(0, 2*x) - 3)/2. - Stefano Spezia, Feb 14 2025

More terms from James Sellers

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

A239310 Numbers of the form A001700(n)*k, n>=1, k>=2.

Original entry on oeis.org

6, 9, 12, 15, 18, 20, 21, 24, 27, 30, 33, 36, 39, 40, 42, 45, 48, 50, 51, 54, 57, 60, 63, 66, 69, 70, 72, 75, 78, 80, 81, 84, 87, 90, 93, 96, 99, 100, 102, 105, 108, 110, 111, 114, 117, 120, 123, 126, 129, 130, 132, 135, 138, 140, 141, 144, 147

Offset: 1

Bob Selcoe, Mar 31 2016

Numbers that are central coefficients T(2k,k) k>=2 in (a,b)-Pascal triangles, where (a,b) represent boundary conditions; i.e., T(2k,k) = (a+b)*A001700(k-1).

			a(n)=50 appears because A001700(2)=10, so T(6,3)=50 in (1,4)- and (2,3)-Pascal triangles.

Cf. A001700.

Cf. A007318 (Pascal's triangle), A029600 ((2,3)-Pascal triangle), A095666 ((1,4)-Pascal triangle).

is(n)=my(k=1,t=3); while(n>=2*t, if(n%t==0, return(1)); k++; t=binomial(2*k+1, k+1)); 0 \\ Charles R Greathouse IV, Apr 04 2016

a(n) ~ kn, where k = 2.441823902640895564.... (This constant exists since A001700 grows exponentially.) - Charles R Greathouse IV, Apr 04 2016

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A096940 Pascal (1,5) triangle.

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A096956 Pascal (1,6) triangle.

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A060488 Number of 4-block ordered tricoverings of an unlabeled n-set.

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A029609 Central numbers in the (2,3)-Pascal triangle A029600.

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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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A239310 Numbers of the form A001700(n)*k, n>=1, k>=2.

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