A051601 -id:A051601 - cp's OEIS frontend

A055796 T(2n+3,n), array T as in A055794.

1, 5, 16, 42, 98, 210, 420, 792, 1419, 2431, 4004, 6370, 9828, 14756, 21624, 31008, 43605, 60249, 81928, 109802, 145222, 189750, 245180, 313560, 397215, 498771, 621180, 767746, 942152, 1148488, 1391280, 1675520, 2006697, 2390829, 2834496, 3344874, 3929770

Offset: 0

Clark Kimberling, May 28 2000

If Y is a 2-subset of an n-set X then, for n>=6, a(n-6) is the number of 6-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A051601.

[(n+1)*(n+2)*(n+3)*(n+4)*(n^2-n+30)/720: n in [0..40]]; // Vincenzo Librandi, Apr 30 2012

seq(binomial(n+4, 6)+binomial(n+4, 4), n=0..33) # Zerinvary Lajos, Jul 24 2006

a=1;b=2;c=3;d=4;s=5;lst={1,s};Do[a+=n;b+=a;c+=b;d+=c;s+=d;AppendTo[lst,s],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, May 24 2009 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{1,5,16,42,98, 210,420},50] (* Vincenzo Librandi, Apr 30 2012 *)
Table[(n+1)(n+2)(n+3)(n+4)(n^2-n+30)/720,{n,0,50}] (* Harvey P. Dale, Feb 12 2013 *)

a(n) = (n+1)(n+2)(n+3)(n+4)(n^2-n+30)/720.
a(n-4) = binomial(n,6) + binomial(n,4) for n>3. - Zerinvary Lajos, Jul 24 2006
G.f.: (1-2*x+2*x^2)/(1-x)^7. - Colin Barker, Feb 22 2012

More terms from Harvey P. Dale, Feb 12 2013

A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

Original entry on oeis.org

1, 3, 3, 9, 6, 9, 27, 15, 15, 27, 81, 42, 30, 42, 81, 243, 123, 72, 72, 123, 243, 729, 366, 195, 144, 195, 366, 729, 2187, 1095, 561, 339, 339, 561, 1095, 2187, 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561, 19683, 9843, 4938, 2556, 1578, 1578, 2556, 4938

Offset: 0

T. D. Noe, Aug 01 2013

All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?

Let b=3 and T(n,k) = A(n-k,k) be the associated reading of the symmetric array A by antidiagonals, then A(n,k) = sum_{r=1..n} b^r*A178300(n-r,k) + sum_{c=1..k} b^c*A178300(k-c,n). Similarly with b=4 and b=5 for A227074 and A227076. - R. J. Mathar, Aug 10 2013

			Triangle:
1,
3, 3,
9, 6, 9,
27, 15, 15, 27,
81, 42, 30, 42, 81,
243, 123, 72, 72, 123, 243,
729, 366, 195, 144, 195, 366, 729,
2187, 1095, 561, 339, 339, 561, 1095, 2187,
6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 3^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227550 A triangle formed like Pascal's triangle, but with factorial(n) on the borders instead of 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 6, 4, 4, 6, 24, 10, 8, 10, 24, 120, 34, 18, 18, 34, 120, 720, 154, 52, 36, 52, 154, 720, 5040, 874, 206, 88, 88, 206, 874, 5040, 40320, 5914, 1080, 294, 176, 294, 1080, 5914, 40320, 362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880, 3628800

Offset: 0

Vincenzo Librandi, Aug 04 2013

A003422 gives the second column (after 0).

			Triangle begins:
       1;
       1,     1;
       2,     2,    2;
       6,     4,    4,    6;
      24,    10,    8,   10,  24;
     120,    34,   18,   18,  34, 120;
     720,   154,   52,   36,  52, 154,  720;
    5040,   874,  206,   88,  88, 206,  874, 5040;
   40320,  5914, 1080,  294, 176, 294, 1080, 5914, 40320;
  362880, 46234, 6994, 1374, 470, 470, 1374, 6994, 46234, 362880;

Vincenzo Librandi, Rows n = 0..70, flattened

Cf. similar triangles with t on the borders: A007318 (t = 1), A028326 (t = 2), A051599 (t = prime(n)), A051601 (t = n), A051666 (t = n^2), A108617 (t = fibonacci(n)), A134636 (t = 2n+1), A137688 (t = 2^n), A227075 (t = 3^n).
Cf. A003422.
Cf. A227791 (central terms), A001563, A074911.

a227550 n k = a227550_tabl !! n !! k
a227550_row n = a227550_tabl !! n
a227550_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([1], a001563_list)
-- Reinhard Zumkeller, Aug 05 2013

function T(n,k)
  if k eq 0 or k eq n then return Factorial(n);
  else return T(n-1,k-1) + T(n-1,k);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 02 2021

t = {}; Do[r = {}; Do[If[k == 0||k == n, m = n!, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

def T(n,k): return factorial(n) if (k==0 or k==n) else T(n-1, k-1) + T(n-1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

From G. C. Greubel, May 02 2021: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = T(n, n) = n!.
Sum_{k=0..n} T(n, k) = 2^n * (1 +Sum_{j=1..n-1} j*j!/2^j) = A140710(n). (End)

A228053 A triangle formed like Pascal's triangle, but with (-1)^(n+1) on the borders instead of 1.

Original entry on oeis.org

-1, 1, 1, -1, 2, -1, 1, 1, 1, 1, -1, 2, 2, 2, -1, 1, 1, 4, 4, 1, 1, -1, 2, 5, 8, 5, 2, -1, 1, 1, 7, 13, 13, 7, 1, 1, -1, 2, 8, 20, 26, 20, 8, 2, -1, 1, 1, 10, 28, 46, 46, 28, 10, 1, 1, -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1, 1, 1, 13, 49, 112, 166, 166, 112

Offset: 0

T. D. Noe, Aug 07 2013

This sequence is almost the same as A026637.

T(n,k) = A026637(n-2,k-1) for n > 3, 1 < k < n-1. - Reinhard Zumkeller, Aug 08 2013

			Triangle begins:
  -1,
  1, 1,
  -1, 2, -1,
  1, 1, 1, 1,
  -1, 2, 2, 2, -1,
  1, 1, 4, 4, 1, 1,
  -1, 2, 5, 8, 5, 2, -1,
  1, 1, 7, 13, 13, 7, 1, 1,
  -1, 2, 8, 20, 26, 20, 8, 2, -1,
  1, 1, 10, 28, 46, 46, 28, 10, 1, 1,
  -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A026637 (many terms in common).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A097073 (row sums).
Cf. A227074 (4^n edges), A227075 (3^n edges), A227076 (5^n edges).

a228053 n k = a228053_tabl !! n !! k
a228053_row n = a228053_tabl !! n
a228053_tabl = iterate (\row@(i:_) -> zipWith (+)
   ([- i] ++ tail row ++ [0]) ([0] ++ init row ++ [- i])) [- 1]
  -- Reinhard Zumkeller, Aug 08 2013

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = (-1)^(n+1), m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.

Original entry on oeis.org

1, 4, 4, 16, 8, 16, 64, 24, 24, 64, 256, 88, 48, 88, 256, 1024, 344, 136, 136, 344, 1024, 4096, 1368, 480, 272, 480, 1368, 4096, 16384, 5464, 1848, 752, 752, 1848, 5464, 16384, 65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536, 262144, 87384, 29160

Offset: 0

T. D. Noe, Aug 06 2013

All rows except the zeroth are divisible by 4. Is there a closed-form formula for these numbers, like for binomial coefficients?

			Triangle begins:
  1,
  4, 4,
  16, 8, 16,
  64, 24, 24, 64,
  256, 88, 48, 88, 256,
  1024, 344, 136, 136, 344, 1024,
  4096, 1368, 480, 272, 480, 1368, 4096,
  16384, 5464, 1848, 752, 752, 1848, 5464, 16384,
  65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A165665 (row sums: 3*4^n - 2*2^n), A227075 (3^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 4^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227076 A triangle formed like Pascal's triangle, but with 5^n on the borders instead of 1.

Original entry on oeis.org

1, 5, 5, 25, 10, 25, 125, 35, 35, 125, 625, 160, 70, 160, 625, 3125, 785, 230, 230, 785, 3125, 15625, 3910, 1015, 460, 1015, 3910, 15625, 78125, 19535, 4925, 1475, 1475, 4925, 19535, 78125, 390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625

Offset: 0

T. D. Noe, Aug 06 2013

All rows except the zeroth are divisible by 5. Is there a closed-form formula for these numbers, as there is for binomial coefficients?

			Triangle begins as:
       1;
       5,     5;
      25,    10,    25;
     125,    35,    35,  125;
     625,   160,    70,  160,  625;
    3125,   785,   230,  230,  785, 3125;
   15625,  3910,  1015,  460, 1015, 3910, 15625;
   78125, 19535,  4925, 1475, 1475, 4925, 19535, 78125;
  390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625;

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A083585 (row sums: (8*5^n - 5*2^n)/3), A227074 (4^n edges), A227075 (3^n edges).
Cf. A000351.

function T(n,k) // T = A227076
  if k eq 0 or k eq n then return 5^n;
  else return T(n-1,k) + T(n-1,k-1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 10 2025

A227076 := proc(n,k)
    if k = 0 or k = n then
        5^n ;
    elif k < 0 or k > n then
        0;
    else
        procname(n-1,k)+procname(n-1,k-1) ;
    end if;
end proc: # R. J. Mathar, Aug 09 2013

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 5^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

from sage.all import *
@CachedFunction
def T(n,k): # T = A227076
    if k==0 or k==n: return pow(5,n)
    else: return T(n-1,k) + T(n-1,k-1)
print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 10 2025

From R. J. Mathar, Aug 09 2013: (Start)
T(n,0) = 5^n.
T(n,1) = 5*A047850(n-1).
T(n,2) = 5*(5^n/80 + 3*n/4 + 51/16).
T(n,3) = 5*(5^n/320 + 45*n/16 + 3*n^2/8 + 819/64). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = 20*(1+(-1)^n)*A009969(floor((n-1)/2)) - (3/5)*[n = 0]. - G. C. Greubel, Jan 10 2025

A300401 Array T(n,k) = n(binomial(k, 2) + 1) + k(binomial(n, 2) + 1) read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 4, 4, 3, 4, 7, 8, 7, 4, 5, 11, 14, 14, 11, 5, 6, 16, 22, 24, 22, 16, 6, 7, 22, 32, 37, 37, 32, 22, 7, 8, 29, 44, 53, 56, 53, 44, 29, 8, 9, 37, 58, 72, 79, 79, 72, 58, 37, 9, 10, 46, 74, 94, 106, 110, 106, 94, 74, 46, 10, 11, 56, 92, 119

Offset: 0

Franck Maminirina Ramaharo, Mar 05 2018

Antidiagonal sums are given by 2*A055795.
Rows/columns n are binomial transform of {n, A152947(n+1), n, 0, 0, 0, ...}.
Some primes in the array are
n = 1: {2, 7, 11, 29, 37, 67, 79, 137, 191, 211, 277, 379, ...} = A055469, primes of the form k*(k + 1)/2 + 1;
n = 3: {3, 7, 37, 53, 479, 653, 1249, 1619, 2503, 3727, 4349, 5737, 7109, 8179, 9803, 11839, 12107, ...};
n = 4: {11, 37, 79, 137, 211, 821, 991, 1597, 1831, 2081, 2347, ...} = A188382, primes of the form 8*(2*k - 1)^2 + 2*(2*k - 1) + 1.

			The array T(n,k) begins
0     1    2    3    4     5     6     7     8     9    10    11  ...
1     2    4    7   11    16    22    29    37    46    56    67  ...
2     4    8   14   22    32    44    58    74    92   112   134  ...
3     7   14   24   37    53    72    94   119   147   178   212  ...
4    11   22   37   56    79   106   137   172   211   254   301  ...
5    16   32   53   79   110   146   187   233   284   340   401  ...
6    22   44   72  106   146   192   244   302   366   436   512  ...
7    29   58   94  137   187   244   308   379   457   542   634  ...
8    37   74  119  172   233   302   379   464   557   658   767  ...
9    46   92  147  211   284   366   457   557   666   784   911  ...
10   56  112  178  254   340   436   542   658   784   920  1066  ...
11   67  134  212  301   401   512   634   767   911  1066  1232  ...
12   79  158  249  352   467   594   733   884  1047  1222  1409  ...
13   92  184  289  407   538   682   839  1009  1192  1388  1597  ...
14  106  212  332  466   614   776   952  1142  1346  1564  1796  ...
15  121  242  378  529   695   876  1072  1283  1509  1750  2006  ...
16  137  274  427  596   781   982  1199  1432  1681  1946  2227  ...
17  154  308  479  667   872  1094  1333  1589  1862  2152  2459  ...
18  172  344  534  742   968  1212  1474  1754  2052  2368  2702  ...
19  191  382  592  821  1069  1336  1622  1927  2251  2594  2956  ...
20  211  422  653  904  1175  1466  1777  2108  2459  2830  3221  ...
...
The inverse binomial transforms of the columns are
0     1    2    3    4     5     6     7     8     9    10    11  ...  A001477
1     1    2    4    7    11    22    29    37    45    56    67  ...  A152947
0     1    2    3    4     5     6     7     8     9    10    11  ...  A001477
0     0    0    0    0     0     0     0     0     0     0     0  ...
0     0    0    0    0     0     0     0     0     0     0     0  ...
0     0    0    0    0     0     0     0     0     0     0     0  ...
...

Miklós Bóna, Introduction to Enumerative Combinatorics, McGraw-Hill, 2007.
L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Reidel Publishing Company, 1974.
R. P. Stanley, Enumerative Combinatorics, second edition, Cambridge University Press, 2011.

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened).
Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv preprint 1410.2657 [math.CO], 2014.
Franck Ramaharo, A generating polynomial for the two-bridge knot with Conway's notation C(n,r), arXiv:1902.08989 [math.CO], 2019.

Cf. A000124, A001477, A006000, A008815, A014206, A051601, A055469, A077028, A081436, A084849, A131074, A134394, A139600, A141387, A179000, A188377, A188382, A273465.

T := (n, k) -> n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1);
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

T[n_, k_] := n (Binomial[k, 2] + 1) + k (Binomial[n, 2] + 1);
Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 07 2018 *)

T(n, k) := n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1)$
for n:0 thru 20 do
  print(makelist(T(n, k), k, 0, 20));

T(n, k) = n*(binomial(k,2) + 1) + k*(binomial(n,2) + 1);
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, Mar 12 2018

T(n,k) = T(k,n) = n*A152947(k+1) + k*A152947(n+1).
T(n,0) = A001477(n).
T(n,1) = A000124(n).
T(n,2) = A014206(n).
T(n,3) = A273465(3*n+2).
T(n,4) = A084849(n+1).
T(n,n) = A179000(n-1,n), n >= 1.
T(2*n,2*n) = 8*A081436(n-1), n >= 1.
T(2*n+1,2*n+1) = 2*A006000(2*n+1).
T(n,n+1) = A188377(n+3).
T(n,n+2) = A188377(n+2), n >= 1.
Sum_{k=0..n} T(k,n-k) = 2*(binomial(n, 4) + binomial(n, 2)).
G.f.: -((2*x*y - y - x)*(2*x*y - y - x + 1))/(((x - 1)*(y - 1))^3).
E.g.f.: (1/2)*(x + y)*(x*y + 2)*exp(x + y).

A055797 T(2n+4,n), array T as in A055794.

Original entry on oeis.org

1, 6, 22, 64, 162, 372, 792, 1584, 3003, 5434, 9438, 15808, 25636, 40392, 62016, 93024, 136629, 196878, 278806, 388608, 533830, 723580, 968760, 1282320, 1679535, 2178306, 2799486, 3567232, 4509384

Offset: 0

Clark Kimberling, May 28 2000

If Y is a 2-subset of an n-set X then, for n>=7, a(n-7) is the number of 7-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A051601.

[Binomial(n,7) + Binomial(n,5): n in [5..40]]; // Vincenzo Librandi, May 01 2012

[seq(binomial(n,7)+binomial(n,5), n=5..34)]; # Zerinvary Lajos, Jul 24 2006

a=1;b=2;c=3;d=4;e=5;f=6;s=7;lst={s};Do[a+=n;b+=a;c+=b;d+=c;e+=d;f+=e;s+=f;AppendTo[lst,s],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, May 24 2009 *)
CoefficientList[Series[(1-2*x+2*x^2)/(1-x)^8,{x,0,30}],x] (* Vincenzo Librandi, May 01 2012 *)
LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{1,6,22,64,162,372,792,1584},30] (* Harvey P. Dale, Mar 11 2023 *)

a(n-5) = binomial(n,7) + binomial(n,5) for n>4. - Zerinvary Lajos, Jul 24 2006
G.f.: (1-2*x+2*x^2)/(1-x)^8. - Colin Barker, Feb 22 2012
a(n) = 8*(n-1) - 28*(n-2) + 56*(n-3) - 70*(n-4) + 56*(n-5) - 28*(n-6) + 8*(n-7) - (n-8). - Vincenzo Librandi, May 01 2012
a(n) = (n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(n^2-n+42)/5040. - R. J. Mathar, Oct 01 2021

A055798 T(2n+5,n), array T as in A055794.

Original entry on oeis.org

1, 7, 29, 93, 255, 627, 1419, 3003, 6006, 11440, 20878, 36686, 62322, 102714, 164730, 257754, 394383, 591261, 870067, 1258675, 1792505, 2516085, 3484845, 4767165, 6446700, 8625006, 11424492, 14991724, 19501108, 25158980, 32208132, 40932804, 51664173

Offset: 0

Clark Kimberling, May 28 2000

If Y is a 2-subset of an n-set X then, for n>=8, a(n-8) is the number of 8-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007

Vincenzo Librandi, 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. A051601.

[Binomial(n,8)-2*Binomial(n-2,7): n in [8..40]]; // Vincenzo Librandi, May 01 2012

CoefficientList[Series[(-2*(z - 1)*z - 1)/(z - 1)^9, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 16 2011 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,7,29,93,255,627,1419,3003,6006},50] (* Vincenzo Librandi, May 01 2012 *)

a(n-8) = binomial(n,8)-2*binomial(n-2,7), n=8,9,10,.... - Milan Janjic, Dec 28 2007
G.f.: (1-2*x+2*x^2)/(1-x)^9. [Colin Barker, Feb 22 2012]
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9). - Vincenzo Librandi, May 01 2012

A055799 T(2n+6,n), array T as in A055794.

Original entry on oeis.org

1, 8, 37, 130, 385, 1012, 2431, 5434, 11440, 22880, 43758, 80444, 142766, 245480, 410210, 667964, 1062347, 1653608, 2523675, 3782350, 5574855, 8090940, 11575785, 16342950, 22789650, 31414656, 42839148, 57830872

Offset: 0

Clark Kimberling, May 28 2000

If Y is a 2-subset of an n-set X then, for n>=9, a(n-9) is the number of 9-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007

Vincenzo Librandi, 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).

Cf. A051601.

[Binomial(n,9)-2*Binomial(n-2,8):n in [9..40]]; // Vincenzo Librandi, May 01 2012

a=1;b=2;c=3;d=4;e=5;f=6;g=7;s=8;lst={1,s};Do[a+=n;b+=a;c+=b;d+=c;e+=d;f+=e;g+=f;s+=g;AppendTo[lst,s],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, May 24 2009 *)
CoefficientList[Series[(1-2*x+2*x^2)/(1-x)^10,{x,0,30}],x] (* Vincenzo Librandi, May 01 2012 *)

a(n-9) = binomial(n,9) - 2*binomial(n-2,8), n=9, 10, ... . - Milan Janjic, Dec 28 2007
G.f.: (1-2*x+2*x^2)/(1-x)^10. - Colin Barker, Feb 21 2012
a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4)+ 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10). - Vincenzo Librandi, May 01 2012

cp's OEIS Frontend

A055796 T(2n+3,n), array T as in A055794.

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A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

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A227550 A triangle formed like Pascal's triangle, but with factorial(n) on the borders instead of 1.

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A228053 A triangle formed like Pascal's triangle, but with (-1)^(n+1) on the borders instead of 1.

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A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.

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A227076 A triangle formed like Pascal's triangle, but with 5^n on the borders instead of 1.

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A300401 Array T(n,k) = n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1) read by antidiagonals.

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A300401 Array T(n,k) = n(binomial(k, 2) + 1) + k(binomial(n, 2) + 1) read by antidiagonals.