A013610 -id:A013610 - cp's OEIS frontend

A081583 Third row of Pascal-(1,2,1) array A081577.

1, 10, 46, 136, 307, 586, 1000, 1576, 2341, 3322, 4546, 6040, 7831, 9946, 12412, 15256, 18505, 22186, 26326, 30952, 36091, 41770, 48016, 54856, 62317, 70426, 79210, 88696, 98911, 109882, 121636, 134200, 147601, 161866, 177022, 193096

Offset: 0

Paul Barry, Mar 23 2003

Equals binomial transform of [1, 9, 27, 27, 0, 0, 0, ...] where (1, 9, 27, 27) = row 3 of triangle A013610. - Gary W. Adamson, Jul 19 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A013610, A038764, A081584, A081577.

[(2+9*n+9*n^3)/2: n in [0..40]]; // Vincenzo Librandi, Aug 09 2013

seq((2+9*n+9*n^3)/2, n=0..40); # G. C. Greubel, May 25 2021

CoefficientList[Series[(1+2x)^3/(1-x)^4, {x,0,50}], x] (* Vincenzo Librandi, Aug 09 2013 *)
LinearRecurrence[{4,-6,4,-1},{1,10,46,136},60] (* Harvey P. Dale, Oct 01 2021 *)

a = lambda n: hypergeometric([-n, -3], [1], 3)
[simplify(a(n)) for n in range(36)] # Peter Luschny, Nov 19 2014

a(n) = (2 + 9*n + 9*n^3)/2.
G.f.: (1+2*x)^3/(1-x)^4.
a(n) = hypergeommetric2F1([-n, -3], [1], 3). - Peter Luschny, Nov 19 2014
E.g.f.: (1/2)*(2 + 18*x + 27*x^2 + 9*x^3)*exp(x). - G. C. Greubel, May 25 2021

A318772 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 3 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 6, 1, 9, 1, 12, 1, 15, 9, 1, 18, 27, 1, 21, 54, 1, 24, 90, 1, 27, 135, 27, 1, 30, 189, 108, 1, 33, 252, 270, 1, 36, 324, 540, 1, 39, 405, 945, 81, 1, 42, 495, 1512, 405, 1, 45, 594, 2268, 1215, 1, 48, 702, 3240, 2835, 1, 51, 819, 4455, 5670, 243, 1, 54, 945, 5940, 10206, 1458

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)
The coefficients in the expansion of 1/(1-x-3*x^4) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.6580980673722..., when n approaches infinity.

			Triangle begins:
  1;
  1;
  1;
  1;
  1,  3;
  1,  6;
  1,  9;
  1, 12;
  1, 15,   9;
  1, 18,  27;
  1, 21,  54;
  1, 24,  90;
  1, 27, 135,   27;
  1, 30, 189,  108;
  1, 33, 252,  270;
  1, 36, 324,  540;
  1, 39, 405,  945,   81;
  1, 42, 495, 1512,  405;
  1, 45, 594, 2268, 1215;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Row sums give A318774.
Cf. A013610, A027465.
Cf. A304236, A304249.
Cf. A317496, A317497.

[3^k*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k]= 3^k(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 21}, {k, 0, Floor[n/4]} ] // Flatten
T[0, 0] = 1; T[n_, k_] := T[n, k] = If[n<0 || k<0, 0, T[n-1, k] + 3T[n-4, k-1]]; Table[T[n, k], {n, 0, 21}, {k, 0, Floor[n/4]}] // Flatten

flatten([[3^k*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^k * (n - 3*k)!/ ((n - 4*k)! k!), where n >= 0 and 0 <= k <= floor(n/4).

A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

Original entry on oeis.org

1, 3, 9, 27, 81, 1, 243, 6, 729, 27, 2187, 108, 6561, 405, 1, 19683, 1458, 9, 59049, 5103, 54, 177147, 17496, 270, 531441, 59049, 1215, 1, 1594323, 196830, 5103, 12, 4782969, 649539, 20412, 90, 14348907, 2125764, 78732, 540, 43046721, 6908733, 295245, 2835, 1, 129140163, 22320522, 1082565, 13608, 15

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)
The coefficients in the expansion of 1/(1-3*x-x^4) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.035744112294..., when n approaches infinity.

			Triangle begins:
          1;
          3;
          9;
         27;
         81,        1;
        243,        6;
        729,       27;
       2187,      108;
       6561,      405,       1;
      19683,     1458,       9;
      59049,     5103,      54;
     177147,    17496,     270;
     531441,    59049,    1215,     1;
    1594323,   196830,    5103,    12;
    4782969,   649539,   20412,    90;
   14348907,  2125764,   78732,   540;
   43046721,  6908733,  295245,  2835,   1;
  129140163, 22320522, 1082565, 13608,  15;
  387420489, 71744535, 3897234, 61236, 135;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Row sums give A052917.
Cf. A013610, A027465.
Cf. A304236, A304249
Cf. A317496, A317497.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3).
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), A317497 (q=3), this sequence (q=4).

[3^(n-4*k)*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^(n-4k)*(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3T[n-1, k] + T[n-4, k-1]]; Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]}]//Flatten

flatten([[3^(n-4*k)*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^(n-4*k) * (n-3*k)!/(k! * (n-4*k)!) where n >= 0 and 0 <= k <= floor(n/4).

A081584 Fourth row of Pascal-(1,2,1) array A081577.

Original entry on oeis.org

1, 13, 79, 307, 886, 2086, 4258, 7834, 13327, 21331, 32521, 47653, 67564, 93172, 125476, 165556, 214573, 273769, 344467, 428071, 526066, 640018, 771574, 922462, 1094491, 1289551, 1509613, 1756729, 2033032, 2340736, 2682136, 3059608

Offset: 0

Paul Barry, Mar 23 2003

Equals binomial transform of [1, 12, 54, 108, 81, 0, 0, 0, ...] where (1, 12, 54, 108, 81) = row 4 of triangle A013610. - Gary W. Adamson, Jul 19 2008

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

Cf. A013610, A081583.

[(8+6*n+81*n^2-18*n^3+27*n^4)/8: n in [0..40]]; // Vincenzo Librandi, Aug 09 2013

seq((8+6*n+81*n^2-18*n^3+27*n^4)/8, n=0..40); # G. C. Greubel, May 26 2021

CoefficientList[Series[(1+2x)^4/(1-x)^5, {x,0,40}], x] (* Vincenzo Librandi, Aug 09 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,13,79,307,886},40] (* Harvey P. Dale, Sep 18 2024 *)

[(8+6*n+81*n^2-18*n^3+27*n^4)/8 for n in (0..40)] # G. C. Greubel, May 26 2021

a(n) = (8 + 6*n + 81*n^2 - 18*n^3 + 27*n^4)/8.
G.f.: (1+2*x)^4/(1-x)^5.
E.g.f.: (1/8)*(8 + 96*x + 216*x^2 + 144*x^3 + 27*x^4)*exp(x). - G. C. Greubel, May 26 2021

A120743 a(n) = (1/2)(1 + 3i)^n + (1/2)(1 - 3i)^n where i = sqrt(-1).

Original entry on oeis.org

1, -8, -26, 28, 316, 352, -2456, -8432, 7696, 99712, 122464, -752192, -2729024, 2063872, 31417984, 42197248, -229785344, -881543168, 534767104, 9884965888, 14422260736, -70005137408, -284232882176, 131585609728, 3105500041216

Offset: 1

Creighton Dement, Jun 11 2007

From R. J. Mathar, Jun 15 2007: (Start)
These are the row sums of the triangle A013610 after every 2nd column is deleted, then every 2nd column reversed in sign, creating an intermediate irregular triangle with entries C(n,2*k)*(-9)^k, k = 0..floor(n/2):
  1;
  1,   -9;
  1,  -27;
  1,  -54,    81;
  1,  -90,   405;
  1, -135,  1215,    -729;
  1, -189,  2835,   -5103;
  1, -252,  5670,  -20412,   6561;
  1, -324, 10206,  -61236,  59049;
  1, -405, 17010, -153090, 295245, -59049; (End)
Floretion Algebra Multiplication Program, FAMP Code: 2tesseq[A*B] with A = + 1.5i' + .5j' + .5k' + .5e and B = 'ji' + e

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Index entries for linear recurrences with constant coefficients, signature (2, -10).

Cf. A006495

[ n eq 1 select 1 else n eq 2 select -8 else 2*Self(n-1) -10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 24 2011

LinearRecurrence[{2,-10}, {1,-8}, 30] (* G. C. Greubel, Nov 09 2018 *)

x='x+O('x^30); Vec((1-10*x)/(1-2*x+10*x^2)) \\ G. C. Greubel, Nov 09 2018

a(n) = 2*a(n-1) - 10*a(n-2).
G.f.: x*(1-10*x)/(10*x^2 - 2*x + 1).
a(n) mod 9 = 1. - Paul Curtz, Apr 20 2011
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(9*k+1)/(x*(9*k+10) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 29 2013
E.g.f.: exp(x)*cos(3*x). - Sergei N. Gladkovskii, May 29 2013
a(n) = A190958(n)-10*A190958(n-1). - R. J. Mathar, Dec 13 2022

a(0)=1: a(n) is main diagonal of A009116(n). - Paul Curtz, Jul 22 2011

Edited by Jon E. Schoenfield, Nov 09 2018

A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 10, 13, 25, 46, 76, 115, 190, 328, 556, 901, 1471, 2455, 4123, 6826, 11239, 18604, 30973, 51451, 85168, 140980, 233899, 388252, 643756, 1066696, 1768393, 2933149, 4864417, 8064505, 13369684, 22169131, 36762382, 60955897, 101064949, 167572342, 277859488, 460727179, 763922026, 1266639052

Offset: 0

Zagros Lalo, Sep 04 2018

The coefficients in the expansion of 1/(1 - x - 3*x^4) are given by the sequence generated by the row sums in triangle A318772.

Coefficients in expansion of 1/(1 - x - 3*x^4) are given by the sum of numbers along "third Layer" skew diagonals pointing top-right in triangle A013610 ((1+3x)^n) and by the sum of numbers along "third Layer" skew diagonals pointing top-left in triangle A027465 ((3+x)^n), see links.

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3 x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n
Index entries for linear recurrences with constant coefficients, signature (1,0,0,3).

Cf. A013610, A027465, A318772, A318773.

Essentially a duplicate of A143454.

[n le 4 select 1 else Self(n-1) +3*Self(n-4): n in [1..51]]; // G. C. Greubel, May 08 2021

CoefficientList[Series[1/(1-x-3x^4), {x, 0, 50}], x]
a[n_]:= a[n]= If[n<4, 1, a[n-1] + 3*a[n-4]]; Table[a[n], {n,0,50}]
LinearRecurrence[{1,0,0,3}, {1,1,1,1}, 51]

my(p=Mod('x,x^4-'x^3-3)); a(n) = vecsum(Vec(lift(p^n))); \\ Kevin Ryde, May 11 2021

def a(n): return 1 if (n<4) else a(n-1) + 3*a(n-4)
[a(n) for n in (0..50)] # G. C. Greubel, May 08 2021

a(n) = a(n-1) + 3*a(n-4) for n >= 0, a(n)=0 for n < 0, with a(0) = a(1) = a(2) = a(3) = 1.

A123187 Triangle of coefficients in expansion of (1+13x)^n.

Original entry on oeis.org

1, 1, 13, 1, 26, 169, 1, 39, 507, 2197, 1, 52, 1014, 8788, 28561, 1, 65, 1690, 21970, 142805, 371293, 1, 78, 2535, 43940, 428415, 2227758, 4826809, 1, 91, 3549, 76895, 999635, 7797153, 33787663, 62748517, 1, 104, 4732, 123032, 1999270, 20792408

Offset: 1

Roger L. Bagula and Gary W. Adamson, Oct 03 2006

T(n,k) equals the number of n-length words on {0,1,...,13} having n-k zeros. - Milan Janjic, Jul 24 2015

			1
1, 13
1, 26, 169
1, 39, 507, 2197
1, 52, 1014, 8788, 28561
1, 65, 1690, 21970, 142805, 371293

Cf. A013609, A013610, A013611, A013621, A038220, A038222.

T:= n-> (p-> seq(coeff(p, x, k), k=0..n))((1+13*x)^n):
seq(T(n), n=0..10);  # Alois P. Heinz, Jul 24 2015

p[0, x] = 1; p[1, x] = 13*x + 1; p[k_, x_] := p[k, x] = (13*x + 1)*p[k - 1, x]; w = Table[CoefficientList[p[n, x], x], {n, 0, 10}]; Flatten[w]

p(k, x) = (13*x + 1)*p(k - 1, x).

T(n,k) = 13^k*C(n,k) = Sum_{i=n-k..n} C(i,n-k)*C(n,i)*12^(n-i). Row sums are 14^n = A001023. G.f.: 1 / [1 - x(1+13y)]. - Mircea Merca, Apr 28 2012

A182042 Triangle T(n,k), read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (3, 0, -3/2, 3/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 3, 0, 6, 9, 0, 9, 27, 27, 0, 12, 54, 108, 81, 0, 15, 90, 270, 405, 243, 0, 18, 135, 540, 1215, 1458, 729, 0, 21, 189, 945, 2835, 5103, 5103, 2187, 0, 24, 252, 1512, 5670, 13608, 20412, 17496, 6561, 0, 27, 324, 2268, 10206, 30618, 61236, 78732, 59049, 19683

Offset: 0

Philippe Deléham, Apr 07 2012

Row sums are 4^n - 1 + 0^n.

Triangle of coefficients in expansion of (1+3*x)^n - 1 + 0^n.

			Triangle begins:
  1;
  0,  3;
  0,  6,   9;
  0,  9,  27,  27;
  0, 12,  54, 108,   81;
  0, 15,  90, 270,  405,  243;
  0, 18, 135, 540, 1215, 1458,  729;
  0, 21, 189, 945, 2835, 5103, 5103, 2187;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000244, A007318, A013610, A193193, A193194.

T:= proc(n, k) option remember;
      if k=n then 3^n
    elif k=0 then 0
    else binomial(n,k)*3^k
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Feb 17 2020

With[{m = 9}, CoefficientList[CoefficientList[Series[(1-2*x+x^2+3*y*x^2)/(1-2*x-3*y*x+x^2+3*y*x^2), {x, 0 , m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 17 2020 *)

T(n,k) = if (k==0, 1, binomial(n,k)*3^k);
matrix(10, 10, n, k, T(n-1,k-1)) \\ to see the triangle \\ Michel Marcus, Feb 17 2020

@CachedFunction
def T(n, k):
    if (k==n): return 3^n
    elif (k==0): return 0
    else: return binomial(n,k)*3^k
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 17 2020

T(n,0) = 0^n; T(n,k) = binomial(n,k)*3^k for k > 0.
G.f.: (1-2*x+x^2+3*y*x^2)/(1-2*x-3*y*x+x^2+3*y*x^2).
T(n,k) = 2*T(n-1,k) + 3*T(n-1,k-1) - T(n-2,k) -3*T(n-2,k-1), T(0,0) = 1, T(1,0) = T(2,0) = 0, T(1,1) = 3, T(2,1) = 6, T(2,2) = 9 and T(n,k) = 0 if k < 0 or if k > n.
T(n,k) = A206735(n,k)*3^k.
T(n,k) = A013610(n,k) - A073424(n,k).

a(48) corrected by Georg Fischer, Feb 17 2020

A267849 Triangular array: T(n,k) is the 2-row creation rook number to place k rooks on a 3 x n board.

Original entry on oeis.org

1, 1, 3, 1, 6, 12, 1, 9, 36, 60, 1, 12, 72, 240, 360, 1, 15, 120, 600, 1800, 2520, 1, 18, 180, 1200, 5400, 15120, 20160, 1, 21, 252, 2100, 12600, 52920, 141120, 181440, 1, 24, 336, 3360, 25200, 141120, 564480, 1451520, 1814400, 1, 27, 432, 5040, 45360, 317520, 1693440, 6531840, 16329600, 19958400

Offset: 0

Ken Joffaniel M. Gonzales, Jan 21 2016

T(n,k) is the number of ways to place k rooks in a 3 x n Ferrers board (or diagram) under the Goldman-Haglund i-row creation rook mode for i=2. All row heights are 3.

			The triangle T(n,k) begins in row n=0 with columns 0<=k<=n:
     1
     1      3
     1      6     12
     1      9     36     60
     1     12     72    240    360
     1     15    120    600   1800   2520
     1     18    180   1200   5400  15120  20160
     1     21    252   2100  12600  52920 141120 181440
     1     24    336   3360  25200 141120 564480 1451520 1814400
     1     27    432   5040  45360 317520 1693440 6531840 16329600 19958400

Jay Goldman and James Haglund, Generalized rook polynomials, J. Combin. Theory A 91 (2000), 509-530.

Cf. A013610 (1-rook coefficients on the 3xn board), A121757 (2-rook coeffs. on the 2xn board), A013609 (1-rook coeffs. on the 2xn board), A013611 (1-rook coeffs. on the 4xn board), A008279 (2-rook coeffs. on the 1xn board), A082030 (row sums?), A049598 (column k=2), A007531 (column k=3 w/o factor 10), A001710 (diagonal?).

T(n,k) = T(n-1,k) + (k+2) T(n-1,k-1) subject to T(0,0)=1, T(n,k)=0 for n

Triangle simplified (reversing rows, offset 0). - R. J. Mathar, May 03 2017

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A081583 Third row of Pascal-(1,2,1) array A081577.

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A318772 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 3 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

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A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

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A081584 Fourth row of Pascal-(1,2,1) array A081577.

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A120743 a(n) = (1/2)*(1 + 3*i)^n + (1/2)*(1 - 3*i)^n where i = sqrt(-1).

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A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

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A120743 a(n) = (1/2)(1 + 3i)^n + (1/2)(1 - 3i)^n where i = sqrt(-1).