A027465 -id:A027465 - cp's OEIS frontend

A027467 Triangle whose (n,k)-th entry is 15^(n-k)*binomial(n,k).

1, 15, 1, 225, 30, 1, 3375, 675, 45, 1, 50625, 13500, 1350, 60, 1, 759375, 253125, 33750, 2250, 75, 1, 11390625, 4556250, 759375, 67500, 3375, 90, 1, 170859375, 79734375, 15946875, 1771875, 118125, 4725, 105, 1, 2562890625, 1366875000, 318937500, 42525000, 3543750, 189000, 6300, 120, 1

Offset: 0

Olivier Gérard

			Triangle begins:
           1;
          15,          1;
         225,         30,         1;
        3375,        675,        45,        1;
       50625,      13500,      1350,       60,       1;
      759375,     253125,     33750,     2250,      75,      1;
    11390625,    4556250,    759375,    67500,    3375,     90,    1;
   170859375,   79734375,  15946875,  1771875,  118125,   4725,  105,   1;
  2562890625, 1366875000, 318937500, 42525000, 3543750, 189000, 6300, 120, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Sequences of the form q^(n-k)*binomial(n, k): A007318 (q=1), A038207 (q=2), A027465 (q=3), A038231 (q=4), A038243 (q=5), A038255 (q=6), A027466 (q=7), A038279 (q=8), A038291 (q=9), A038303 (q=10), A038315 (q=11), A038327 (q=12), A133371 (q=13), A147716 (q=14), this sequence (q=15).

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

Table[Binomial[n,k]15^(n-k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Dec 31 2017 *)

flatten([[(15)^(n-k)*binomial(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 12 2021

Numerators of lower triangle of (a[i,j])^4 where a[i,j] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i.

Sum_{k=0..n} T(n,k)*x^k = (15 + x)^n.

Simpler definition from Philippe Deléham, Nov 10 2008

A038243 Triangle whose (i,j)-th entry is 5^(i-j)*binomial(i,j).

Original entry on oeis.org

1, 5, 1, 25, 10, 1, 125, 75, 15, 1, 625, 500, 150, 20, 1, 3125, 3125, 1250, 250, 25, 1, 15625, 18750, 9375, 2500, 375, 30, 1, 78125, 109375, 65625, 21875, 4375, 525, 35, 1, 390625, 625000, 437500, 175000, 43750, 7000, 700, 40, 1, 1953125, 3515625, 2812500, 1312500, 393750, 78750, 10500, 900, 45, 1

Offset: 0

N. J. A. Sloane

Mirror image of A013612. - Zerinvary Lajos, Nov 25 2007
T(i,j) is the number of i-permutations of 6 objects a,b,c,d,e,f, with repetition allowed, containing j a's. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (5+x)^n - N-E. Fahssi, Apr 13 2008
Also the convolution triangle of A000351. - Peter Luschny, Oct 09 2022

			Triangle begins as:
       1;
       5,      1;
      25,     10,      1;
     125,     75,     15,      1;
     625,    500,    150,     20,     1;
    3125,   3125,   1250,    250,    25,    1;
   15625,  18750,   9375,   2500,   375,   30,   1;
   78125, 109375,  65625,  21875,  4375,  525,  35,  1;
  390625, 625000, 437500, 175000, 43750, 7000, 700, 40, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
B. N. Cyvin, J. Brunvoll, and S. J. Cyvin, Isomer enumeration of unbranched catacondensed polygonal systems with pentagons and heptagons, Match, No. 34 (Oct 1996), 109-121.

Cf. A000351, A000400 (row sums), A036071, A050982, A053464, A081135, A081143.

Sequences of the form q^(n-k)*binomial(n, k): A007318 (q=1), A038207 (q=2), A027465 (q=3), A038231 (q=4), this sequence (q=5), A038255 (q=6), A027466 (q=7), A038279 (q=8), A038291 (q=9), A038303 (q=10), A038315 (q=11), A038327 (q=12), A133371 (q=13), A147716 (q=14), A027467 (q=15).

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

for i from 0 to 8 do seq(binomial(i, j)*5^(i-j), j = 0 .. i) od; # Zerinvary Lajos, Dec 21 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 5^(n-1)); # Peter Luschny, Oct 09 2022

With[{q=5}, Table[q^(n-k)*Binomial[n,k], {n,0,12}, {k,0,n}]//Flatten] (* G. C. Greubel, May 12 2021 *)

flatten([[5^(n-k)*binomial(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 12 2021

See A038207 and A027465 and replace 2 and 3 in analogous formulas with 5. - Tom Copeland, Oct 26 2012

A304249 Triangle T(n,k) = 3*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1 and T(n,k) = 0 for n < 0 or k < 0, read by rows.

Original entry on oeis.org

1, 3, 9, 1, 27, 6, 81, 27, 1, 243, 108, 9, 729, 405, 54, 1, 2187, 1458, 270, 12, 6561, 5103, 1215, 90, 1, 19683, 17496, 5103, 540, 15, 59049, 59049, 20412, 2835, 135, 1, 177147, 196830, 78732, 13608, 945, 18, 531441, 649539, 295245, 61236, 5670, 189, 1

Offset: 0

Zagros Lalo, May 08 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n), cf. formula.
The coefficients in the expansion of 1/(1-3x-x^2) are given by the sequence generated by the row sums.
The sequence of the row sums are the "Bronze Fibonacci numbers" A006190, and the limit of their ratio is 3.30277563773... (Bronze ratio), see A098316.

			Triangle begins:
          1;
          3;
          9,         1;
         27,         6;
         81,        27,         1;
        243,       108,         9;
        729,       405,        54,        1;
       2187,      1458,       270,       12;
       6561,      5103,      1215,       90,        1;
      19683,     17496,      5103,      540,       15;
      59049,     59049,     20412,     2835,      135,       1;
     177147,    196830,     78732,    13608,      945,      18;
     531441,    649539,    295245,    61236,     5670,     189,      1;
    1594323,   2125764,   1082565,   262440,    30618,    1512,     21;
    4782969,   6908733,   3897234,  1082565,   153090,   10206,    252,    1;
   14348907,  22320522,  13817466,  4330260,   721710,   61236,   2268,   24;
   43046721,  71744535,  48361131, 16888014,  3247695,  336798,  17010,  324,  1;
  129140163, 229582512, 167403915, 64481508, 14073345, 1732104, 112266, 3240, 27;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 86, 363.

Zagros Lalo, Left justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1+3x)^n

Row sums give A006190.
Cf. A013610, A098316.
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), this sequence (q=2), A317497 (q=3), A318773 (q=4).

[3^(n-2*k)*Binomial(n-k,k): k in [0..Floor(n/2)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[0, 0] = 1; T[n_, k_]:= If[n<0 || k<0, 0, 3T[n-1, k] + T[n-2, k-1]];
Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}]//Flatten
With[{q=2}, Table[3^(n-q*k)*Binomial[n-(q-1)*k, k], {n,0,24}, {k,0,Floor[n/q]}] ]//Flatten (* G. C. Greubel, May 12 2021 *)

T(n,k)=if( n>0 && k>0, 3*T(n-1, k) + T(n-2, k-1), !n && !k)
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

flatten([[3^(n-2*k)*binomial(n-k,k) for k in (0..n//2)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = A013610(n-k, n-2k). - M. F. Hasler, Jun 01 2018

A099097 Riordan array (1, 3+x).

Original entry on oeis.org

1, 0, 3, 0, 1, 9, 0, 0, 6, 27, 0, 0, 1, 27, 81, 0, 0, 0, 9, 108, 243, 0, 0, 0, 1, 54, 405, 729, 0, 0, 0, 0, 12, 270, 1458, 2187, 0, 0, 0, 0, 1, 90, 1215, 5103, 6561, 0, 0, 0, 0, 0, 15, 540, 5103, 17496, 19683, 0, 0, 0, 0, 0, 1, 135, 2835, 20412, 59049, 59049, 0, 0, 0, 0, 0, 0, 18, 945, 13608, 78732, 196830, 177147

Offset: 0

Paul Barry, Sep 25 2004

Row sums are A006190(n+1). Diagonal sums are A052931. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).

Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/3, -1/3, 0, 0, 0, 0, 0, ...] DELTA [3, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2008

			Triangle begins:
  1;
  0, 3;
  0, 1, 9;
  0, 0, 6, 27;
  0, 0, 1, 27,  81;
  0, 0, 0,  9, 108, 243;
  ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A027465.

Diagonals are of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Table[3^(2*k-n)*Binomial[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 19 2021 *)

flatten([[3^(2*k-n)*binomial(k, n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 19 2021

Triangle: T(n, k) = binomial(k, n-k)*3^k*(1/3)^(n-k).
G.f. of column k: (3*x + x^2)^k.
G.f.: 1/(1 - 3*y*x - y*x^2). - Philippe Deléham, Nov 21 2011
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A006190(n+1), A135030(n+1), A181353(n+1) for x = 0,1,2,3 respectively. - Philippe Deléham, Nov 21 2011

A317496 Triangle T(n,k) = T(n-1,k) + 3*T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1, T(n,k) = 0 for n or k < 0, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 6, 1, 9, 1, 12, 9, 1, 15, 27, 1, 18, 54, 1, 21, 90, 27, 1, 24, 135, 108, 1, 27, 189, 270, 1, 30, 252, 540, 81, 1, 33, 324, 945, 405, 1, 36, 405, 1512, 1215, 1, 39, 495, 2268, 2835, 243, 1, 42, 594, 3240, 5670, 1458, 1, 45, 702, 4455, 10206, 5103, 1, 48, 819, 5940, 17010, 13608, 729

Offset: 0

Zagros Lalo, Jul 31 2018

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n) and  along a "second layer" of 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-3x^3) 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 1.863706527819..., when n approaches infinity.

			Triangle begins:
  1;
  1;
  1;
  1,  3;
  1,  6;
  1,  9;
  1, 12,   9;
  1, 15,  27;
  1, 18,  54;
  1, 21,  90,   27;
  1, 24, 135,  108;
  1, 27, 189,  270;
  1, 30, 252,  540,    81;
  1, 33, 324,  945,   405;
  1, 36, 405, 1512,  1215;
  1, 39, 495, 2268,  2835,   243;
  1, 42, 594, 3240,  5670,  1458;
  1, 45, 702, 4455, 10206,  5103;
  1, 48, 819, 5940, 17010, 13608, 729;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 364-366.

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

Row sums give A084386.
Cf. A013610, A027465, A304236, A304249, A317497.
Sequences of the form 3^k*binomial(n-(q-1)*k, k): A013610 (q=1), A304236 (q=2), this sequence (q=3), A318772 (q=4).

Flat(List([0..20],n->List([0..Int(n/3)],k->3^k/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Aug 01 2018

[3^k*Binomial(n-2*k,k): k in [0..Floor(n/3)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^k*(n-2*k)!/((n-3*k)!*k!); Table[T[n, k], {n, 0, 18}, {k, 0, Floor[n/3]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, T[n-1, k] + 3T[n-3, k-1]]; Table[T[n, k], {n, 0, 18}, {k, 0, Floor[n/3]}]//Flatten

flatten([[3^k*binomial(n-2*k,k) for k in (0..n//3)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^k * (n-2*k)!/ (k! * (n-3*k)!) where n is a nonnegative integer and k = 0..floor(n/3).

A317497 Triangle T(n,k) = 3*T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3) 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, 1, 81, 6, 243, 27, 729, 108, 1, 2187, 405, 9, 6561, 1458, 54, 19683, 5103, 270, 1, 59049, 17496, 1215, 12, 177147, 59049, 5103, 90, 531441, 196830, 20412, 540, 1, 1594323, 649539, 78732, 2835, 15, 4782969, 2125764, 295245, 13608, 135, 14348907, 6908733, 1082565, 61236, 945, 1

Offset: 0

Zagros Lalo, Jul 31 2018

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and  along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in 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/(1-3x-x^3) are given by the sequence generated by the row sums.
The row sums give A052541.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.1038034027355..., when n approaches infinity.

			Triangle begins:
         1;
         3;
         9;
        27,        1;
        81,        6;
       243,       27;
       729,      108,       1;
      2187,      405,       9;
      6561,     1458,      54;
     19683,     5103,     270,      1;
     59049,    17496,    1215,     12;
    177147,    59049,    5103,     90;
    531441,   196830,   20412,    540,    1;
   1594323,   649539,   78732,   2835,   15;
   4782969,  2125764,  295245,  13608,  135;
  14348907,  6908733, 1082565,  61236,  945,  1;
  43046721, 22320522, 3897234, 262440, 5670, 18;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 364-366

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

Row sums give A052541.
Cf. A013610, A027465, A317496, A304236, A304249.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3), A036217 (column 4).
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), this sequence (q=3), A318773 (q=4).

Flat(List([0..16],n->List([0..Int(n/3)],k->3^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Aug 01 2018

[3^(n-3*k)*Binomial(n-2*k,k): k in [0..Floor(n/3)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^(n-3k)(n-2k)!/((n-3k)! k!); Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3 T[n-1, k] + T[n-3, k-1]]; Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]}]//Flatten

flatten([[3^(n-3*k)*binomial(n-2*k,k) for k in (0..n//3)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^(n-3*k) * (n-2*k)!/(k! * (n-3*k)!) where n is a nonnegative integer and k = 0..floor(n/3).

A102741 a(n) = 3^4 * binomial(n+3, 4).

Original entry on oeis.org

81, 405, 1215, 2835, 5670, 10206, 17010, 26730, 40095, 57915, 81081, 110565, 147420, 192780, 247860, 313956, 392445, 484785, 592515, 717255, 860706, 1024650, 1210950, 1421550, 1658475, 1923831, 2219805, 2548665, 2912760, 3314520, 3756456, 4241160, 4771305, 5349645

Offset: 1

Zerinvary Lajos, Aug 06 2008

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A027465.

Sequences of the form 3^m*binomial(n+m-1, m): A008585 (m=1), A027468 (m=2), A134171 (m=3), this sequence (m=4), A113335 (m=5).

[3^4*Binomial(n+3,4): n in [1..30]]; // G. C. Greubel, May 17 2021

Maple
```
seq(binomial(n+3,4)*3^4, n=1..27);
```

With[{c=3^4},Table[c Binomial[n+3,4],{n,40}]]  (* Harvey P. Dale, Mar 12 2011 *)

[3^4*binomial(n+3,4) for n in (1..30)] # G. C. Greubel, May 17 2021

G.f.: 81*x/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
E.g.f.: (27/8)*x*(24 + 36*x + 12*x^2 + x^3)*exp(x). - G. C. Greubel, May 17 2021
From Amiram Eldar, Aug 28 2022: (Start)
Sum_{n>=1} 1/a(n) = 4/243.
Sum_{n>=1} (-1)^(n+1)/a(n) = 32*log(2)/81 - 64/243. (End)

A127543 Triangle T(n,k), 0<=k<=n, read by rows given by :[ -1,1,1,1,1,1,1,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, -1, 1, -1, 1, -2, 0, 2, -1, 1, -6, 2, 1, 3, -1, 1, -18, 5, 7, 2, 4, -1, 1, -57, 17, 19, 13, 3, 5, -1, 1, -186, 56, 64, 36, 20, 4, 6, -1, 1, -622, 190, 212, 124, 56, 28, 5, 7, -1, 1, -2120, 654, 722, 416, 198, 79, 37, 6, 8, -1, 1, -7338, 2282, 2494, 1434, 673, 287, 105, 47, 7, 9, -1, 1

Offset: 0

Philippe Deléham, Apr 01 2007

Riordan array (2/(3-sqrt(1-4*x)), (1-sqrt(1-4*x))/(3-sqrt(1-4*x))). - Philippe Deléham, Jan 27 2014

			Triangle begins:
    1;
   -1,  1;
    0, -1,  1;
   -1,  1, -1,  1;
   -2,  0,  2, -1,  1;
   -6,  2,  1,  3, -1,  1;
  -18,  5,  7,  2,  4, -1,  1;
  -57, 17, 19, 13,  3,  5, -1, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

A065600[n_, k_]:= If[k==n, 1, Sum[j*Binomial[k+j, j]*Binomial[2*(n-k-j), n-k]/(n-k-j), {j,0, Floor[(n-k)/2]}]];
A127543[n_, k_]:= A065600[n-1,k-1] - A065600[n-1,k];
Table[A127543[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 17 2021 *)

def A065600(n,k): return 1 if (k==n) else sum( j*binomial(k+j, j)*binomial(2*(n-k-j), n-k)/(n-k-j) for j in (0..(n-k)//2) )
def A127543(n,k): return A065600(n-1, k-1) - A065600(n-1, k)
flatten([[A127543(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 17 2021

T(n,k) = A065600(n-1,k-1) - A065600(n-1,k).
Sum_{k=0..n} T(n,k)*x^k = A127053(n), A126985(n), A127016(n), A127017(n), A126987(n), A126986(n), A126982(n), A126984(n), A126983(n), A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for n= -8,-7,...,8,9 respectively.
Sum_{j>=0} T(n,j)*A007318(j,k) = A106566(n,k).
Sum_{j>=0} T(n,j)*A038207(j,k) = A039599(n,k).
Sum_{j>=0} T(n,j)*A027465(j,k) = A116395(n,k).

A134171 a(n) = (9/2)(n-1)(n-2)*(n-3).

Original entry on oeis.org

0, 0, 0, 27, 108, 270, 540, 945, 1512, 2268, 3240, 4455, 5940, 7722, 9828, 12285, 15120, 18360, 22032, 26163, 30780, 35910, 41580, 47817, 54648, 62100, 70200, 78975, 88452, 98658, 109620, 121365, 133920, 147312, 161568, 176715, 192780, 209790, 227772, 246753

Offset: 1

N. J. A. Sloane, Jan 30 2008

Number of n permutations (n>=3) of 4 objects u, v, z, x with repetition allowed, containing n-3=0 u's. Example: if n=3 then n-3 =zero u, a()=27 because we have vzx, vxz, zvx, zxv, xvz, xzv, vvv, zzz, xxx, vvx, vxv, xvv, xxv, xvx, vxx, vvz, vzv, zvv, zzv, zvz, vzz, xzz, zxz, zzx, xxz, xzx, zxx. A027465 formatted as a triangular array: diagonal: 27, 108, 270, 540, 945, 1512. - Zerinvary Lajos, Aug 06 2008

G. C. Greubel, Table of n, a(n) for n = 1..1000
D. Zvonkine, Home Page.
D. Zvonkine, Counting ramified coverings and intersection theory on Hurwitz spaces II (local structure of Hurwitz spaces and combinatorial results), Moscow Mathematical Journal, Vol. 7, No. 1 (2007), 135-162.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A008585, A027465, A027468. - Zerinvary Lajos, Aug 06 2008

Cf. A000292, A102741, A113335.

[(9/2)*(n-1)*(n-2)*(n-3) : n in [1..50]]; // Wesley Ivan Hurt, May 29 2016

seq(27*binomial(n-1, 3), n=1..30); # Zerinvary Lajos, May 18 2008

LinearRecurrence[{4,-6,4,-1}, {0,0,0,27}, 50] (* G. C. Greubel, May 29 2016 *)

a(n) = 27 * binomial(n-1,3). - Zerinvary Lajos, Aug 06 2008
From Chai Wah Wu, May 29 2016: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4.
G.f.: 27*x^4/(1-x)^4. (End)
E.g.f.: 27 + (9/2*(x^3-3*x^2+6*x-6))*exp(x). - G. C. Greubel, May 17 2021
a(n) = 27 * A000292(n-3) for n >= 3. - Alois P. Heinz, May 17 2021
From Amiram Eldar, Sep 24 2022: (Start)
Sum_{n>=4} 1/a(n) = 1/18.
Sum_{n>=4} (-1)^n/a(n) = 4*log(2)/9 - 5/18. (End)

A136215 Triangle T, read by rows, where T(n,k) = A007559(n-k)*C(n,k) where A007559 equals the triple factorials in column 0.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 28, 12, 3, 1, 280, 112, 24, 4, 1, 3640, 1400, 280, 40, 5, 1, 58240, 21840, 4200, 560, 60, 6, 1, 1106560, 407680, 76440, 9800, 980, 84, 7, 1, 24344320, 8852480, 1630720, 203840, 19600, 1568, 112, 8, 1, 608608000, 219098880, 39836160

Offset: 0

Paul D. Hanna, Feb 07 2008

Comments from Peter Bala, Jul 10 2008: (Start) This array is the particular case P(1,3) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below
n\k|0....................1...............2.........3.....4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
See A094587 for some general properties of these arrays.
Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159 and P(2,3) = A136216. (End)
The generalized Pascal matrix that Bala refers to is itself a special case of application of the formalism of A133314 to fundamental matrices derived from infinitesimal generators described in A133314, of which the fundamental Pascal (A007318), unsigned Lah (A105278) and associated Laguerre (A135278) matrices are special examples. The formalism gives, among other relations, the inverse of T as TI(n,k) = b(n-k)*C(n,k) where the sequence b is given by the list partition transform (A133314) of A007559; i.e., b = LPT(A007559) = (1,-A008544)= (1,-1,-2,-10,-80,...). The formalism of A132382 may also be applied with the double factorial A001147 replaced by the triple factorial A007559 (see also A133480). - Tom Copeland, Aug 18 2008
From Peter Bala, Aug 29 2013: (Start)
Exponential Riordan array [1/(1 - 3*y)^(1/3), y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = sum {k = 0..n} binomial(n,k)*y^(n-k)*R(k,x).
Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = product {k = 0..n-1} (x + 3*k) with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x).
For example, row 3 is (28, 12, 3, 1) so P(3,x + 1) = (x + 1)*(x + 4)*(x + 7) = 28 + 12*x + 3*x*(x + 3) + x*(x + 3)*(x + 6). (End)

			Column k of T = column 0 of U^(k+1), while
column k of U = column 0 of T^(3k+1) where U = A136214 and
column k of V = column 0 of T^(3k+2) where V = A112333.
This triangle T begins:
        1;
        1,      1;
        4,      2,     1;
       28,     12,     3,    1;
      280,    112,    24,    4,   1;
     3640,   1400,   280,   40,   5,  1;
    58240,  21840,  4200,  560,  60,  6, 1;
  1106560, 407680, 76440, 9800, 980, 84, 7, 1; ...
Triangle U = A136214 begins:
     1;
     1,    1;
     4,    4,   1;
    28,   28,   7,   1;
   280,  280,  70,  10,  1;
  3640, 3640, 910, 130, 13, 1; ...
with triple factorials A007559 in column 0.
Triangle V = A112333 begins:
      1;
      2,    1;
     10,    5,    1;
     80,   40,    8,   1;
    880,  440,   88,  11,  1;
  12320, 6160, 1232, 154, 14, 1; ...
with triple factorials A008544 in column 0.

G. C. Greubel, Rows n=0..100 of triangle, flattened
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3
Wikipedia, Sheffer sequence

Cf. A136216 (matrix square); A007559, A008544; A136212, A136213.

Cf. A094587.

T[n_, k_]:= Binomial[n, k]*If[n - k == 0, 1, Product[3*j + 1, {j, 0, n - k - 1}]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 10 2018 *)

T(n,k)=binomial(n,k)*if(n-k==0,1,prod(j=0,n-k-1,3*j+1))

Column k of T = column 0 of U^(k+1) (matrix power) for k>=0 where U = A136214. Matrix square equals A136216, where A136216(n,k) = A008544(n-k)*C(n,k) where A008544 are also triple factorials.
From Peter Bala, Jul 10 2008: (Start)
T(n,k) = (3*n-3*k-2)*T(n-1,k) + T(n-1,k-1).
E.g.f. exp(x*y)/(1-3*y)^(1/3) = 1 + (1+x)*y + (4+2*x+x^2)*y^2/2! + ... . (End)

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A134171 a(n) = (9/2)(n-1)(n-2)*(n-3).