A093560 -id:A093560 - cp's OEIS frontend

A051836 a(n) = n(n+1)(n+2)(n+3)(3*n+2)/120.

0, 1, 8, 33, 98, 238, 504, 966, 1716, 2871, 4576, 7007, 10374, 14924, 20944, 28764, 38760, 51357, 67032, 86317, 109802, 138138, 172040, 212290, 259740, 315315, 380016, 454923, 541198, 640088, 752928, 881144, 1026256, 1189881, 1373736, 1579641, 1809522, 2065414

Offset: 0

Barry E. Williams, Dec 12 1999

5-dimensional version of pentagonal-based pyramidal numbers. - Ben Creech (mathroxmysox(AT)yahoo.com)
If Y is a 3-subset of an n-set X then, for n>=7, a(n-6) is the number of 7-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
Antidiagonal sums of the convolution array A213548. - Clark Kimberling, Jun 17 2012
After 0, convolution of nonzero triangular numbers (A000217) and nonzero pentagonal numbers (A000326). - Bruno Berselli, Jun 27 2013
a(n) is also the number of odd chordless cycles in the graph complement of the (n+1)-Andrásfai graph. - Eric W. Weisstein, Apr 14 2017

			By the fourth comment: A000217(1..6) and A000326(1..6) give the term a(6) = 1*21+5*15+12*10+22*6+35*3+51*1 = 504. - _Bruno Berselli_, Jun 27 2013

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Andrásfai Graph.
Eric Weisstein's World of Mathematics, Chordless Cycle.
Eric Weisstein's World of Mathematics, Graph Complement.
Index to sequences related to pyramidal numbers.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Partial sums of A001296.
Cf. A093560 ((3, 1) Pascal, column m=5).
Cf. A000217, A000326, A213548.

[0] cat [Binomial(n+4, n)*(3*n+5)/5: n in [0..40]]; // Vincenzo Librandi, Jul 04 2017

with (combinat):a[0]:=0:for n from 1 to 50 do a[n]:=stirling2(n+2,n)+a[n-1] od: seq(a[n], n=0..34); # Zerinvary Lajos, Mar 17 2008

Table[n(n + 1)(n + 2)(n + 3)(3n + 2)/120, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Apr 08 2011 *)
CoefficientList[Series[x (1 + 2 x) / (1 - x)^6, {x, 0, 33}], x] (* Vincenzo Librandi, Jul 04 2017 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,8,33,98,238},40] (* Harvey P. Dale, Jun 01 2018 *)

a(n)=n*(n+1)*(n+2)*(n+3)*(3*n+2)/120 \\ Charles R Greathouse IV, Oct 07 2015

[((3*n+2)/(n+4))*binomial(n+4,5) for n in range(41)] # G. C. Greubel, Dec 27 2023

a(n) = C(n+4, n)*(3n+5)/5.
G.f.: x*(1+2*x)/(1-x)^6. (adapted by Vincenzo Librandi, Jul 04 2017)
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 135*sqrt(3)*Pi/14 - 1215*log(3)/14 + 925/21.
Sum_{n>=1} (-1)^(n+1)/a(n) = 135*sqrt(3)*Pi/7 - 880*log(2)/7 - 355/21. (End)
E.g.f.: (1/5!)*x*(120 + 360*x + 240*x^2 + 50*x^3 + 3*x^4)*exp(x). - G. C. Greubel, Dec 27 2023

Simpler definition from Ben Creech (mathroxmysox(AT)yahoo.com), Nov 13 2005

A093562 (5,1) Pascal triangle.

Original entry on oeis.org

1, 5, 1, 5, 6, 1, 5, 11, 7, 1, 5, 16, 18, 8, 1, 5, 21, 34, 26, 9, 1, 5, 26, 55, 60, 35, 10, 1, 5, 31, 81, 115, 95, 45, 11, 1, 5, 36, 112, 196, 210, 140, 56, 12, 1, 5, 41, 148, 308, 406, 350, 196, 68, 13, 1, 5, 46, 189, 456, 714, 756, 546, 264, 81, 14, 1, 5, 51, 235, 645, 1170

Offset: 0

Wolfdieter Lang, Apr 22 2004

This is the fifth member, d=5, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-1, for d=1..4.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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)=(1+4*z)/(1-(1+x)*z).
The SW-NE diagonals give A022095(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 4. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
The array F(5;n,m) gives in the columns m >= 1 the figurate numbers based on A016861, including the heptagonal numbers A000566 (see the W. Lang link).
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

			Triangle begins
  [1];
  [5,  1];
  [5,  6,  1];
  [5, 11,  7,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A007283(n-1), n>=1, 1 for n=0. A082505(n+1), alternating row sums are 1 for n=0, 4 for n=2 and 0 else.
Column sequences give for m=1..9: A016861, A000566 (heptagonal), A002413, A002418, A027800, A051946, A050484, A052255, A055844.
A007318, A093563 (d=6), A228196, A228576.

a093562 n k = a093562_tabl !! n !! k
a093562_row n = a093562_tabl !! n
a093562_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [5, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093562(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<2))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(5;n-m, m) for 0<= m <= n, otherwise 0, with F(5;0, 0)=1, F(5;n, 0)=5 if n>=1 and F(5;n, m):=(5*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+4*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=5 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 4*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(5 + 11*x + 7*x^2/2! + x^3/3!) = 5 + 16*x + 34*x^2/2! + 60*x^3/3! + 95*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A093565 (8,1) Pascal triangle.

Original entry on oeis.org

1, 8, 1, 8, 9, 1, 8, 17, 10, 1, 8, 25, 27, 11, 1, 8, 33, 52, 38, 12, 1, 8, 41, 85, 90, 50, 13, 1, 8, 49, 126, 175, 140, 63, 14, 1, 8, 57, 175, 301, 315, 203, 77, 15, 1, 8, 65, 232, 476, 616, 518, 280, 92, 16, 1, 8, 73, 297, 708, 1092, 1134, 798, 372, 108, 17, 1, 8, 81, 370, 1005

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(8;n,m) gives in the columns m>=1 the figurate numbers based on A017077, including the decagonal numbers A001107,(see the W. Lang link).
This is the eighth member, d=8, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-4, for d=1..7.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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)=(1+7*z)/(1-(1+x)*z).
The SW-NE diagonals give A022098(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 7. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
  [1];
  [8,  1];
  [8,  9,  1];
  [8, 17, 10,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
W. Lang, First 10 rows and array of figurate numbers .

Row sums: A005010(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 7 for n=2 and 0 else.
The column sequences give for m=1..9: A017077, A001107 (decagonal), A007585, A051797, A051878, A050404, A052226, A056001, A056122.
Cf. A093644 (d=9).

a093565 n k = a093565_tabl !! n !! k
a093565_row n = a093565_tabl !! n
a093565_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [8, 1]
-- Reinhard Zumkeller, Aug 31 2014

a(n, m)=F(8;n-m, m) for 0<= m <= n, otherwise 0, with F(8;0, 0)=1, F(8;n, 0)=8 if n>=1 and F(8;n, m):=(8*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=8 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+7*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 7*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(8 + 17*x + 10*x^2/2! + x^3/3!) = 8 + 25*x + 52*x^2/2! + 90*x^3/3! + 140*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A093564 (7,1) Pascal triangle.

Original entry on oeis.org

1, 7, 1, 7, 8, 1, 7, 15, 9, 1, 7, 22, 24, 10, 1, 7, 29, 46, 34, 11, 1, 7, 36, 75, 80, 45, 12, 1, 7, 43, 111, 155, 125, 57, 13, 1, 7, 50, 154, 266, 280, 182, 70, 14, 1, 7, 57, 204, 420, 546, 462, 252, 84, 15, 1, 7, 64, 261, 624, 966, 1008, 714, 336, 99, 16, 1, 7, 71, 325, 885

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(7;n,m) gives in the columns m>=1 the figurate numbers based on A016993, including the 9-gonal numbers A001106, (see the W. Lang link).
This is the seventh member, d=7, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-3, for d=1..6.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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)=(1+6*z)/(1-(1+x)*z).
The SW-NE diagonals give A022097(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
  [1];
  [7,  1];
  [7,  8,  1];
  [7, 15,  9,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch. 5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
W. Lang, First 10 rows and array of figurate numbers .

Row sums: A000079(n+2), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 6 for n=2 and 0 otherwise.
The column sequences give for m=1..9: A016993, A001106 (9-gonal), A007584, A051740, A051877, A050403, A027818, A034266, A055994.
Cf. A093565 (d=8).

a093564 n k = a093564_tabl !! n !! k
a093564_row n = a093564_tabl !! n
a093564_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [7, 1]
-- Reinhard Zumkeller, Sep 01 2014

N:= 20: # to get the first N rows
T:=Matrix(N,N):
T[1,1]:= 1:
for m from 2 to N do
T[m,1]:= 7:
T[m,2..m]:= T[m-1,1..m-1] + T[m-1,2..m];
od:
for m from 1 to N do
convert(T[m,1..m],list)
od; # Robert Israel, Dec 28 2014

a(n, m)=F(7;n-m, m) for 0<= m <= n, otherwise 0, with F(7;0, 0)=1, F(7;n, 0)=7 if n>=1 and F(7;n, m):=(7*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=7 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+6*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 6*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 15*x + 9*x^2/2! + x^3/3!) = 7 + 22*x + 46*x^2/2! + 80*x^3/3! + 125*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A051923 Partial sums of A051836.

Original entry on oeis.org

1, 9, 42, 140, 378, 882, 1848, 3564, 6435, 11011, 18018, 28392, 43316, 64260, 93024, 131784, 183141, 250173, 336490, 446292, 584430, 756470, 968760, 1228500, 1543815, 1923831, 2378754, 2919952, 3560040, 4312968, 5194112, 6220368, 7410249, 8783985, 10363626

Offset: 0

Barry E. Williams, Dec 19 1999

If Y is a 3-subset of an n-set X then, for n >= 8, a(n-8) is the number of 8-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007

a(n) is the n-th antidiagonal sum of the convolution array A213551. - Clark Kimberling, Jun 17 2012

			From the third formula: a(4) = 15+60+108+120+75 = 378. - _Bruno Berselli_, Sep 04 2013

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000217, A027801, A051836.

Cf. A093560 ((3, 1) Pascal, column m=6).

[Binomial(n+5, 5)*(n+2)/2: n in [0..40]]; // Vincenzo Librandi, Dec 27 2018

CoefficientList[Series[(1 + 2 x)/(1 - x)^7, {x, 0, 25}], x]  (* Harvey P. Dale, Mar 13 2011 *)
Nest[Accumulate,Range[1,120,3],5] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
Table[Binomial[n + 5, 5] (n + 2) / 2, {n, 0, 35}] (* Vincenzo Librandi, Dec 27 2018 *)

a(n) = binomial(n+5, 5)*(n+2)/2.
G.f.: (1+2*x)/(1-x)^7.
a(n) = Sum_{k=1..n+1} k*A000217(k)*A000217(n-k+2). - Bruno Berselli, Sep 04 2013
From Amiram Eldar, Jan 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 1205/18 - 20*Pi^2/3.
Sum_{n>=0} (-1)^n/a(n) = 10*Pi^2/3 - 320*log(2)/3 + 755/18. (End)

A106516 A Pascal-like triangle based on 3^n.

Original entry on oeis.org

1, 3, 1, 9, 4, 1, 27, 13, 5, 1, 81, 40, 18, 6, 1, 243, 121, 58, 24, 7, 1, 729, 364, 179, 82, 31, 8, 1, 2187, 1093, 543, 261, 113, 39, 9, 1, 6561, 3280, 1636, 804, 374, 152, 48, 10, 1, 19683, 9841, 4916, 2440, 1178, 526, 200, 58, 11, 1, 59049, 29524, 14757, 7356, 3618, 1704, 726, 258, 69, 12, 1

Offset: 0

Paul Barry, May 05 2005

Row sums are A027649. Antidiagonal sums are A106517.
From Wolfdieter Lang, Jan 09 2015: (Start)
Alternating row sums give A025192. The A-sequence of this Riordan lower triangular matrix is [1, 1, repeat(0, )] (leading to the Pascal recurrence for T(n,k) for n >= k >= 1. The Z-sequence is [3, repeat(0, )] (leading to the recurrence T(n,0) = 3*T(n-1,0), n >= 1. For A- and Z-sequences see the W. Lang link under A006232.
The inverse of this Riordan matrix is Tinv = ((1 - 2*x)/(1 + x), x/(1 + x)) given as a signed version of A093560: Tinv(n,m) = (-1)^(n-m)*A093560(n,m). (End)

			The triangle T(n,k) begins:
n\k     0     1     2    3    4    5   6   7  8  9 10 ...
0:      1
1:      3     1
2:      9     4     1
3:     27    13     5    1
4:     81    40    18    6    1
5:    243   121    58   24    7    1
6:    729   364   179   82   31    8   1
7:   2187  1093   543  261  113   39   9   1
8:   6561  3280  1636  804  374  152  48  10  1
9:  19683  9841  4916 2440 1178  526 200  58 11  1
10: 59049 29524 14757 7356 3618 1704 726 258 69 12  1
... reformatted and extended. - _Wolfdieter Lang_, Jan 06 2015
----------------------------------------------------------
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/ 1        \/1           \/1        \       /1         \
| 3  1     ||0  1        ||0 1      |      | 3  1      |
| 9  4 1   ||0  3  1     ||0 0 1    |... = | 9  7  1   |
|27 13 5 1 ||0  9  4 1   ||0 0 3 1  |      |27 37 12 1 |
|...       ||0 27 13 5 1 ||0 0 9 4 1|      |...        |
|...       ||...         ||...      |      |...        |
= A143495. - _Peter Bala_, Dec 23 2014

Columns 1, 2, 3, 4, 5: A003462, A000340, A052150, A097786, A097787.

Cf. A055248, A143495.

a106516[n_] := Block[{a, k},
a[x_] := Flatten@ Last@ Reap[For[k = -1, k < x, Sow[Binomial[x, k] +
2 Sum[3^(i - 1)*Binomial[x - i, k], {i, 1, x}]], k++]]; Flatten@Array[a, n, 0]]; a106516[11] (* Michael De Vlieger, Dec 23 2014 *)

Riordan array (1/(1-3x), x/(1-x)); Number triangle T(n, 0)=A000244(n), T(n, k)=T(n-1, k-1)+T(n-1, k); T(n, k)=sum{j=0..n, binomial(n, k+j)2^j}.
From Peter Bala, Jul 16 2013: (Start)
T(n,k) = binomial(n,k) + 2*sum {i = 1..n} 3^(i-1)*binomial(n-i,k).
O.g.f.: (1 - t)/( (1 - 3*t)*(1 - (1 + x)*t) ) = 1 + (3 + x)*t + (9 + 4*x + x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(x - 2)*( x*(x + 1)^n - 2*3^n ). (End)
Closed-form formula for arbitrary left and right borders of Pascal-like triangle see A228196. - Boris Putievskiy, Aug 19 2013
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 3*T(n-2,k-1), T(0,0)=1, T(1,0)=3, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 26 2013
From Peter Bala, Dec 23 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(27 + 13*x + 5*x^2/2! + x^3/3!) = 27 + 40*x + 58*x^2/2! + 82*x^3/3! + 113*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143495 (but with a different offset). See the Example section. Cf. A055248. (End)
n-th row polynomial R(n, x) = (2*3^n - x*(1 + x)^n)/(2 - x). - Peter Bala, Mar 05 2025

A053367 Partial sums of A050494.

Original entry on oeis.org

1, 11, 63, 255, 825, 2277, 5577, 12441, 25740, 50050, 92378, 163098, 277134, 455430, 726750, 1129854, 1716099, 2552517, 3725425, 5344625, 7548255, 10508355, 14437215, 19594575, 26295750, 34920756, 45924516, 59848228, 77331980, 99128700, 126119532, 159330732, 199952181, 249357615

Offset: 0

Barry E. Williams, Jan 06 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A050494.

Cf. A093560 ((3, 1) Pascal, column m=8).

[(3*n+8)*Binomial(n+7,7)/8: n in [0..30]]; // G. C. Greubel, May 25 2018

LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {1, 11, 63, 255, 825, 2277, 5577, 12441, 25740}, 30] (* or *) Table[(3*n+8)* Binomial[n+7,7]/8, {n,0,30}] (* G. C. Greubel, May 25 2018 *)

a(n)=binomial(n+7, 7)*(3*n+8)/8 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = binomial(n+7, 7)*(3n+8)/8.

G.f.: (1+2*x)/(1-x)^9.

Terms a(24) onward added by G. C. Greubel, May 25 2018

A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4x + 3)(x + 1)^(2n) + (x^2 - 1)(x^2 + 3*x + 3).

Original entry on oeis.org

0, 1, 3, 3, 1, 0, 7, 14, 9, 2, 0, 13, 37, 43, 26, 8, 1, 0, 19, 72, 129, 141, 98, 42, 10, 1, 0, 25, 119, 291, 463, 504, 378, 192, 63, 12, 1, 0, 31, 178, 553, 1156, 1716, 1848, 1452, 825, 330, 88, 14, 1, 0, 37, 249, 939, 2432, 4576, 6435, 6864, 5577, 3432, 1573

Offset: 0

Franck Maminirina Ramaharo, Jul 18 2018

The triangle is related to the Kauffman bracket polynomial evaluated at the shadow diagram of the two-bridge knot with Conway's notation C(2n,3).

			The triangle T(n,k) begins:
n\k| 0   1    2    3     4     5     9     7     8     9    10   11   12  13 14
-------------------------------------------------------------------------------
0  | 0   1    3    3     1
1  | 0   7   14    9     2
2  | 0  13   37   43    26     8     1
3  | 0  19   72  129   141    98    42    10     1
4  | 0  25  119  291   463   504   378   192    63    12     1
5  | 0  31  178  553  1156  1716  1848  1452   825   330    88   14    1
6  | 0  37  249  939  2432  4576  6435  6864  5577  3432  1573  520  117  16  1
...

Ji-Young Ham and Joongul Lee, An explicit formula for the A-polynomial of the knot with Conway’s notation C(2n,3), Journal of Knot Theory and Its Ramifications 25 (2016), 1-9.
Ryo Hanaki, On scannable properties of the original knot from a knot shadow, Topology and its Applications 194 (2015), 296-305.
Bin Lu and Jianyuan K. Zhong, The Kauffman Polynomials of 2-bridge Knots, arXiv:math/0606114 [math.GT], 2006.

Row sums: 4*A004171.

Cf. A093560, A137396, A299989, A300184, A300192, A300453, A300454, A316659.

T := proc (n, k) if k = 1 then 6*n + 1 else binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) end if end proc:
for n from 0 to 12 do seq(T(n, k), k = 0 .. max(4, 2*(n + 1))) od;

row[n_] := CoefficientList[(x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3), x];
Array[row, 12, 0] // Flatten

T(n, k) := binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) - kron_delta(1, k)$
for n:0 thru 12 do print(makelist(T(n, k), k, 0, max(4, 2*(n + 1))));

T(n,1) = A016921(n) and T(n,k) = C(2*n+3,k+1) + (C(2*n+1,k)*(2*k - 2*n) + C(4,k)*(2*k - 3))/(k + 1) for k > 1.
T(n,2) = A173247(2*n+1) = A300401(2*n,3).
T(n,3) = 2*A099721(n) + 3.
T(n,4) = A244730(n) - A002412(n) + 1.
T(n,k) = A093560(2*n,k) for n > 2 and k > 4.
G.f.: (x^2 + 4*x + 3)/(1 - y*(x + 1)^2) + (x^4 + 3*x^3 + 2*x^2 - 3*x - 3)/(1 - y).

A050494 Partial sums of A051923.

Original entry on oeis.org

1, 10, 52, 192, 570, 1452, 3300, 6864, 13299, 24310, 42328, 70720, 114036, 178296, 271320, 403104, 586245, 836418, 1172908, 1619200, 2203630, 2960100, 3928860, 5157360, 6701175, 8625006, 11003760, 13923712, 17483752, 21796720, 26990832, 33211200, 40621449

Offset: 0

Barry E. Williams, Dec 26 1999

If Y is a 3-subset of an n-set X then, for n>=9, a(n-9) is the number of 9-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1)

Cf. A051923.

Cf. A093560 ((3, 1) Pascal, column m=7).

Table[Binomial[n+6, 6]*(3*n+7)/7, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2012 *)

a(n)=C(n+6, 6)*(3n+7)/7.

G.f.: (1+2*x)/(1-x)^8.

A053310 a(n) = (n+3)*binomial(n+8, 8)/3.

Original entry on oeis.org

1, 12, 75, 330, 1155, 3432, 9009, 21450, 47190, 97240, 189618, 352716, 629850, 1085280, 1812030, 2941884, 4657983, 7210500, 10935925, 16280550, 23828805, 34337160, 48774375, 68368950, 94664700, 129585456, 175509972, 235358200

Offset: 0

Barry E. Williams, Mar 06 2000

If Y is a 3-subset of an n-set X then, for n>=11, a(n-11) is the number of 11-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 189, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Partial sums of A053367.
Cf. A053367, A053347, A000581.
Cf. A093560 ((3, 1) Pascal, column m=9).

[(n+3)*Binomial(n+8, 8)/3: n in [0..30]]; // G. C. Greubel, May 24 2018

CoefficientList[Series[(1+2*x)/(1-x)^10, {x, 0, 50}], x] (* G. C. Greubel, May 24 2018 *)
Table[(n+3) Binomial[n+8,8]/3,{n,0,30}] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,12,75,330,1155,3432,9009,21450,47190,97240},30] (* Harvey P. Dale, Feb 25 2021 *)

for(n=0, 30, print1((n+3)*binomial(n+8, 8)/3, ", ")) \\ G. C. Greubel, May 24 2018

G.f.: (1+2*x)/(1-x)^10.

a(n) = binomial(n+8,n+2)*binomial(n+3,n)/28. - Zerinvary Lajos, May 12 2006

cp's OEIS Frontend

A051836 a(n) = n*(n+1)*(n+2)*(n+3)*(3*n+2)/120.

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A093562 (5,1) Pascal triangle.

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A093565 (8,1) Pascal triangle.

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A093564 (7,1) Pascal triangle.

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A051923 Partial sums of A051836.

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A106516 A Pascal-like triangle based on 3^n.

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A053367 Partial sums of A050494.

A051836 a(n) = n(n+1)(n+2)(n+3)(3*n+2)/120.

A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4x + 3)(x + 1)^(2n) + (x^2 - 1)(x^2 + 3*x + 3).