A000580 -id:A000580 - cp's OEIS frontend

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

1, 4, 1, 7, 5, 1, 10, 12, 6, 1, 13, 22, 18, 7, 1, 16, 35, 40, 25, 8, 1, 19, 51, 75, 65, 33, 9, 1, 22, 70, 126, 140, 98, 42, 10, 1, 25, 92, 196, 266, 238, 140, 52, 11, 1, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1

Offset: 0

Gary W. Adamson, Dec 01 2007

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.
Row sums give A036563(n+2), n >= 0.
From Wolfdieter Lang, Mar 23 2015: (Start)
This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).
This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.
The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.
The alternating row sums are A122553(n) = {1, repeat(3)}.
The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).
The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)
When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.
Cf. A036563, A125232.
Cf. A110813. - Philippe Deléham, Feb 08 2009

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).
T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009
From Wolfdieter Lang, Mar 23 2015: (Start)
O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).
O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).
T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.
T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).
(End)

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015

A213808 Triangle of numbers C^(7)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 7 appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6427, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12861, 24229, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24300, 48520, 91828

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 20 2012

For k <= 6, the triangle coincides with triangle A213745.

			Triangle begins
n/k |  0     1     2     3     4     5     6     7     8
----+---------------------------------------------------
  0 |  1
  1 |  1     1
  2 |  1     2     3
  3 |  1     3     6    10
  4 |  1     4    10    20    35
  5 |  1     5    15    35    70   126
  6 |  1     6    21    56   126   252   462
  7 |  1     7    28    84   210   462   924  1716
  8 |  1     8    36   120   330   792  1716  3432  6427

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

Cf. A007318, A005725, A059481, A111808, A187925, A213742, A213743, A213744, A000217, A000292, A000332, A000389, A000579, A000580.

Table[Sum[(-1)^r*Binomial[n, r]*Binomial[n - 8*r + k - 1, n - 1], {r, 0, Floor[k/8]}], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Nov 25 2017 *)

for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, sum(r=0, floor(k/8), (-1)^r*binomial(n,r)*binomial(n-8*r + k-1,n-1))), ", "))) \\ G. C. Greubel, Nov 25 2017

T(n,k) = Sum_{r=0..floor(k/8)} (-1)^r*C(n,r)*C(n-8*r+k-1, n-1).

T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n > 1, T(n,3)=A000292(n) for n >= 3, T(n,4)=A000332(n) for n >= 7, T(n,5)=A000389(n) for n >= 9, T(n,6)=A000579(n) for n >= 11, T(n,7)=A000580(n) for n >= 13.

A253944 a(n) = 3*binomial(n+1,7).

Original entry on oeis.org

3, 24, 108, 360, 990, 2376, 5148, 10296, 19305, 34320, 58344, 95472, 151164, 232560, 348840, 511632, 735471, 1038312, 1442100, 1973400, 2664090, 3552120, 4682340, 6107400, 7888725, 10097568, 12816144, 16138848, 20173560, 25043040, 30886416, 37860768

Offset: 6

Serhat Bulut, Jan 20 2015

For a set of integers {1,2,...,n}, a(n) is the sum of the 2 smallest elements of each subset with 6 elements, which is 3*C(n+1,7) (for n>=6), hence a(n) = 3*C(n+1,7) = 3*A000580(n+1).

			For A={1,2,3,4,5,6,7}, subsets with 6 elements are {1,2,3,4,5,6}, {1,2,3,4,5,7}, {1,2,3,4,6,7}, {1,2,3,5,6,7}, {1,2,4,5,6,7}, {1,3,4,5,6,7}, {2,3,4,5,6,7}.
Sum of 2 smallest elements of each subset:
a(7) = (1+2)+(1+2)+(1+2)+(1+2)+(1+2)+(1+3)+(2+3) = 24 = 3*C(7+1,7) = 3*A000580(7+1).

G. C. Greubel, Table of n, a(n) for n = 6..1000
Serhat Bulut and Oktay Erkan Temizkan, Subset Sum Problem
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000580.

[3*Binomial(n+1, 7): n in [6..40]]; // Vincenzo Librandi, Feb 13 2015

Drop[Plus @@ Flatten[Part[#, 1 ;; 2] & /@ Subsets[Range@ #, {6}]] & /@
  Range@ 28, 5] (* Michael De Vlieger, Jan 20 2015 *)
3 Binomial[Range[7, 29], 7] (* Michael De Vlieger, Feb 13 2015, after Alonso del Arte at A253946 *)

a(n)=3*binomial(n+1,7) \\ Charles R Greathouse IV, Feb 04 2015

def A253944(n): return 3*binomial(n+1,7)
print([A253944(n) for n in range(6,51)]) # G. C. Greubel, Apr 03 2025

a(n) = 3*C(n+1,7) = 3*A000580(n+1).
a(n) = 3*C(n+1,7) = n*(n^6 - 14*n^5 + 70*n^4 - 140*n^3 + 49*n^2 + 154*n - 120)/1680.
From G. C. Greubel, Apr 03 2025: (Start)
G.f.: 3*x^6/(1-x)^8.
E.g.f.: (3/7!)*x^6*(x+7)*exp(x). (End)

More terms from Vincenzo Librandi, Feb 13 2015

A266733 a(n) = 21*binomial(n+6,7).

Original entry on oeis.org

0, 21, 168, 756, 2520, 6930, 16632, 36036, 72072, 135135, 240240, 408408, 668304, 1058148, 1627920, 2441880, 3581424, 5148297, 7268184, 10094700, 13813800, 18648630, 24864840, 32776380, 42751800, 55221075, 70682976, 89713008, 112971936, 141214920, 175301280

Offset: 0

Alan Shore and N. J. A. Sloane, Jan 06 2016

Total number of pips on a set of hexominoes (6-celled linear dominoes) with up to n pips in each cell.

Colin Barker, Table of n, a(n) for n = 0..1000
Steve Butler and Pavel Karasik, A note on nested sums, J. Int. Seq. (2010) Vol. 13, Issue 4, Art. No. 10.4.4. See p=6 in the last equation on page 3.
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 9.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Row 6 of array in A129533.

Table[21 Binomial[n+6,7],{n,0,40}] (* Harvey P. Dale, Jan 13 2021 *)

a(n) = (n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n))/240 \\ Colin Barker, Jan 08 2016

concat(0, Vec(21*x/(1-x)^8 + O(x^40))) \\ Colin Barker, Jan 08 2016

a(n) = 21*A000580(n+6).
From Colin Barker, Jan 08 2016: (Start)
a(n) = n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)/240.
a(n) = 8*a(n-1)-28*a(n-2)+56*a(n-3)-70*a(n-4)+56*a(n-5)-28*a(n-6)+8*a(n-7)-a(n-8) for n>7.
G.f.: 21*x / (1-x)^8.
(End)

A116690 a(n) = C(n,8) + C(n,7) + C(n,6) + C(n,5) + C(n,4) + C(n,3) + C(n,2) + C(n,1).

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 63, 127, 255, 510, 1012, 1980, 3796, 7098, 12910, 22818, 39202, 65535, 106761, 169765, 263949, 401929, 600369, 880969, 1271625, 1807780, 2533986, 3505698, 4791322, 6474540, 8656936, 11460948, 15033172, 19548045

Offset: 0

Jonathan Vos Post, Mar 15 2006

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).

[n*(n+1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6 ) /40320: n in [0..30]]; // G. C. Greubel, Nov 25 2017

seq(sum(binomial(n,k),k=1..8),n=0..33); # Zerinvary Lajos, Dec 14 2007

Table[n*(n + 1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6)/40320, {n, 0, 50}] (* G. C. Greubel, Nov 25 2017 *)
Table[Total[Binomial[n,Range[8]]],{n,0,40}] (* Harvey P. Dale, Aug 14 2023 *)

for(n=0,30, print1(n*(n+1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6 ) /40320, ", ")) \\ G. C. Greubel, Nov 25 2017

[binomial(n,2)+binomial(n,4)+binomial(n,6)+binomial(n,8) for n in range(1, 35)] # Zerinvary Lajos, May 17 2009

[binomial(n,2)+binomial(n,4)+binomial(n,6)+binomial(n,8)+binomial(n,1)+binomial(n,3)+binomial(n,5)+binomial(n,7)for n in range(0, 34)] # Zerinvary Lajos, May 17 2009

a(n) = A000581(n) + A000580(n) + A000579(n) + A000389(n) + A000332(n) + A000292(n) + A000217(n) + n. a(n) = A000581(n) + A116082(n).
G.f. ( -x*(2*x^2 - 2*x + 1)*(2*x^4 - 4*x^3 + 6*x^2 - 4*x + 1) ) / (x-1)^9. - R. J. Mathar, Oct 21 2011
a(n) = n*(n+1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6)/40320. - G. C. Greubel, Nov 25 2017

A130813 If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n) is equal to the number of 7-subsets of X containing none of X_i, (i=1,...n).

Original entry on oeis.org

128, 1024, 4608, 15360, 42240, 101376, 219648, 439296, 823680, 1464320, 2489344, 4073472, 6449664, 9922560, 14883840, 21829632, 31380096, 44301312, 61529600, 84198400, 113667840, 151557120, 199779840, 260582400, 336585600, 430829568

Offset: 7

Milan Janjic, Jul 16 2007

nonn

Number of n permutations (n>=7) of 3 objects u,v,z, with repetition allowed, containing n-7 u's. Example: if n=7 then n-7 =(0) zero u, a(1)=128. - Zerinvary Lajos, Aug 05 2008

a(n) is the number of 6-dimensional elements in an n-cross polytope where n>=7. - Patrick J. McNab, Jul 06 2015

Milan Janjic, Two Enumerative Functions
Eric Weisstein's World of Mathematics, Cross Polytope

Cf. A038207, A000079, A001787, A001788, A001789, A003472, A054849, A002409, A054851, A140325, A140354, A046092, A130809, A130810, A130811, A130812. - Zerinvary Lajos, Aug 05 2008

Cf. A000580.

[Binomial(n,n-7)*2^7: n in [7..40]]; // Vincenzo Librandi, Jul 09 2015

a:=n->binomial(2*n,7)+binomial(n,2)*binomial(2*n-4,3)-n*binomial(2*n-2,5)-(2*n-6)*binomial(n,3);
seq(binomial(n,n-7)*2^7,n=7..32); # Zerinvary Lajos, Dec 07 2007
seq(binomial(n+6, 7)*2^7, n=1..22); # Zerinvary Lajos, Aug 05 2008

Table[Binomial[n, n - 7] 2^7, {n, 7, 40}] (* Vincenzo Librandi, Jul 09 2015 *)

a(n) = binomial(2*n,7) + binomial(n,2)*binomial(2*n-4,3) - n*binomial(2*n-2,5) - (2*n-6)*binomial(n,3).
a(n) = C(n,n-7)*2^7, n>=7. - Zerinvary Lajos, Dec 07 2007
G.f.: 128*x^7/(1-x)^8. - Colin Barker, Mar 18 2012
a(n) = 128*A000580(n). a(n+1) = 2*(n+1)*a(n)/(n-6) for n >= 7. - Robert Israel, Jul 08 2015

A253947 a(n) = 6*binomial(n+1,7).

Original entry on oeis.org

6, 48, 216, 720, 1980, 4752, 10296, 20592, 38610, 68640, 116688, 190944, 302328, 465120, 697680, 1023264, 1470942, 2076624, 2884200, 3946800, 5328180, 7104240, 9364680, 12214800, 15777450, 20195136, 25632288, 32277696, 40347120, 50086080, 61772832, 75721536

Offset: 6

Serhat Bulut, Jan 20 2015

For a set of integers {1,2,...,n}, a(n) is the sum of the 3 smallest elements of each subset with 6 elements, which is 6*binomial(n+1,7) (for n>=6), hence a(n) = 6*binomial(n+1,7) = 6*A000580(n+1).

			For A={1,2,3,4,5,6,7}, the subsets with 6 elements are {1,2,3,4,5,6}, {1,2,3,4,5,7}, {1,2,3,4,6,7}, {1,2,3,5,6,7}, {1,2,4,5,6,7}, {1,3,4,5,6,7}, and {2,3,4,5,6,7}.
Sum of 3 smallest elements of each subset: a(7) = (1+2+3) + (1+2+3) + (1+2+3) + (1+2+3) + (1+2+4) + (1+3+4) + (2+3+4) = 48 = 6*binomial(7+1,7) = 6*A000580(7+1).

Colin Barker, Table of n, a(n) for n = 6..1000
Serhat Bulut, Oktay Erkan Temizkan, Subset Sum Problem
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000580 (binomial(n, 7)).

[6*Binomial(n+1, 7): n in [6..40]]; // Vincenzo Librandi, Feb 13 2015

Drop[Plus @@ Flatten[Part[#, 1 ;; 3] & /@ Subsets[Range@ #, {6}]] & /@
  Range@ 30, 5] (* Michael De Vlieger, Jan 20 2015 *)
6 Binomial[Range[7, 31], 7] (* Michael De Vlieger, Feb 13 2015, after Alonso del Arte at A253946 *)

Vec(6*x^6/(1-x)^8 + O(x^100)) \\ Colin Barker, Apr 03 2015

a(n) = 6*binomial(n+1,7) = 6*A000580(n+1).
G.f.: 6*x^6 / (1-x)^8. - Colin Barker, Apr 03 2015
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8). - Wesley Ivan Hurt, Sep 03 2022

More terms from Vincenzo Librandi, Feb 13 2015

A256861 a(n) = n(n + 1)(n + 2)(n + 3)(n + 4)*(n^2 - n + 6)/720.

Original entry on oeis.org

1, 8, 42, 168, 546, 1512, 3696, 8184, 16731, 32032, 58058, 100464, 167076, 268464, 418608, 635664, 942837, 1369368, 1951642, 2734424, 3772230, 5130840, 6888960, 9140040, 11994255, 15580656, 20049498, 25574752, 32356808, 40625376, 50642592, 62706336

Offset: 1

Luciano Ancora, Apr 14 2015

This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(n+3)*(n+4)*(k*(n-1)+6)/120, where b(n,k) is the n-th hypersolid number in 6 dimensions generated from an arithmetical progression with the first term 1 and common difference k (see Sardelis et al. paper).

D. A. Sardelis and T. M. Valahas, On Multidimensional Pythagorean Numbers, arXiv:0805.4070 [math.GM], 2008.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000580.

Cf. similar sequences listed in A256859.

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (6 - n + n^2)/720, {n, 40}]
Table[Times@@(n+Range[0,4])(n^2-n+6)/720,{n,40}] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{1,8,42,168,546,1512,3696,8184},40] (* Harvey P. Dale, Sep 25 2019 *)

vector(40, n, n*(n+1)*(n+2)*(n+3)*(n+4)*(n^2-n+6)/720) \\ Bruno Berselli, Apr 15 2015

G.f.: x*(1 + 6*x^2)/(1 - x)^8.

a(n) = 6*A000580(n+4) + A000580(n+6). [Bruno Berselli, Apr 15 2015]

A270950 Number of distinct cardinalities of orbits of lattice points under the automorphism group of the n-dimensional integer lattice.

Original entry on oeis.org

0, 1, 2, 5, 9, 12, 20, 29, 40, 53, 76, 99, 132, 172, 216, 270, 341, 424, 532, 660, 810, 983, 1210, 1446, 1750, 2111, 2508, 2975, 3569, 4197, 4948, 5807, 6817, 7963, 9351, 10863, 12604, 14598, 16892, 19439, 22472, 25780, 29588, 33892, 38800, 44206, 50463, 57297, 65086, 73919, 83842, 94510

Offset: 0

Philippe A.J.G. Chevalier, Mar 26 2016

A finite number of orbits partition hypercubic shells of infinity norm s in the n-dimensional integer lattice. The number of orbits is given by C(n+s-1,s). The number of distinct cardinalities of the orbits of lattice points under the automorphism group of the n-dimensional integer lattice is found under the condition that n <= s.

A new connection was discovered using the partition of the dimension 'n'. These partitions create a base set of cardinalities. Each of these cardinalities can be subjected to the process of prime factorization. The prime factorization yields the exponents of the primes that form lattice points in a new integer lattice of dimension 'n'. These lattice points become elements of a set A. The unique summands of a specific partition of 'n' give the multipliers of the base vector (1,0^n) that need to be subtracted from the specific partition representative element of set A. The cardinality of the set A increases until all the specific partitions of 'n' have been processed. This augmented set A* has the correct cardinality. This method is much faster than the brute force technique. - Philippe A.J.G. Chevalier, Jun 24 2022

			For n=0 the a(0)=0.
For n=3 we have the following distinct cardinalities of the orbits 6, 8, 12, 24, 48 and thus a(3)=5.
For n=4 we have the distinct cardinalities of the orbits 8, 16, 24, 32, 48, 64, 96, 192, 384 and thus a(4)=9.
For n=5 we have the distinct cardinalities of the orbits 10, 32, 40, 160, 240, 320, 480, 640, 960, 1920, 3840 and thus a(5)=12.

Philippe A.J.G. Chevalier, Table of n, a(n) for n = 0..54

Cf. A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287.

a(17) corrected and a(18)-a(51) from Philippe A.J.G. Chevalier, Jun 24 2022

A336174 Number of non-symmetric binary n X n matrices M over the reals such that M^2 is the transpose of M.

Original entry on oeis.org

0, 0, 0, 2, 16, 80, 360, 1680, 8064, 39872, 209920, 1168640, 6779520, 41403648, 265434624, 1765487360, 12227461120, 88163164160, 656547803136, 5054718763008, 40261284495360, 330010833797120, 2783003768258560, 24166721457815552, 215318925878132736, 1966855934150246400

Offset: 0

Torlach Rush, Jul 10 2020

nonn

We classify the (0,1) n X n matrices M_n by k, the number of 1's.
Let [T(n,k), n >= 0, k=0..n], be the lower triangular matrix where T(n,k) is the number of M^2 matrices equal to the transpose of M for n and k. Then:
T(n,n) = A001471(n).
Column sequences k=3..7 (without leading 0's) are:
T(n,3) = A001471(3) * A000292(n+1).
T(n,4) = A001471(4) * A000332(n+4).
T(n,5) = A001471(5) * A000389(n+5).
T(n,6) = A001471(6) * A000579(n+6).
T(n,7) = A001471(7) * A000580(n+7).
Row sums of T(n,k) generate known terms of this sequence and the next term a(10) evaluates to 209920 (see conjectured formula below).

			a(3) = 2 because [0,1,0]    [0,1,0]    [0,0,1]
                 [0,0,1]  * [0,0,1]  = [1,0,0]
                 [1,0,0]    [1,0,0]    [0,1,0],
             and [0,0,1]    [0,0,1]    [0,1,0]
                 [1,0,0]  * [1,0,0]  = [0,0,1]
                 [0,1,0]    [0,1,0]    [1,0,0].

Cf. A000292, A000332, A000389, A000579, A000580, A001471, A344912.

a := n -> 2^n*(add(n!/(24^k*k!*(n-3*k)!), k=0..n/3) - 1): seq(a(n), n=0..25);
# Alternative:
gf := exp(x*(x^2+6)/3) - exp(2*x): ser := series(gf,x,32):
seq(n!*coeff(ser,x,n), n = 0..25); # Peter Luschny, Jun 05 2021

m(n, t) = matrix(n, n, i, j, (t>>(i*n+j-n-1))%2)
a(n) = sum(t = 0, 2^n^2-1, m(n, t)^2 == m(n, t)~) - 2^n
for(n = 0, 9, print1(a(n), ", "))

a(n) = Sum_{k=0..n} A001471(k) * binomial(n, k). [Previously conjectured, for a proof see the link in A344912.]
From Peter Luschny, Jun 05 2021: (Start)
a(n) = 2^n*(add(n!/(24^k * k! * (n - 3*k)!), k=0..n/3) - 1).
a(n) = 2^n*(hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [], -9/8) - 1).
a(n) = [x^n] exp(x*(x^2 + 6)/3) - exp(2*x). (End)
D-finite with recurrence (-n+3)*a(n) +4*(n-2)*a(n-1) +4*(-n+1)*a(n-2) +(n-1)*(n-2)*(n-3)*a(n-3) -2*(n-1)*(n-2)*(n-3)*a(n-4)=0. - R. J. Mathar, Jul 27 2022

More terms from Peter Luschny, Jun 05 2021

cp's OEIS Frontend

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

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A213808 Triangle of numbers C^(7)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 7 appearances allowed.

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A253944 a(n) = 3*binomial(n+1,7).

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A266733 a(n) = 21*binomial(n+6,7).

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A116690 a(n) = C(n,8) + C(n,7) + C(n,6) + C(n,5) + C(n,4) + C(n,3) + C(n,2) + C(n,1).

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A130813 If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n) is equal to the number of 7-subsets of X containing none of X_i, (i=1,...n).

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A253947 a(n) = 6*binomial(n+1,7).

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A256861 a(n) = n(n + 1)(n + 2)(n + 3)(n + 4)*(n^2 - n + 6)/720.