A033996 -id:A033996 - cp's OEIS frontend

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).

0, 0, 0, 0, 1, 0, 0, 8, 2, 0, 0, 49, 24, 3, 0, 0, 288, 242, 48, 4, 0, 0, 1681, 2400, 675, 80, 5, 0, 0, 9800, 23762, 9408, 1444, 120, 6, 0, 0, 57121, 235224, 131043, 25920, 2645, 168, 7, 0, 0, 332928, 2328482, 1825200, 465124, 58080, 4374, 224, 8, 0

Offset: 0

Seiichi Manyama, Dec 23 2018

			Square array begins:
   0, 0,   0,    0,      0,       0,        0, ...
   0, 1,   8,   49,    288,    1681,     9800, ...
   0, 2,  24,  242,   2400,   23762,   235224, ...
   0, 3,  48,  675,   9408,  131043,  1825200, ...
   0, 4,  80, 1444,  25920,  465124,  8346320, ...
   0, 5, 120, 2645,  58080, 1275125, 27994680, ...
   0, 6, 168, 4374, 113568, 2948406, 76545000, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-5 give A000004, A001477, A033996, A322675, A322677, A322745.
Rows 0-9 give A000004, A001108, A132596, A007654(n+1), A132584, A322707, A322708, A322709, A132592, A132593.
Main diagonal gives A322746.
Cf. A173175 (A(n,2*n)), A322790.

Unprotect[Power]; 0^0 := 1; Protect[Power]; Table[(-1 + Sum[Binomial[2 k, 2 j] (# + 1)^(k - j)*#^j, {j, 0, k}])/2 &[n - k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jan 01 2019 *)
nmax = 9; row[n_] := LinearRecurrence[{4n+3, -4n-3, 1}, {0, n, 4n(n+1)}, nmax+1]; T = Array[row, nmax+1, 0]; A[n_, k_] := T[[n+1, k+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 06 2019 *)

def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  (0..n).map{|i| (0..k).inject(-1){|s, j| s + ncr(2 * k, 2 * j) * (i + 1) ** (k - j) * i ** j} / 2}
end
def A322699(n)
  a = []
  (0..n).each{|i| a << A(i, n - i)}
  ary = []
  (0..n).each{|i|
    (0..i).each{|j|
      ary << a[i - j][j]
    }
  }
  ary
end
p A322699(10)

sqrt(A(n,k)+1) + sqrt(A(n,k)) = (sqrt(n+1) + sqrt(n))^k.
sqrt(A(n,k)+1) - sqrt(A(n,k)) = (sqrt(n+1) - sqrt(n))^k.
A(n,0) = 0, A(n,1) = n and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) + 2*n for k > 1.
A(n,k) = (T_{k}(2*n+1) - 1)/2 where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = (2*n+1-1)/2 = n.

A188147 T(n,k)=Number of n-step self-avoiding walks on a kXk square summed over all starting positions.

Original entry on oeis.org

1, 4, 0, 9, 8, 0, 16, 24, 8, 0, 25, 48, 44, 8, 0, 36, 80, 104, 80, 0, 0, 49, 120, 188, 232, 104, 0, 0, 64, 168, 296, 456, 432, 128, 0, 0, 81, 224, 428, 752, 972, 800, 112, 0, 0, 100, 288, 584, 1120, 1712, 2112, 1248, 112, 0, 0, 121, 360, 764, 1560, 2652, 4008, 4152, 1976, 40

Offset: 1

R. H. Hardin Mar 22 2011

Table starts
.1.4...9...16....25....36.....49.....64.....81....100....121.....144.....169
.0.8..24...48....80...120....168....224....288....360....440.....528.....624
.0.8..44..104...188...296....428....584....764....968...1196....1448....1724
.0.8..80..232...456...752...1120...1560...2072...2656...3312....4040....4840
.0.0.104..432...972..1712...2652...3792...5132...6672...8412...10352...12492
.0.0.128..800..2112..4008...6472...9504..13104..17272..22008...27312...33184
.0.0.112.1248..4152..8752..14932..22672..31972..42832..55252...69232...84772
.0.0.112.1976..8160.19312..35024..55104..79528.108296.141408..178864..220664
.0.0..40.2640.14520.39792..78168.128688.191068.265280.351324..449200..558908
.0.0...0.3696.26000.82032.175312.303328.464304.657848.883928.1142544.1433696

			Some n=3 solutions for 3X3
..1..0..0....0..0..0....0..0..3....0..0..0....0..0..0....0..0..0....0..0..0
..2..0..0....0..3..0....0..0..2....0..3..2....0..1..0....0..0..3....0..0..0
..3..0..0....1..2..0....0..0..1....0..0..1....3..2..0....0..1..2....1..2..3

R. H. Hardin, Table of n, a(n) for n = 1..264

Row 2 is A033996(n-1)

Empirical: T(1,k) = k^2
Empirical: T(2,k) = 4*k^2 - 4*k
Empirical: T(3,k) = 12*k^2 - 24*k + 8 for k>1
Empirical: T(4,k) = 36*k^2 - 100*k + 56 for k>2
Empirical: T(5,k) = 100*k^2 - 360*k + 272 for k>3
Empirical: T(6,k) = 284*k^2 - 1228*k + 1152 for k>4
Empirical: T(7,k) = 780*k^2 - 3960*k + 4432 for k>5
Empirical: T(8,k) = 2172*k^2 - 12500*k + 16096 for k>6
Empirical: T(9,k) = 5916*k^2 - 38192*k + 55600 for k>7
Empirical: T(10,k) = 16268*k^2 - 115548*k + 186528 for k>8

A262221 a(n) = 25n(n + 1)/2 + 1.

Original entry on oeis.org

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651

Offset: 0

Bruno Berselli, Sep 15 2015

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of numbers of the form (k*n*(n+1)-(-1)^k+1)/2.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A008607 (first differences), A011379, A034856, A054254, A080956, A123296, A186349, A255184.
Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

Magma
```
[25*n*(n+1)/2+1: n in [0..50]];
```

Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)

PARI
```
vector(50, n, n--; 25*n*(n+1)/2+1)
```
Sage
```
[25*n*(n+1)/2+1 for n in (0..50)]
```

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024

A187172 T(n,k) is the number of n-step left-handed knight's tours (moves only out two, left one) on a k X k board summed over all starting positions.

Original entry on oeis.org

1, 4, 0, 9, 0, 0, 16, 8, 0, 0, 25, 24, 0, 0, 0, 36, 48, 16, 0, 0, 0, 49, 80, 60, 8, 0, 0, 0, 64, 120, 128, 48, 0, 0, 0, 0, 81, 168, 220, 176, 16, 0, 0, 0, 0, 100, 224, 336, 384, 136, 0, 0, 0, 0, 0, 121, 288, 476, 664, 456, 88, 0, 0, 0, 0, 0, 144, 360, 640, 1016, 1024, 496, 16, 0, 0, 0, 0, 0

Offset: 1

R. H. Hardin, Mar 06 2011

Table starts
4 9 16 25  36  49   64   81   100   121    144    169    196    225    256
0 8 24 48  80 120  168  224   288   360    440    528    624    728    840
0 0 16 60 128 220  336  476   640   828   1040   1276   1536   1820   2128
0 0  8 48 176 384  664 1016  1440  1936   2504   3144   3856   4640   5496
0 0  0 16 136 456 1024 1804  2784  3964   5344   6924   8704  10684  12864
0 0  0  0  88 496 1440 3064  5344  8208  11640  15640  20208  25344  31048
0 0  0  0  16 368 1600 4284  8760 15104  23144  32764  43944  56684  70984
0 0  0  0   0 280 1784 5944 14072 27104  45288  68400  96048 128064 164424
0 0  0  0   0  88 1440 6828 19840 43668  80624 131576 196192 273592 363080
0 0  0  0   0   8 1088 7896 27984 70344 142816 250728 396808 580696 800584

			One of 98568 n=51 solutions for 16 X 16:
   0  1  0  0  0  0  4  0  0  0  0  7  0  0  0  0
   0  0  0  0  3  0  0  0  0  6  0  0  0  0  9  0
   0  0  2  0  0  0  0  5  0  0  0  0  8  0  0  0
  51  0  0  0  0 16  0  0  0  0 13  0  0  0  0 10
   0  0  0 17  0  0  0  0 14  0  0  0  0 11  0  0
   0 50  0  0  0  0 15  0  0  0  0 12  0  0  0  0
   0  0  0  0 18  0  0  0  0 21  0  0  0  0 24  0
   0  0 49  0  0  0  0 20  0  0  0  0 23  0  0  0
  48  0  0  0  0 19  0  0  0  0 22  0  0  0  0 25
   0  0  0 46  0  0  0  0 35  0  0  0  0 26  0  0
   0 47  0  0  0  0 36  0  0  0  0 27  0  0  0  0
   0  0  0  0 45  0  0  0  0 34  0  0  0  0 29  0
   0  0 44  0  0  0  0 37  0  0  0  0 28  0  0  0
  43  0  0  0  0 40  0  0  0  0 33  0  0  0  0 30
   0  0  0 41  0  0  0  0 38  0  0  0  0 31  0  0
   0 42  0  0  0  0 39  0  0  0  0 32  0  0  0  0

R. H. Hardin, Table of n, a(n) for n = 1..310

Row 2 is A033996(n-2).

Empirical:
  T(1,k) =      k^2;
  T(2,k) =    4*k^2 -    12*k +      8;
  T(3,k) =   12*k^2 -    64*k +     80 for k >  3;
  T(4,k) =   36*k^2 -   260*k +    440 for k >  5;
  T(5,k) =  100*k^2 -   920*k +   1984 for k >  7;
  T(6,k) =  284*k^2 -  3100*k +   7944 for k >  9;
  T(7,k) =  780*k^2 -  9880*k +  29384 for k > 11;
  T(8,k) = 2172*k^2 - 30972*k + 103944 for k > 13.

A333714 Squares visited by a chess king moving on a square-spiral numbered board where the king moves to the adjacent unvisited square containing the spiral number with the most divisors. In case of a tie it chooses the square with the highest spiral number.

Original entry on oeis.org

1, 8, 24, 48, 80, 120, 168, 224, 288, 360, 440, 528, 624, 728, 840, 960, 1088, 1224, 1368, 1520, 1680, 1848, 2024, 2208, 2400, 2600, 2808, 3024, 3248, 3480, 3720, 3968, 4224, 4488, 4760, 5040, 5328, 5624, 5928, 6240, 6560, 6888, 7224, 7568, 7920, 8280, 8648, 9024, 9408, 9800, 10200, 10608

Offset: 1

Scott R. Shannon, Jul 02 2020

This sequence gives the numbers of the squares visited by a chess king moving on a square-spiral numbered board where the king starts on the 1 numbered square and at each step moves to an adjacent unvisited square, out of the eight adjacent neighboring squares, which contains the number with the most divisors. If two or more adjacent squares exist with the same highest number of divisors then the square with the highest spiral number is chosen. Given both of these rules tend to force the king to squares with larger numbers, and thus move away from the central 1 starting square, it is remarkable that the king is eventually trapped. Note that if the king simply moves to the highest available number the sequence will be infinite as the king will step along the southeast diagonal from square 1 forever.
The sequence is finite. After 1113 steps the square with number 855481 is visited, after which all adjacent neighboring squares have been visited.
Due to the king's preference for squares with the most divisors it will avoid prime numbers unless no other choice exists. Of the 1113 visited squares only once does it visit a square with a prime number, at a(308) = 108223. This is due to a(307) = 106913 having square 108223 as its sole neighboring unvisited square. This is the only time in the sequence where only one unvisited adjacent neighbor is available.
As even numbers >= 6 will always contain 4 or more divisors the king will tend to visit more even numbers than odd numbers; in the 1113 visited squares 929 contain an even number while only 184 contain an odd number.
As the even numbers are diagonally adjacent in the square spiral the king's path will be dominated by diagonal steps, often taking many diagonal steps in succession - see the attached link image. In fact after the first downward step to 8 the next 110 steps are along the southeast diagonal, stepping to successively larger even numbers. This sequence is finally broken on the 112th step when the square with number 50624, with 28 divisors, is the next square in the southeast direction. However the square with number 50622, with 32 divisors, is in the southwest direction so is the next square chosen. It is not until the 166th step, to the square with number 108230, that the path takes a step to a lower number than the one it is currently on.
The largest visited square is a(1050) = 942676. The visited square with the maximum number of divisors is a(680) = 388080, which has 180 divisors. The lowest unvisited square is 2.

			The board is numbered with the square spiral:
.
  17--16--15--14--13   .
   |               |   .
  18   5---4---3  12   29
   |   |       |   |   |
  19   6   1---2  11   28
   |   |           |   |
  20   7---8---9--10   27
   |                   |
  21--22--23--24--25--26
.
a(1) = 1, the starting square for the king.
a(2) = 8. The eight unvisited squares around a(1) the king can move to are numbered 2,3,4,5,6,7,8,9. Of these 6 and 8 both have the maximum four divisors, and of those 8 is the largest.
a(3) = 24. The seven unvisited squares around a(2) = 8 the king can move to are numbered 9,2,6,7,22,23,24. Of these 24 has eight divisors, the largest number.
a(113) = 50622. The seven unvisited squares around a(112) = 49728 the king can move to are numbered 50622, 49727, 50623, 48841, 50624, 49729, 48842. Of these 50622 has thirty-two divisors, the largest number. This is the step that breaks the sequence of 110 steps to the southeast direction starting from a(2) = 8.
a(308) = 108223. This is the first and only time a prime number is visited; a(307) = 106913 has square 108223 as the sole unvisited adjacent neighbor.
a(1114) = 855481. The two unvisited squares around a(1113) = 859184 the king can move to are numbered 862894 and 855481. Of these 855481 has eight divisors, the largest number. However square 855481 is surrounded by the eight squares with numbers 859183, 855480, 851785, 859184, 851786, 859185, 855482, 851787 all of which have been previously visited, so the king is trapped.

Scott R. Shannon, Table of n, a(n) for n = 1..1114
Scott R. Shannon, Image showing the 1113 steps of the king's path. A green dot marks the starting 1 square and a red dot the final square with number 855481. The red dot is surrounded by eight blue dots to show the occupied neighboring squares. A yellow dots marks the smallest unvisited square with number 2.

Cf. A333713 (choose lowest spiral number in case of tie), A335816, A316667, A330008, A329520, A326922, A328928, A328929, A033996.

A113689 Number of semiprimes in clumps of size > 1 through n^2 in the semiprime spiral.

Original entry on oeis.org

0, 0, 2, 6, 9, 13, 17, 21, 23, 31, 37, 45, 54, 59, 72, 77, 83, 93, 104, 116, 125, 140, 150, 164, 180, 188, 203, 219, 236, 255, 272, 287, 301, 317, 334, 354, 378, 403, 419, 430, 450, 475, 498, 521, 542, 560, 588, 608, 626, 652, 677, 698

Offset: 1

Jonathan Vos Post, Nov 05 2005

Write the integers 1, 2, 3, 4, ... in a counterclockwise square spiral. Analogous to Ulam coloring in the primes in the spiral and discovering unexpectedly many connected diagonals, we construct a semiprime spiral by coloring in all semiprimes (A001358). Each integer has 8 adjacent integers in the spiral, horizontally, vertically and diagonally. Curious extended clumps coagulate, slightly denser towards the origin, of semiprimes connected by adjacency. This sequence, A113689, gives an enumeration of the number of semiprimes in clumps of size > 1 through n^2, not looking past the square boundary. A113688 gives isolated semiprimes in the semiprime spiral, namely those semiprimes none of whose adjacent integers in the spiral are semiprimes.

			a(3) = 2 because there is one visible clump through 3^2 = 9, {4,6}, which two semiprimes are diagonally connected.
a(4) = 6 because there are 6 semiprimes in the 2 visible clumps through 4^2 = 16, {4, 6, 14, 15}, {9, 10}.
a(5) = 9 because there are 9 semiprimes in the 3 visible clumps through 5^2 = 25, {4, 6, 14, 15}, {9, 10, 25}, {21, 22}.
......................
... 17 16 15 14 13 ...
... 18  5  4  3 12 ...
... 19  6  1  2 11 ...
... 20  7  8  9 10 ...
... 21 22 23 24 25 ...
......................

S. M. Ellerstein, The square spiral, J. Recreational Mathematics 29 (#3, 1998) 188; 30 (#4, 1999-2000), 246-250.

M. Stein and S. M. Ulam, An Observation on the Distribution of Primes, Amer. Math. Monthly 74, 43-44, 1967.
M. Stein, S. M. Ulam and M. B. Wells, A Visual Display of Some Properties of the Distribution of Primes, Amer. Math. Monthly 71, 516-520, 1964.
Eric Weisstein's World of Mathematics, Prime Spiral.
Eric Weisstein's World of Mathematics, Semiprime

Cf. A001107, A001358, A002939, A002943, A004526, A005620, A007742, A033951-A033954, A033988, A033989-A033991, A033996, A063826, A113688.

Corrected and extended by Alois P. Heinz, Jan 02 2011

A322677 a(n) = 16n(n+1)(2n+1)^2.

Original entry on oeis.org

0, 288, 2400, 9408, 25920, 58080, 113568, 201600, 332928, 519840, 776160, 1117248, 1560000, 2122848, 2825760, 3690240, 4739328, 5997600, 7491168, 9247680, 11296320, 13667808, 16394400, 19509888, 23049600, 27050400, 31550688, 36590400, 42211008, 48455520

Offset: 0

Seiichi Manyama, Dec 23 2018

			(sqrt(2) - sqrt(1))^4 = (sqrt(9) - sqrt(8))^2 = sqrt(289) - sqrt(288). So a(1) = 288.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^k: A033996(n) (k=2), A322675 (k=3), this sequence (k=4).

Cf. A180324, A185096, A339483.

A322677[n_] := 16*n*(n + 1)*(2*n + 1)^2; Array[A322677, 50, 0] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 288, 2400, 9408, 25920}, 50] (* Paolo Xausa, Aug 26 2025 *)

PARI
```
{a(n) = 16*n*(n+1)*(2*n+1)^2}
```

concat(0, Vec(96*x*(3 + x)*(1 + 3*x) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Dec 23 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^4.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^4.
a(n) = A033996(A033996(n)).
Sum_{n>=1} 1/a(n) = (5 - Pi^2/2)/16 = 0.004074862465957543161422156253870277... - Vaclav Kotesovec, Dec 23 2018
From Colin Barker, Dec 23 2018: (Start)
G.f.: 96*x*(3 + x)*(1 + 3*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4. (End)
From Elmo R. Oliveira, Aug 20 2025: (Start)
E.g.f.: 16*x*(2 + x)*(9 + 24*x + 4*x^2)*exp(x).
a(n) = 96*A180324(n) = 32*A339483(n) = 8*A185096(n). (End)

A068857 a(0) = 0, a(1) = 8; for n>=2: a(n) = smallest multiple of a(n-1) which is of the form 2k*(2k+2).

Original entry on oeis.org

0, 8, 24, 48, 288, 16128, 11950848, 4636929024, 88106288385024, 8038489644431643930624, 15177535939786079616000991061008232448, 40096515501441989312471498490435884509054125751527350190658560000

Offset: 0

Amarnath Murthy, Mar 12 2002

nonn

			24 = 4*6 is a member and the smallest multiple of 24 which is of the form 2k(2k+2) is 48 = 6*8.

Chai Wah Wu, Table of n, a(n) for n = 0..14

Cf. A000217, A033996, A068776.

m = 0; {0} ~Join~ Rest@ NestList[(m++; While[! Divisible[Set[k, # (# + 2) &[2 m]], #], m++]; k) &, 1, 8] (* Michael De Vlieger, Mar 18 2024 *)

from itertools import islice
from sympy import sqrt_mod_iter
def A068857_gen(): # generator of terms
    yield 0
    a = 8
    while True:
        yield a
        b = a+1
        for d in sqrt_mod_iter(1,a):
            if d==1 or d**2-1 == a:
                d += a
            if d&1 and d < b:
                b = d
        a = b**2-1
A068857_list = list(islice(A068857_gen(),11)) # Chai Wah Wu, May 05 2024

a(n) = 8 * A068776(n-1) for n>=1.

More terms from Sascha Kurz, Mar 23 2002
a(8) onward corrected by Sean A. Irvine, Mar 18 2024
a(10)-a(11) from Alois P. Heinz, Mar 19 2024

A131104 Rectangular array read by antidiagonals: a(n, k) is the number of ways to put k labeled objects into n labeled boxes so that there is one box with exactly one object (n, k >= 1).

Original entry on oeis.org

1, 2, 0, 3, 0, 0, 4, 0, 6, 0, 5, 0, 18, 8, 0, 6, 0, 36, 24, 10, 0, 7, 0, 60, 48, 120, 12, 0, 8, 0, 90, 80, 420, 396, 14, 0, 9, 0, 126, 120, 1000, 1512, 1092, 16, 0, 10, 0, 168, 168, 1950, 3720, 6804, 2736, 18, 0, 11, 0, 216, 224, 3360, 7380, 23240, 31008, 6480, 20, 0, 12, 0

Offset: 1

David Wasserman, Jun 14 2007, Jun 15 2007

Problem suggested by Brandon Zeidler. Columns 3 through 5 are A028896, A033996, 10*A007586.

			Array begins:
1 0 0 0 0 0 0
2 0 6 8 10 12 14
3 0 18 24 120 396 1092

Cf. A131103, A131105, A131106, A131107.

a(n, 1) = n. For k > 1, a(n, k) = sum_{j=1..min(floor((k-1)/2), n-1)} A008299(k-1, j)*n!*k*/(n-j-1)!.

A253145 Triangular numbers (A000217) omitting the term 1.

Original entry on oeis.org

0, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275

Offset: 0

Paul Curtz, Mar 23 2015

The full triangle of the inverse Akiyama-Tanigawa transform applied to (-1)^n*A062510(n)=3*(-1)^n*A001045(n) yielding a(n) is
0,      3,    6,  10,   15,  21,  28, 36, ...
-3,    -6,  -12, -20,  -30, -42, -56, ...    essentially -A002378
3,     12,   24,  40,   60,  84, ...         essentially  A046092
-9,   -24,  -48, -80, -120, ...              essentially -A033996
15,    48,   96, 160, ...
-33,  -96, -192, ...
63,   192, ...
-129, ...
etc.
First column: (-1)^n*A062510(n).
The following columns are multiples of A122803(n)=(-2)^n. See A007283(n), A091629(n), A020714(n+1), A110286, A175805(n), 4*A005010(n).
An autosequence of the first kind is a sequence whose main diagonal is A000004 = 0's.
b(n) = 0, 0 followed by a(n) is an autosequence of the first kind.
The successive differences of b(n) are
0,   0,  0, 3, 6, 10, 15, 21, ...
0,   0,  3, 3, 4,  5,  6,  7, ...  see A194880(n)
0,   3,  0, 1, 1,  1,  1,  1, ...
3,  -3,  1, 0, 0,  0,  0,  0, ...
-6,  4, -1, 0, 0,  0,  0,  0, ...
10, -5,  1, 0, 0,  0,  0,  0, ...
-15, 6, -1, 0, 0,  0,  0,  0, ...
21, -7,  1, 0, 0,  0,  0,  0, ...
The inverse binomial transform (first column) is the signed sequence. This is general.
Also generalized hexagonal numbers without 1. - Omar E. Pol, Mar 23 2015

Muniru A Asiru, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A179865, A001045, A062510, A002378, A046092, A033996, A122803, A007283, A091629, A020714, A110286, A161680, A175805, A005010, A194880, A001840, A022003, A255935.

Concatenation([0],List([1..50],n->(n+1)*(n+2)/2)); # Muniru A Asiru, Oct 31 2018

Prepend[Table[(n + 1) (n + 2)/2, {n, 49}], 0] (* Michael De Vlieger, Mar 23 2015 *)

a(n)=if(n,(n+1)*(n+2)/2,0) \\ Charles R Greathouse IV, Mar 23 2015

Inverse Akiyama-Tanigawa transform of (-1)^n*A062510(n).
a(n) = (n+1)*(n+2)/2 for n > 0. - Charles R Greathouse IV, Mar 23 2015
a(n+1) = 3*A001840(n+1) + A022003(n).
a(n) = A161680(n+2) for n >= 1. - Georg Fischer, Oct 30 2018
From Stefano Spezia, May 28 2025: (Start)
G.f.: x*(3 - 3*x + x^2)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 4*x + x^2)/2 - 1. (End)

cp's OEIS Frontend

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j).

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A188147 T(n,k)=Number of n-step self-avoiding walks on a kXk square summed over all starting positions.

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A262221 a(n) = 25*n*(n + 1)/2 + 1.

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A187172 T(n,k) is the number of n-step left-handed knight's tours (moves only out two, left one) on a k X k board summed over all starting positions.

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A333714 Squares visited by a chess king moving on a square-spiral numbered board where the king moves to the adjacent unvisited square containing the spiral number with the most divisors. In case of a tie it chooses the square with the highest spiral number.

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A113689 Number of semiprimes in clumps of size > 1 through n^2 in the semiprime spiral.

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A322677 a(n) = 16*n*(n+1)*(2*n+1)^2.

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A068857 a(0) = 0, a(1) = 8; for n>=2: a(n) = smallest multiple of a(n-1) which is of the form 2k*(2k+2).

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A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).

A262221 a(n) = 25n(n + 1)/2 + 1.

A322677 a(n) = 16n(n+1)(2n+1)^2.