A045991 -id:A045991 - cp's OEIS frontend

A219605 Square array T(n,k), read by antidiagonals: T(n,2k) = T(n,2k-1)n, T(n,2k+1) = T(n,2*k)+n, T(n,0) = 1.

1, 1, 1, 0, 2, 1, 0, 2, 3, 1, 0, 3, 6, 4, 1, 0, 3, 8, 12, 5, 1, 0, 4, 16, 15, 20, 6, 1, 0, 4, 18, 45, 24, 30, 7, 1, 0, 5, 36, 48, 96, 35, 42, 8, 1, 0, 5, 38, 144, 100, 175, 48, 56, 9, 1, 0, 6, 76, 147, 400, 180, 288, 63, 72, 10, 1, 0, 6, 78, 441, 404, 900, 294, 441

Offset: 0

Philippe Deléham, Apr 12 2013

			Square array begins:
1..1....0....0....0....0....0....0.....0.....0...
1..2....2....3....3....4....4....5.....5.....5...
1..3....6....8...16...18...36...38....76....78...
1..4...12...15...45...48..144..147...441...444...
1..5...20...24...96..100..400..404..1616..1620...
1..6...30...35..175..180..900..905..4525..4530...
1..7...42...48..288..294.1764.1770.10620.10626...
1..8...56...63..441..448.3136.3143.22001.22008...
1..9...72...80..640..648.5184.5192.41536.41544...
1.10...90...99..891..900.8100.8109.72971.72980...
...

t[n_, k_] /; n < 0 || k < 0 = 0; t[n_, 0] = 1; t[n_, 1] = n+1; t[0, k_ /; k > 1] = 0; t[n_?Positive, k_?EvenQ] := t[n, k] = t[n, k-1]*n; t[n_?Positive, k_?OddQ] := t[n, k] = t[n, k-1] + n; Table[t[n-k, k], {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Apr 19 2013 *)

T(n,0) = A000012(n).
T(n,1) = A000027(n).
T(n,2) = A002378(n+1).
T(n,3) = A005563(n).
T(n,4) = A152618(n+1).
T(n,5) = A045991(n+1).
T(n,6) = A035287(n+1).
T(0,k) = A019590(k+1).
T(1,k) = A008619(k+1).
T(2,k) = A123208(k).

A244309 a(n) = F(n)^3 - F(n)^2, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

0, 0, 0, 4, 18, 100, 448, 2028, 8820, 38148, 163350, 697048, 2965248, 12595048, 53440504, 226608900, 960530634, 4070452764, 17246835648, 73069580980, 309555981900, 1311374255620, 5555264316910, 23532984885744, 99688652356608, 422291386890000

Offset: 0

Colin Barker, Jun 25 2014

			a(4) is 18 because F(4)^3 - F(4)^2 = 3^3 - 3^2 = 18.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,2,-22,-4,14,-1,-1).

Cf. A000045, A045991, A244310, A056570, A007598.

[Fibonacci(n)^3 - Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Jun 26 2014

CoefficientList[Series[2 x^3 (x^2 - x + 2)/((x + 1) (x^2 - 3 x + 1) (x^2 - x - 1) (x^2 + 4 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 26 2014 *)
Table[#^3 - #^2 &@ Fibonacci@ n, {n, 0, 25}] (* Michael De Vlieger, Mar 27 2016 *)
LinearRecurrence[{5,2,-22,-4,14,-1,-1},{0,0,0,4,18,100,448},30] (* Harvey P. Dale, Aug 22 2020 *)

vector(50, n, fibonacci(n-1)^3-fibonacci(n-1)^2)

G.f.: 2*x^3*(x^2-x+2) / ((x+1)*(x^2-3*x+1)*(x^2-x-1)*(x^2+4*x-1)).
a(n) = A045991(A000045(n)). - Michel Marcus, Jun 25 2014
a(n) = (F(3*n) - 3*(-1)^n*F(n))/5 - (L(2*n) - 2*(-1)^n)/5, where F=A000045 and L=A000032. - Ehren Metcalfe, Mar 26 2016

A244310 a(n) = L(n)^3 - L(n)^2, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

4, 0, 18, 48, 294, 1210, 5508, 23548, 101614, 433200, 1845738, 7840998, 33282564, 141149320, 598366458, 2535856048, 10745092894, 45524786370, 192866785668, 817050731748, 3461224027254, 14662350247600, 62111682111618, 263111844646798, 1114566304573444

Offset: 0

Colin Barker, Jun 25 2014

			a(3) is 48 because L(3)^3 - L(3)^2 = 4^3 - 4^2 = 48.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,2,-22,-4,14,-1,-1).

Cf. A000032, A045991, A244309.

[Lucas(n)^3 - Lucas(n)^2: n in [0..30]]; // Vincenzo Librandi, Jun 26 2014

CoefficientList[Series[2 (x^6 - 8 x^5 + 17 x^4 + 23 x^3 + 5 x^2 - 10 x + 2)/((x + 1) (x^2 - 3 x + 1) (x^2 - x - 1) (x^2 + 4 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 26 2014 *)
Table[LucasL[n]^3 - LucasL[n]^2, {n,0,50}] (* G. C. Greubel, Oct 13 2018 *)

lucas(n) = if(n==0, 2, 2*fibonacci(n-1)+fibonacci(n))
vector(50, n, lucas(n-1)^3-lucas(n-1)^2)

G.f.: 2*(x^6-8*x^5+17*x^4+23*x^3+5*x^2-10*x+2) / ((x+1)*(x^2-3*x+1)*(x^2-x-1)*(x^2+4*x-1)).

a(n) = A045991(A000032(n)). - Michel Marcus, Jun 25 2014

A286932 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of continued fraction 1/(1 + kx/(1 + kx^2/(1 + kx^3/(1 + kx^4/(1 + k*x^5/(1 + ...)))))).

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -2, 1, 0, 1, -3, 4, 0, 0, 1, -4, 9, -4, -1, 0, 1, -5, 16, -18, 0, 1, 0, 1, -6, 25, -48, 27, 8, -1, 0, 1, -7, 36, -100, 128, -27, -24, 1, 0, 1, -8, 49, -180, 375, -320, -27, 48, 0, 0, 1, -9, 64, -294, 864, -1375, 704, 243, -64, -1, 0, 1, -10, 81, -448, 1715, -4104, 4875, -1280, -810, 48, 2, 0

Offset: 0

Ilya Gutkovskiy, May 16 2017

			G.f. of column k: A(x) = 1 - k*x + k^2*x^2 - (k - 1)*k^2*x^3 + (k - 2)*k^3*x^4 - k^3*(k^2 - 3*k + 1)*x^5 + ...
Square array begins:
  1,  1,  1,   1,    1,     1,  ...
  0, -1, -2,  -3,   -4,    -5,  ...
  0,  1,  4,   9,   16,    25,  ...
  0,  0, -4, -18,  -48,  -100,  ...
  0, -1,  0,  27,  128,   375,  ...
  0,  1,  8, -27, -320, -1375,  ...

Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Columns k=0..1 give: A000007, A007325.
Rows n=0..3 give: A000012, A001489, A000290, A045991 (gives absolute value).
Main diagonal gives A291335.
Cf. A286509.

Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[k x^i, 1, {i, 1, n}]), {x, 0, n}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten

G.f. of column k: 1/(1 + k*x/(1 + k*x^2/(1 + k*x^3/(1 + k*x^4/(1 + k*x^5/(1 + ...)))))), a continued fraction.

G.f. of column k (for k > 0): (Sum_{j>=0} k^j*x^(j*(j+1))/Product_{i=1..j} (1 - x^i)) / (Sum_{j>=0} k^j*x^(j^2)/Product_{i=1..j} (1 - x^i)).

A322550 Table read by ascending antidiagonals: T(n, k) is the minimum number of cubes necessary to fill a right square prism with base area n^2 and height k.

Original entry on oeis.org

1, 4, 2, 9, 1, 3, 16, 18, 12, 4, 25, 4, 1, 2, 5, 36, 50, 48, 36, 20, 6, 49, 9, 75, 1, 45, 3, 7, 64, 98, 4, 100, 80, 2, 28, 8, 81, 16, 147, 18, 1, 12, 63, 4, 9, 100, 162, 192, 196, 180, 150, 112, 72, 36, 10, 121, 25, 9, 4, 245, 1, 175, 2, 3, 5, 11, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44, 12

Offset: 1

Stefano Spezia, Dec 15 2018

			The table T starts in row n = 1 with columns k >= 1 as:
   1     2     3     4     5     6     7     8     9 ...
   4     1    12     2    20     3    28     4    36 ...
   9    18     1    36    45     2    63    72     3 ...
  16     4    48     1    80    12   112     2   144 ...
  25    50    75   100     1   150   175   200   225 ...
  36     9     4    18   180     1   252    36    12 ...
  49    98   147   196   245   294     1   392   441 ...
  64    16   192     4   320    48   448     1   576 ...
  81   162     9   324   405    18   567   648     1 ...
...
The triangle X(n, k) begins
  n\k|   1     2     3     4     5     6     7     8     9
  ---+----------------------------------------------------
   1 |   1
   2 |   4     2
   3 |   9     1     3
   4 |  16    18    12     4
   5 |  25     4     1     2     5
   6 |  36    50    48    36    20     6
   7 |  49     9    75     1    45     3     7
   8 |  64    98     4   100    80     2    28     8
   9 |  81    16   147    18     1    12    63     4     9
...

Stefano Spezia, First 150 antidiagonals of the table, flattened

Cf. A000012 (main diagonal of the table), A000027 (1st row of the table or diagonal of the triangle), A000290 (k=1), A000578, A011379 (superdiagonal of the table), A045991 (subdiagonal of the table), A050873, A119619, A320043 (row sums of the triangle).

Flat(List([1..12], n->List([1..n], k->(n+1-k)^2*k/GcdInt(n+1-k,k)^3)));

[[(n+1-k)^2*k/Gcd(n+1-k,k)^3: k in [1..n]]: n in [1..12]]; // triangle output

a := (n, k) -> (n+1-k)^2*k/gcd(n+1-k, k)^3: seq(seq(a(n, k), k = 1 .. n), n = 1 .. 12)

T[n_,k_]:=n^2*k/GCD[n,k]^3; Flatten[Table[T[n-k+1,k], {n, 12}, {k, n}]]

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 1 thru n do disp(sjoin(makelist((i+1-j)^2*j/gcd(i+1-j,j)^3, j, 1, i), " ")); display_triangle(12);

T(n, k) = (n+1-k)^2*k/gcd(n+1-k,k)^3;
tabl(nn) = for(i=1, nn, for(j=1, i, print1(T(i, j), ", ")); print);
tabl(12) \\ triangle output

T(n, k) = n^2*k/gcd(n, k)^3.
T(n, k) = A000290(n)*k/A000578(A050873(n,k)).
X(n, k) = T(n + 1 - k, k).
X(2*n - 1, n) = A000012(n).
Product_{k=1..n} X(n, k)^(1/3) = A119619(n+1). - Stefano Spezia, Jun 24 2024

A342239 Table read by upward antidiagonals: T(n,k) is the number of strings of length k over an n-letter alphabet that are bifix free; n >= 1, k >= 1.

Original entry on oeis.org

1, 2, 0, 3, 2, 0, 4, 6, 4, 0, 5, 12, 18, 6, 0, 6, 20, 48, 48, 12, 0, 7, 30, 100, 180, 144, 20, 0, 8, 42, 180, 480, 720, 414, 40, 0, 9, 56, 294, 1050, 2400, 2832, 1242, 74, 0, 10, 72, 448, 2016, 6300, 11900, 11328, 3678, 148, 0, 11, 90, 648, 3528, 14112, 37620, 59500, 45132, 11034, 284, 0

Offset: 1

Peter Kagey, Mar 06 2021

			Table begins:
n\k | 1  2   3    4     5      6       7        8         9
----+------------------------------------------------------
  1 | 1  0   0    0     0      0       0        0         0
  2 | 2  2   4    6    12     20      40       74       148
  3 | 3  6  18   48   144    414    1242     3678     11034
  4 | 4 12  48  180   720   2832   11328    45132    180528
  5 | 5 20 100  480  2400  11900   59500   297020   1485100
  6 | 6 30 180 1050  6300  37620  225720  1353270   8119620
  7 | 7 42 294 2016 14112  98490  689430  4823994  33767958
  8 | 8 56 448 3528 28224 225344 1802752 14418488 115347904

Peter Kagey, Antidiagonals n = 1..100, flattened

Rows: A003000 (n=2), A019308 (n=3), A019309 (n=4).

Columns: A002378 (k=1), A045991 (k=2), A047927 (k=3).

T(n,0)    = n.
T(n,2k)   = n*T(n,2k-1) - T(n,k).
T(n,2k+1) = n*T(n,2k).

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

Original entry on oeis.org

4, 9, 8, 16, 18, 12, 25, 32, 27, 16, 36, 50, 48, 36, 20, 49, 72, 75, 64, 45, 24, 64, 98, 108, 100, 80, 54, 28, 81, 128, 147, 144, 125, 96, 63, 32, 100, 162, 192, 196, 180, 150, 112, 72, 36, 121, 200, 243, 256, 245, 216, 175, 128, 81, 40, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44

Offset: 1

Bernard Schott, Mar 28 2022

When m and k are both positive integers and k | m, with m/k = n, then T(n,k) = S(m,k) = (m+k) + (m-k) + (m*k) + (m/k) = S(n*k,k) = n*(k+1)^2, problem proposed by Yakov Perelman.

All terms are nonsquarefree (A013929).

			Table begins:
  n \ k |   1      2      3      4      5      6      7      8      9     10
  ----------------------------------------------------------------------------
     1  |   4      9     16      25    36     49     64     81    100    121
     2  |   8     18     32      50    72     98    128    162    200    242
     3  |  12     27     48      75   108    147    192    243    300    363
     4  |  16     36     64     100   144    196    256    324    400    484
     5  |  20     45     80     125   180    245    320    405    500    605
     6  |  24     54     96     150   216    294    384    486    600    726
     7  |  28     63    112     175   252    343    448    567    700    847
     8  |  32     72    128     200   288    392    512    648    800    968
     9  |  36     81    144     225   324    441    576    729    900   1089
    10  |  40     90    160     250   360    490    640    810   1000   1210
  ............................................................................
T(3,4) = 75 = 3*(4+1)^2 corresponds to S(3*4,4) = S(12,4) = (12+4) + (12-4) + (12*4) + 12/4 = 75.
S(10,5) = (10+5) + (10-5) + (10*5) + (10/5) = T(10/5,5) = T(2,5) = 72.

I. Perelman, L'Algèbre Récréative, Chapitre IV, Les équations de Diophante, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101-102.
Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.

Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.
Wikipedia, Yakov Perelman.

Cf. A013929.
Cf. A000290 \ {0,1} (row 1), A001105 \ {0,2} (row 2), A033428 \ {0,3} (row 3), A016742 \ {0,4} (row 4), A033429 \ {0,5} (row 5), A033581 \ {0,6} (row 6).
Cf. A008586 \ {0} (column 1), A008591 \ {0} (column 2), A008598 \ {0} (column 3), A008607 \ {0} (column 4), A044102 \ {0} (column 5).
Cf. A045991 \ {0} (diagonal).

T[n_, k_] := n*(k + 1)^2; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Mar 29 2022 *)

T(n,k) = n*(k+1)^2.
T(n,n) = (n+1)^3 - (n+1)^2 = A045991(n+1) for n >= 1.
G.f.: x*(1 + y)/((1 - x)^2*(1 - y)^3). - Stefano Spezia, Mar 31 2022

A353435 Array read by descending antidiagonals: T(n,m) is the number of sequences of length n >= 0 with elements in 0..m-1 such that the Hankel matrix of any odd number of consecutive terms is invertible over the ring of integers modulo m >= 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 4, 0, 1, 1, 4, 4, 4, 0, 1, 1, 2, 16, 0, 4, 0, 1, 1, 6, 4, 48, 0, 0, 0, 1, 1, 4, 36, 0, 144, 0, 0, 0, 1, 1, 6, 16, 180, 0, 320, 0, 0, 0, 1, 1, 4, 36, 0, 900, 0, 720, 0, 0, 0, 1, 1, 10, 16, 108, 0, 3744, 0, 1312, 0, 0, 0, 1

Offset: 0

Pontus von Brömssen, Apr 21 2022

T(n,m) is divisible by T(2,m) = A127473(n) for n >= 2, because if r and s are coprime to m, the sequence (x_1, ..., x_n) satisfies the conditions if and only if the sequence (r*s^0*x_1 mod m, ..., r*s^(n-1)*x_n mod m) does.

			Array begins:
  n\m| 1  2  3  4    5  6        7  8   9 10
  ---+--------------------------------------
   0 | 1  1  1  1    1  1        1  1   1  1
   1 | 1  1  2  2    4  2        6  4   6  4
   2 | 1  1  4  4   16  4       36 16  36 16
   3 | 1  0  4  0   48  0      180  0 108  0
   4 | 1  0  4  0  144  0      900  0 324  0
   5 | 1  0  0  0  320  0     3744  0   0  0
   6 | 1  0  0  0  720  0    15552  0   0  0
   7 | 1  0  0  0 1312  0    54216  0   0  0
   8 | 1  0  0  0 2400  0   189468  0   0  0
   9 | 1  0  0  0 3232  0   550728  0   0  0
  10 | 1  0  0  0 4560  0  1604088  0   0  0
  11 | 1  0  0  0 4656  0  3895560  0   0  0
  12 | 1  0  0  0 4928  0  9467856  0   0  0
  13 | 1  0  0  0 4368  0 19185516  0   0  0

Cf. A035287, A045991, A350364, A353433, A353436.
Rows: A000012 (n=0), A000010 (n=1), A127473 (n=2).
Columns: A000012 (m=1), A130716 (m=2), A166926 (m=4 and m=6).

For fixed n, T(n,m) is multiplicative with T(n,p^e) = T(n,p)*p^(n*(e-1)).
T(n,m) = A353436(n,m) if m is prime.
T(3,m) = (m-1)^2*(m-2) = A045991(m-1) if m is prime.
T(4,m) = (m-1)^2*(m-2)^2 = A035287(m-1) if m is prime.
Empirically: T(5,m) = (m-1)^2*(m-3)*(m^2-4*m+5) if m >= 3 is prime.
T(n,2) = 0 for n >= 3.
T(n,3) = 0 for n >= 5.
T(n,5) = 0 for n >= 23.

A360205 Triangle read by rows. T(n, k) = (-1)^(n-k)(k+1)binomial(n, k)*pochhammer(1-n, n-k).

Original entry on oeis.org

1, 0, 2, 0, 4, 3, 0, 12, 18, 4, 0, 48, 108, 48, 5, 0, 240, 720, 480, 100, 6, 0, 1440, 5400, 4800, 1500, 180, 7, 0, 10080, 45360, 50400, 21000, 3780, 294, 8, 0, 80640, 423360, 564480, 294000, 70560, 8232, 448, 9, 0, 725760, 4354560, 6773760, 4233600, 1270080, 197568, 16128, 648, 10

Offset: 0

Peter Luschny, Feb 08 2023

A refinement of the number of partial permutations of an n-set (A002720).

Also the coefficients of a shifted derivative of the unsigned Lah polynomials (A271703).

			Triangle T(n, k) starts:
[0] 1;
[1] 0,     2;
[2] 0,     4,      3;
[3] 0,    12,     18,      4;
[4] 0,    48,    108,     48,      5;
[5] 0,   240,    720,    480,    100,     6;
[6] 0,  1440,   5400,   4800,   1500,   180,    7;
[7] 0, 10080,  45360,  50400,  21000,  3780,  294,   8;
[8] 0, 80640, 423360, 564480, 294000, 70560, 8232, 448, 9;

Cf. A052849 (column 1), A045991 (subdiagonal), A002720 (row sums), A271703.

Cf. A069138 (Stirling2 counterpart), A360174 (Stirling1 counterpart).

T := (n, k) -> (-1)^(n - k)*(k + 1)*binomial(n, k)*pochhammer(1 - n, n - k):
seq(seq(T(n, k), k = 0..n), n = 0..9);

A054026 a(n) is the number of sets of natural numbers [a,b,c,d,e] that can be produced with the numbers [0..n] such that the values of all the distinct parenthesized expressions of a-b-c-d-e are different.

Original entry on oeis.org

0, 0, 0, 0, 300, 1296, 4116, 9984, 21384, 40800, 72600, 120960, 192660, 294000, 434700, 623616, 873936, 1197504, 1611504, 2131200, 2778300, 3571920, 4538820, 5702400, 7095000, 8744736, 10690056, 12964224, 15612324, 18673200, 22199100, 26234880, 30840480, 36067200, 41983200, 48646656, 56134476

Offset: 0

Asher Auel, Jan 27 2000

There are 14 ways to put parentheses in the expression a - b - c - d - e: ((a - (b - c)) - d) - e, (((a - b) - c) - d) - e, ((a - b) - (c - d)) - e, etc. This sequence describes how many sets of natural numbers [a,b,c,d,e] can be produced with the numbers {0,1,2,3,...,n} such that the values of all the distinct expressions are different.

It can be shown that in the set of expressions obtained this way, for any number of variables, a is always positive, b is always negative, and the other variables appear with every possible combination of signs. Therefore, the valid k-tuples of numbers in [0..n] are precisely those such that every subset of {c,d,e,...}, including the empty subset, has a distinct sum. For 5 variables, there are n*(n-1)*(n-2) ways to choose distinct, nonzero values for c, d, and e. For each k, there are floor((n-1)/2) ways to choose distinct numbers x and y in [0..n] such that x + y = k. Summing over all k in [0..n], allowing arbitrary permutations of {x,y,k}, and allowing a and b to be any value gives the formula below. - Charlie Neder, Jan 13 2019

			For example, no such sets can be produced with only 0's, only 0's and 1's, only 0's and 1's and 2's, only 1's and 2's and 3's; with {0,1,2,3,4}, 300 such sets can be produced.

Charlie Neder, Table of n, a(n) for n = 0..1000
Index entries for sequences related to parenthesizing
Index entries for linear recurrences with constant coefficients, signature (3,0,-8,6,6,-8,0,3,-1).

Cf. A045991 (similar for a - b - c), A047929 (similar for a - b - c - d).

LinearRecurrence[{3,0,-8,6,6,-8,0,3,-1},{0,0,0,0,300,1296,4116,9984,21384},40] (* Harvey P. Dale, May 25 2025 *)

a(n) = (1+n)^2*(3*(-1)^n+4*n^3-18*n^2+20*n-3)/4; \\ Jinyuan Wang, Jun 27 2020

a(n) = (n+1)^2 * (n*(n-1)*(n-2) - 6*A002620(n-1)). - Charlie Neder, Jan 13 2019

a(9)-a(36) from Charlie Neder, Jan 13 2019

Incorrect formula removed by Jinyuan Wang, Jun 27 2020

cp's OEIS Frontend

A219605 Square array T(n,k), read by antidiagonals: T(n,2*k) = T(n,2*k-1)*n, T(n,2*k+1) = T(n,2*k)+n, T(n,0) = 1.

Views

Author

Keywords

Examples

Programs

Mathematica

Formula

A244309 a(n) = F(n)^3 - F(n)^2, where F(n) is the n-th Fibonacci number (A000045).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A244310 a(n) = L(n)^3 - L(n)^2, where L(n) is the n-th Lucas number (A000032).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A286932 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of continued fraction 1/(1 + k*x/(1 + k*x^2/(1 + k*x^3/(1 + k*x^4/(1 + k*x^5/(1 + ...)))))).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A322550 Table read by ascending antidiagonals: T(n, k) is the minimum number of cubes necessary to fill a right square prism with base area n^2 and height k.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

Maxima

PARI

Formula

A342239 Table read by upward antidiagonals: T(n,k) is the number of strings of length k over an n-letter alphabet that are bifix free; n >= 1, k >= 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A353435 Array read by descending antidiagonals: T(n,m) is the number of sequences of length n >= 0 with elements in 0..m-1 such that the Hankel matrix of any odd number of consecutive terms is invertible over the ring of integers modulo m >= 1.

Views

A219605 Square array T(n,k), read by antidiagonals: T(n,2k) = T(n,2k-1)n, T(n,2k+1) = T(n,2*k)+n, T(n,0) = 1.

A286932 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of continued fraction 1/(1 + kx/(1 + kx^2/(1 + kx^3/(1 + kx^4/(1 + k*x^5/(1 + ...)))))).

A360205 Triangle read by rows. T(n, k) = (-1)^(n-k)(k+1)binomial(n, k)*pochhammer(1-n, n-k).