A095660 -id:A095660 - 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

A006503 a(n) = n(n+1)(n+8)/6.

Original entry on oeis.org

0, 3, 10, 22, 40, 65, 98, 140, 192, 255, 330, 418, 520, 637, 770, 920, 1088, 1275, 1482, 1710, 1960, 2233, 2530, 2852, 3200, 3575, 3978, 4410, 4872, 5365, 5890, 6448, 7040, 7667, 8330, 9030, 9768, 10545, 11362, 12220, 13120, 14063, 15050, 16082, 17160, 18285

Offset: 0

N. J. A. Sloane

If Y is a 3-subset of an n-set X then, for n>=4, a(n-4) is the number of 3-subsets of X having at most one element in common with Y. - Milan Janjic, Nov 23 2007

The coefficient of x^3 in (1-x-x^2)^{-n} is the coefficient of x^3 in (1+x+2x^2+3x^3)^n. Using the multinomial theorem one then finds that a(n)=n(n+1)(n+8)/3!. - Sergio Falcon, May 22 2008

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..5000
Margaret Bayer, Mark Denker, Marija Jelić Milutinović, Rowan Rowlands, Sheila Sundaram, and Lei Xue, Topology of Cut Complexes of Graphs, arXiv:2304.13675 [math.CO], 2023.
G. E. Bergum and V. E. Hoggatt, Jr., Numerator polynomial coefficient array for the convolved Fibonacci sequence, Fib. Quart., 14 (1976), 43-44. (Annotated scanned copy)
G. E. Bergum and V. E. Hoggatt, Jr., Numerator polynomial coefficient array for the convolved Fibonacci sequence, Fib. Quart., 14 (1976), 43-48.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
P. Moree, Convoluted convolved Fibonacci numbers, arXiv:math/0311205 [math.CO], 2003.
P. Moree, Convoluted Convolved Fibonacci Numbers, Journal of Integer Sequences, Vol. 7 (2004), Article 04.2.2.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

a(n) = A095660(n+2, 3): fourth column of (1, 3)-Pascal triangle.
Cf. A000027, A000096, A006504.
Cf. A000292, A002378.
Row n=3 of A144064.

A006503:=-(-3+2*z)/(z-1)**4; # [Simon Plouffe in his 1992 dissertation.]

Clear["Global`*"] a[n_] := n(n + 1)(n + 8)/3! Do[Print[n, " ", a[n]], {n, 1, 25}] (* Sergio Falcon, May 22 2008 *)
Table[n(n+1)(n+8)/6,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,3,10,22},50] (* Harvey P. Dale, Jan 27 2016 *)

x='x+O('x^50); concat([0], Vec(x*(3-2*x)/(1-x)^4)) \\ G. C. Greubel, May 11 2017

a(n) = n*(n+1)*(n+8)/6.
G.f.: x*(3-2*x)/(1-x)^4.
a(n) = A000292(n) + A002378(n). - Reinhard Zumkeller, Sep 24 2008
a(n) = 4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4) with a(0)=0, a(1)=3, a(2)=10, a(3)=22. - Harvey P. Dale, Jan 27 2016

Better description from Jeffrey Shallit, Aug 1995

A096940 Pascal (1,5) triangle.

Original entry on oeis.org

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

A000574 Coefficient of x^5 in expansion of (1 + x + x^2)^n.

Original entry on oeis.org

3, 16, 51, 126, 266, 504, 882, 1452, 2277, 3432, 5005, 7098, 9828, 13328, 17748, 23256, 30039, 38304, 48279, 60214, 74382, 91080, 110630, 133380, 159705, 190008, 224721, 264306, 309256, 360096, 417384, 481712, 553707, 634032, 723387, 822510

Offset: 3

N. J. A. Sloane

If Y is a 3-subset of an n-set X then, for n>=7, a(n-4) is the number of 5-subsets of X having at most one element in common with Y. - Milan Janjic, Nov 23 2007

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 3..1000
L. Carlitz et al., Permutations and sequences with repetitions by number of increases, J. Combin. Theory, 1 (1966), 350-374.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Trinomial Coefficient
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000389, A005581, A005712, A005714-A005716, A111808.
Column m=5 of (1, 3) Pascal triangle A095660.
Cf. A000217, A055998.

[3*Binomial(n+2,5)-2*Binomial(n+1,5): n in [3..50]]; // Vincenzo Librandi, Jun 10 2012

A000574:=-(-3+2*z)/(z-1)**6; # conjectured by Simon Plouffe in his 1992 dissertation
seq(3*binomial(n+2,5)-2*binomial(n+1,5),n=3..100); # Robert Israel, Aug 04 2015
A000574 := n -> GegenbauerC(`if`(5A000574(n)), n=3..20); # Peter Luschny, May 10 2016

CoefficientList[Series[(3-2*x)/(1-x)^6,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)

x='x+O('x^50); Vec(x^3*(3-2*x)/(1-x)^6) \\ G. C. Greubel, Nov 22 2017

G.f.: x^3*(3-2*x)/(1-x)^6.
a(n) = 3*binomial(n+2,5) - 2*binomial(n+1,5).
a(n) = A111808(n,5) for n>4. - Reinhard Zumkeller, Aug 17 2005
a(n) = binomial(n+1, 4)*(n+12)/5 = 3*b(n-3)-2*b(n-4), with b(n)=binomial(n+5, 5); cf. A000389.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Vincenzo Librandi, Jun 10 2012
a(n) = 3*binomial(n, 3) + 4*binomial(n, 4) + binomial(n, 5). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 5 if 5Peter Luschny, May 10 2016
a(n) = Sum_{i=1..n-1} A000217(i)*A055998(n-1-i). - Bruno Berselli, Mar 05 2018
E.g.f.: exp(x)*x^3*(60 + 20*x + x^2)/120. - Stefano Spezia, Jul 09 2023

More terms from Vladeta Jovovic, Oct 02 2000

A100320 A Catalan transform of (1 + 2x)/(1 - 2x).

Original entry on oeis.org

1, 4, 12, 40, 140, 504, 1848, 6864, 25740, 97240, 369512, 1410864, 5408312, 20801200, 80233200, 310235040, 1202160780, 4667212440, 18150270600, 70690527600, 275693057640, 1076515748880, 4208197927440, 16466861455200, 64495207366200, 252821212875504, 991837065896208

Offset: 0

Paul Barry, Nov 14 2004

A Catalan transform of (1 + 2*x)/(1 - 2*x) under the mapping g(x) -> g(x*c(x)). (Here c(x) is the g.f. of A000108.) The original sequence can be retrieved by g(x) -> g(x*(1-x)).
Hankel transform is A144704. - Paul Barry, Sep 19 2008
Central terms of the triangle in A124927. - Reinhard Zumkeller, Mar 04 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.

Cf. A000108, A010684, A028329, A039599, A095660, A124927, A132046, A144704.

a100320 n = a124927 (2 * n) n  -- Reinhard Zumkeller, Mar 04 2012

[4*Binomial(2*n-1, n) - 3*0^n: n in [0..40]]; // G. C. Greubel, Feb 01 2023

a[0]= 1; a[n_]:= 2 Binomial[2 n, n];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 31 2018 *)

def A100320(n): return 4*binomial(2*n-1, n) - 3*0^n
[A100320(n) for n in range(41)] # G. C. Greubel, Feb 01 2023

G.f.: (1 + 2*x*c(x))/(1 - 2*x*c(x)), where c(x) is the g.f. of A000108.
a(n) = 4*binomial(2*n-1, n) - 3*0^n.
a(n) = binomial(2*n, n)*(4*2^(n-1) - 0^n)/2^n.
a(n) = Sum_{j=0..n} Sum_{k=0..n} C(2*n, n-k)*((2*k + 1)/(n + k + 1))*C(k, j)*(-1)^(j-k)*(4*2^(j-1) - 0^j).
a(n) = A028329(n), n > 0. - R. J. Mathar, Sep 02 2008
a(n) = T(2*n,n), where T(n,k) = A132046(n,k). - Paul Barry, Sep 19 2008
a(n) = Sum_{k=0..n} A039599(n,k)*A010684(k). - Philippe Deléham, Oct 29 2008
a(n) = A095660(2*n,n) for n > 0. - Reinhard Zumkeller, Apr 08 2012
G.f.: G(0) - 1, where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) + (k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
a(n) = [x^n] (1 + 2*x)/(1 - x)^(n+1). - Ilya Gutkovskiy, Oct 12 2017
a(n) = 2*(2*n-1)*a(n-1)/n. - G. C. Greubel, Feb 01 2023
E.g.f.: 2*exp(2*x)*BesselI(0, 2*x) - 1. - Stefano Spezia, May 11 2024

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A139359 Number L([n],m) of ways the labeled parts of each integer partition of n can be distributed into m nonempty labeled boxes.

Original entry on oeis.org

1, 2, 2, 3, 6, 6, 5, 16, 36, 24, 7, 46, 150, 240, 120, 11, 114, 546, 1560, 1800, 720, 15, 614, 2058, 8400, 16800, 15120, 5040, 22, 1366, 6984, 40848, 126000, 191520, 141120, 40320, 30, 12516, 73488, 192816, 834120, 1905120, 2328480, 1451520, 362880

Offset: 1

Thomas Wieder, Apr 14 2008

nonn

This formula is related to a formula given by Riordan, see Riordan, 1958, page 94. Furthermore, this formula is related to the distribution of labeled elements into labeled boxes, as described by A019538.
The first column is equal to A000041 = number of partitions of n (the partition numbers).
The main diagonal is equal to the A000142 = Factorial numbers: n!
The second diagonal is equal to A001286 = Lah numbers: (n-1)*n!/2.
The third diagonal is equal to A019538 = Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).
If we normalize the m-th column by m! we get the triangle
  1
  2 1
  3 3 1
  5 8 6 1
  7 23 25 10 1
  11 57 91 65 15 1
  15 307 343 350 140 21 1
  22 683 1164 1702 1050 266 28 1
  30 6258 12248 8034 6951 2646 462 36 1
In this triangle we observe:
The second diagonal is equal to A000217 = Triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n.
The third diagonal is composed of numbers belonging to A095660 = Pascal (1,3) triangle.

			Triangle begins:
  1
  2 2
  3 6 6
  5 16 36 24
  7 46 150 240 120
  11 114 546 1560 1800 720
  15 614 2058 8400 16800 15120 5040
  22 1366 6984 40848 126000 191520 141120 40320
  30 12516 73488 192816 834120 1905120 2328480 1451520 362880
  ...

John Riordan: Introduction to Combinatorics, John Wiley & Sons, New York, 1958, ISBN 0-486-42536-3.

Thomas Wieder, Further comments on this sequence

Cf. A019538, A137383.

A187801 Pascal's triangle construction method applied to {1,1,2} as an initial term.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 2, 1, 3, 5, 5, 2, 1, 4, 8, 10, 7, 2, 1, 5, 12, 18, 17, 9, 2, 1, 6, 17, 30, 35, 26, 11, 2, 1, 7, 23, 47, 65, 61, 37, 13, 2, 1, 8, 30, 70, 112, 126, 98, 50, 15, 2, 1, 9, 38, 100, 182, 238, 224, 148, 65, 17, 2, 1, 10, 47, 138, 282, 420, 462, 372

Offset: 2

Jakub Jaroslaw Ciaston, Jan 06 2013

A neighborhood decomposition of triangle graph applied to each node gives three identical sequences (independent of start point) {1,3}.
For star graph (depend of start point) generated sequences are: one time {1,3} and three times {1,1,2}.
Triangle of expansion of (1+x+2*x^2)*(1+x)^n. - Philippe Deléham, Mar 10 2013

			Triangle begins:
1,1,2;
1,2,3,2;
1,3,5,5,2;
1,4,8,10,7,2;
1,5,12,18,17,9,2;
1,6,17,30,35,26,11,2;
1,7,23,47,65,61,37,13,2;
1,8,30,70,112,126,98,50,15,2;
1,9,38,100,182,238,224,148,65,17,2;
1,10,47,138,282,420,462,372,213,82,19,2;
1,11,57,185,420,702,882,834,585,295,101,21,2;
1,12,68,242,605,1122,1584,1716,1419,880,396,122,23,2;
1,13,80,310,847,1727,2706,3300,3135,2299,1276,518,145,25,2;
From _Philippe Deléham_, Mar 10 2013: (Start)
Row 2: 1+x+2*x^2
Row 3: (1+x+2*x^2)*(1+x) = 1+2*x+3*x^2+2*x^3
Row 4: (1+x+2*x^2)*(1+x)^2 = 1+3*x+5*x^2+5*x^3+2*x^4
Row 5: (1+x+2*x^2)*(1+x)^3 = 1+4*x+8*x^2+10*x^3+7*x^4+2*x^5
(End)

T. D. Noe, Rows n = 2..50 of triangle, flattened
Eric W. Weisstein, MathWorld: Triangle Graph
Eric W. Weisstein, MathWorld: Star Graph

Cf. A187801, A095660, A107232, A029635, A095660.

c = {1, 1, 2}; Join[{c}, t = Table[c = Append[c, 0]; c = c + RotateRight[c], {9}]]; Flatten[t] (* T. D. Noe, Mar 11 2013 *)

For the selection of the initial term: neighborhood decomposition of graph.
For sequence: Pascal's triangle construction method applied to selected initial term.
Row sums: A000079(n+2) = (4, 8, 16, 32, 64, ...). - Philippe Deléham, Mar 10 2013

cp's OEIS Frontend

A095666 Pascal (1,4) triangle.

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A006503 a(n) = n*(n+1)*(n+8)/6.

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PARI

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A096940 Pascal (1,5) triangle.

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A096956 Pascal (1,6) triangle.

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A000574 Coefficient of x^5 in expansion of (1 + x + x^2)^n.

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A100320 A Catalan transform of (1 + 2*x)/(1 - 2*x).

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A006503 a(n) = n(n+1)(n+8)/6.

A100320 A Catalan transform of (1 + 2x)/(1 - 2x).