A081135 -id:A081135 - cp's OEIS frontend

A081142 12th binomial transform of (0,0,1,0,0,0,...).

0, 0, 1, 36, 864, 17280, 311040, 5225472, 83607552, 1289945088, 19349176320, 283787919360, 4086546038784, 57954652913664, 811365140791296, 11234286564802560, 154070215745863680, 2095354934143746048

Offset: 0

Paul Barry, Mar 08 2003

Starting at 1, the three-fold convolution of A001021 (powers of 12).

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (36,-432,1728).

Cf. A001021.

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), this sequence (q=12), A027476 (q=15).

List([0..20],n->12^(n-2)*Binomial(n,2)); # Muniru A Asiru, Nov 24 2018

[12^(n-2)* Binomial(n, 2): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011

seq(coeff(series(x^2/(1-12*x)^3,x,n+1), x, n), n = 0 .. 20); # Muniru A Asiru, Nov 24 2018

LinearRecurrence[{36,-432,1728},{0,0,1},30] (* or *) Table[(n-1) (n-2) 3^(n-3) 2^(2n-7),{n,20}] (* Harvey P. Dale, Jul 25 2013 *)

vector(20, n, n--; 2^(2*n-5)*3^(n-2)*n*(n-1)) \\ G. C. Greubel, Nov 23 2018

[2^(2*n-5)*3^(n-2)*n*(n-1) for n in range(20)] # G. C. Greubel, Nov 23 2018

a(n) = 36*a(n-1) - 432*a(n-2) + 1728*a(n-3), a(0) = a(1) = 0, a(2) = 1.
a(n) = 12^(n-2)*binomial(n, 2).
G.f.: x^2/(1 - 12*x)^3.
a(n) = 2^(2*n-5)*3^(n-2)*n*(n-1). - Harvey P. Dale, Jul 25 2013
E.g.f.: (1/2)*exp(12*x)*x^2. - Franck Maminirina Ramaharo, Nov 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 24 - 264*log(12/11).
Sum_{n>=2} (-1)^n/a(n) = 312*log(13/12) - 24. (End)

A081143 5th binomial transform of (0,0,0,1,0,0,0,0,......).

Original entry on oeis.org

0, 0, 0, 1, 20, 250, 2500, 21875, 175000, 1312500, 9375000, 64453125, 429687500, 2792968750, 17773437500, 111083984375, 683593750000, 4150390625000, 24902343750000, 147857666015625, 869750976562500, 5073547363281250

Offset: 0

Paul Barry, Mar 08 2003

Starting at 1, four-fold convolution of A000351 (powers of 5).

With a different offset, number of n-permutations (n=4)of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly three u's. Example: a(4)=20 because we have uuuv, uuvu, uvuu, vuuu, uuuw, uuwu, uwuu, wuuu, uuuz, uuzu, uzuu, zuuu, uuux, uuxu, uxuu, xuuu, uuuy, uuyu, uyuu and yuuu. - Zerinvary Lajos, Jun 03 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (20,-150,500,-625).

Cf. A038846, A036216, A001789, A081144, A081135.

[5^(n-3) * Binomial(n, 3): n in [0..25]]; // Vincenzo Librandi, Aug 06 2013

seq(binomial(n,3)*5^(n-3), n=0..25); # Zerinvary Lajos, Jun 03 2008

CoefficientList[Series[x^3/(1-5x)^4, {x, 0, 30}], x] (* Vincenzo Librandi, Aug 06 2013 *)
LinearRecurrence[{20,-150,500,-625}, {0,0,0,1}, 30] (* Harvey P. Dale, Dec 24 2015 *)

vector(31, n, my(m=n-1); 5^(m-3)*binomial(m,3)) \\ G. C. Greubel, Mar 05 2020

[lucas_number2(n, 5, 0)*binomial(n,3)/5^3 for n in range(0, 22)] # Zerinvary Lajos, Mar 12 2009

a(n) = 20*a(n-1) - 150*a(n-2) + 500*a(n-3) - 625*a(n-4), with a(0)=a(1)=a(2)=0, a(3)=1.
a(n) = 5^(n-3)*binomial(n,3).
G.f.: x^3/(1-5*x)^4.
E.g.f.: x^3*exp(x)/6. - G. C. Greubel, Mar 05 2020
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=3} 1/a(n) = 240*log(5/4) - 105/2.
Sum_{n>=3} (-1)^(n+1)/a(n) = 540*log(6/4) - 195/2. (End)

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

A081130 Square array of binomial transforms of (0,0,1,0,0,0,...), read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 3, 0, 0, 0, 1, 6, 6, 0, 0, 0, 1, 9, 24, 10, 0, 0, 0, 1, 12, 54, 80, 15, 0, 0, 0, 1, 15, 96, 270, 240, 21, 0, 0, 0, 1, 18, 150, 640, 1215, 672, 28, 0, 0, 0, 1, 21, 216, 1250, 3840, 5103, 1792, 36, 0, 0, 0, 1, 24, 294, 2160, 9375, 21504, 20412, 4608, 45, 0

Offset: 0

Paul Barry, Mar 08 2003

Rows, of the square array, are three-fold convolutions of sequences of powers.

			The array begins as:
  0,  0,  0,   0,   0,    0, ...
  0,  0,  0,   0,   0,    0, ...
  0,  1,  1,   1,   1,    1, ... A000012
  0,  3,  6,   9,  12,   15, ... A008585
  0,  6, 24,  54,  96,  150, ... A033581
  0, 10, 80, 270, 640, 1250, ... A244729
The antidiagonal triangle begins as:
  0;
  0, 0;
  0, 0, 0;
  0, 0, 1, 0;
  0, 0, 1, 3,  0;
  0, 0, 1, 6,  6,  0;
  0, 0, 1, 9, 24, 10, 0;

G. C. Greubel, Antidiadoganal rows n = 0..50, flattened

Main diagonal: A081131.
Rows: A000012 (n=2), A008585 (n=3), A033581 (n=4), A244729 (n=5).
Columns: A000217 (k=1), A001788 (k=2), A027472 (k=3), A038845 (k=4), A081135 (k=5), A081136 (k=6), A027474 (k=7), A081138 (k=8), A081139 (k=9), A081140 (k=10), A081141 (k=11), A081142 (k=12), A027476 (k=15).

[k eq n select 0 else (n-k)^(k-2)*Binomial(k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, May 14 2021

Table[If[k==n, 0, (n-k)^(k-2)*Binomial[k, 2]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 14 2021 *)

T(n, k)=if (k==0, 0, k^(n-2)*binomial(n, 2));
seq(nn) = for (n=0, nn, for (k=0, n, print1(T(k, n-k), ", ")); );
seq(12) \\ Michel Marcus, May 14 2021

flatten([[0 if (k==n) else (n-k)^(k-2)*binomial(k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 14 2021

T(n, k) = k^(n-2)*binomial(n, 2), with T(n, 0) = 0 (square array).
T(n, n) = A081131(n).
Rows have g.f. x^3/(1-k*x)^n.
From G. C. Greubel, May 14 2021: (Start)
T(k, n-k) = (n-k)^(k-2)*binomial(k,2) with T(n, n) = 0 (antidiagonal triangle).
Sum_{k=0..n} T(n, n-k) = A081197(n). (End)

Term a(5) corrected by G. C. Greubel, May 14 2021

A084902 a(n) = 5^(n-1)n(n+1)/2.

Original entry on oeis.org

0, 1, 15, 150, 1250, 9375, 65625, 437500, 2812500, 17578125, 107421875, 644531250, 3808593750, 22216796875, 128173828125, 732421875000, 4150390625000, 23345947265625, 130462646484375, 724792480468750, 4005432128906250

Offset: 0

Paul Barry, Jun 10 2003

Binomial transform of A084901. Fourth binomial transform of heptagonal numbers A000566. Fifth binomial transform of (0,1,5,0,0,0,...).
Number of n-permutations of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly two u's. Example: a(2)=15 because we have uuw, uuv, uuz, uux, uuy, uwu, uvu, uzu, uxu, uyu, wuu, vuu, zuu, xuu, yuu. - Zerinvary Lajos, Dec 30 2007
A shifted version of A081135. - R. J. Mathar, Sep 17 2008

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (15,-75,125).

Cf. A000566, A038243, A081135, A084901.

[5^(n-1)*Binomial(n+1,2): n in [0..30]]; // G. C. Greubel, May 17 2021

Table[5^(n-1)n(n+1)/2,{n,0,30}] (* or *) LinearRecurrence[{15,-75,125},{0,1,15},30] (* Harvey P. Dale, Sep 18 2018 *)

a(n)=5^(n-1)*n*(n+1)/2 \\ Charles R Greathouse IV, Oct 07 2015

[5^(n-1)*binomial(n+1,2) for n in (0..30)] # G. C. Greubel, May 17 2021

G.f.: x/(1 - 5*x)^3.
E.g.f.: (x/2)*(2 + 5*x)*exp(5*x). - G. C. Greubel, May 17 2021
a(n) = 15*a(n-1) - 75*a(n-2) + 125*a(n-3). - Wesley Ivan Hurt, May 17 2021

A082151 A transform of C(n,2).

Original entry on oeis.org

0, 0, 1, 12, 102, 760, 5295, 35364, 228956, 1445616, 8936685, 54252220, 324214242, 1911205608, 11132579003, 64170616020, 366497915640, 2076171038176, 11676266706969, 65242364726124, 362433045180830, 2002838101907160, 11015341078090503, 60321223747375492

Offset: 0

Paul Barry, Apr 07 2003

Represents the mean of the second and fourth binomial transforms of C(n,2). Binomial transform of A082150

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (24,-237,1232,-3555,5400,-3375).

Cf. A027472, A081135, A000217.

[Binomial(n,2)*(3^(n-2) + 5^(n-2))/2: n in [0..30]]; // G. C. Greubel, Feb 10 2018

CoefficientList[Series[(x^2/(1-5*x)^3 + x^2/(1-3*x)^3)/2, {x,0,50}], x] (* or *) Table[Binomial[n,2]*(3^(n-2) + 5^(n-2))/2, {n,0,30}] (* G. C. Greubel, Feb 10 2018 *)
LinearRecurrence[{24,-237,1232,-3555,5400,-3375},{0,0,1,12,102,760},30] (* Harvey P. Dale, Apr 10 2023 *)

for(n=0,30, print1(binomial(n,2)*(3^(n-2) + 5^(n-2))/2, ", ")) \\ G. C. Greubel, Feb 10 2018

a(n) = C(n, 2)*(3^(n-2) + 5^(n-2))/2.
G.f.: (x^2/(1-5*x)^3 + x^2/(1-3*x)^3)/2.
a(n) = x^2*(76*x^3 - 51*x^2 + 12*x - 1)/((1-3*x)^3*(5*x-1)^3).
E.g.f.: x^2*exp(4*x)*cosh(x)/2.

A305837 Triangle read by rows: T(0,0) = 1; T(n,k) = 5*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 5, 25, 1, 125, 10, 625, 75, 1, 3125, 500, 15, 15625, 3125, 150, 1, 78125, 18750, 1250, 20, 390625, 109375, 9375, 250, 1, 1953125, 625000, 65625, 2500, 25, 9765625, 3515625, 437500, 21875, 375, 1, 48828125, 19531250, 2812500, 175000, 4375, 30, 244140625, 107421875, 17578125, 1312500, 43750, 525, 1

Offset: 0

Shara Lalo, Jun 11 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013612 ((1+5*x)^n).
The coefficients in the expansion of 1/(1-5x-x^2) 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 5.1925824035..., a metallic mean (see A098318), when n approaches infinity.

			Triangle begins:
            1;
            5;
           25,           1;
          125,          10;
          625,          75,          1;
         3125,         500,         15;
        15625,        3125,        150,         1;
        78125,       18750,       1250,        20;
       390625,      109375,       9375,       250,        1;
      1953125,      625000,      65625,      2500,       25;
      9765625,     3515625,     437500,     21875,      375,      1;
     48828125,    19531250,    2812500,    175000,     4375,     30;
    244140625,   107421875,   17578125,   1312500,    43750,    525,     1;
   1220703125,   585937500,  107421875,   9375000,   393750,   7000,    35;
   6103515625,  3173828125,  644531250,  64453125,  3281250,  78750,   700,  1;
  30517578125, 17089843750, 3808593750, 429687500, 25781250, 787500, 10500, 40;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 92, 380, 382.

Shara Lalo, Left-justified triangle
Shara Lalo, Skew diagonals in triangle A013612

Row sums give A052918.
Cf. A000351 (column 0), A053464 (column 1), A081135 (column 2), A081143 (column 3), A036071 (column 4).
Cf. A013612.
Cf. A098318.

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 5 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

G.f.: 1/(1 - 5*t*x - t^2).

A128799 a(n) = n(n-1)5^n.

Original entry on oeis.org

0, 0, 50, 750, 7500, 62500, 468750, 3281250, 21875000, 140625000, 878906250, 5371093750, 32226562500, 190429687500, 1110839843750, 6408691406250, 36621093750000, 207519531250000, 1167297363281250, 6523132324218750

Offset: 0

Mohammad K. Azarian, Apr 07 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (15,-75,125).

Cf. A007758, A036289, A081135.

[(n^2-n)*5^n: n in [0..20]]; // Vincenzo Librandi, Feb 10 2013

seq(5^n*n*(n-1), n=0..30); # G. C. Greubel, May 17 2021

CoefficientList[Series[50x^2/(1-5x)^3, {x, 0, 30}] ,x] (* Vincenzo Librandi, Feb 10 2013 *)
Table[n(n-1)5^n,{n,0,30}] (* or *) LinearRecurrence[{15,-75,125},{0,0,50},30] (* Harvey P. Dale, Nov 05 2019 *)

[2*5^n*binomial(n,2) for n in (0..30)] # G. C. Greubel, May 17 2021

G.f.: 50*x^2/(1 - 5*x)^3. - Vincenzo Librandi, Feb 10 2013
a(n) = 50*A081135(n). - R. J. Mathar, Apr 26 2015
E.g.f.: 25*x^2*exp(5*x). - G. C. Greubel, May 17 2021
a(n) = 15*a(n-1) - 75*a(n-2) + 125*a(n-3). - Wesley Ivan Hurt, May 17 2021

A362353 Triangle read by rows: T(n,k) = (-1)^(n-k)binomial(n, k)(k+3)^n, for n >= 0, and k = 0,1, ..., n. Coefficients of certain Sidi polynomials.

Original entry on oeis.org

1, -3, 4, 9, -32, 25, -27, 192, -375, 216, 81, -1024, 3750, -5184, 2401, -243, 5120, -31250, 77760, -84035, 32768, 729, -24576, 234375, -933120, 1764735, -1572864, 531441, -2187, 114688, -1640625, 9797760, -28824005, 44040192, -33480783, 10000000, 6561, -524288, 10937500, -94058496, 403536070, -939524096, 1205308188, -800000000, 214358881

Offset: 0

Harlan J. Brothers and Wolfdieter Lang, Apr 27 2023

This is the member N = 2 of a family of signed triangles with row sums n! = A000142(n): T(N; n, k) = (-1)^(n-k)*binomial(n, k)*(k + N + 1)^n, for integer N, n >= 0 and k = 0, 1, ..., n. The row polynomials PS(N; n, z) = Sum_{k=0..n} T(N; n, k)*z^k = ((-1)^n/z^N)*D_{n,N+1,n}(z) in [Sidi 1980].
For N = -1, 0 and 1 see A258773(n, k), A075513(n+1, k) and (-1)^(n-k) * A154715(n, k), respectively.
The column sequences, for k = 0, 1, ..., 6 and n >= k, are A141413(n+2), (-1)^(n+1)*A018215(n) = 4*(-1)^(n+1)*A002697(n), 5^2*(-1)^n*A081135(n), (-1)^(n+1)*A128964(n-1) = 6^3*(-1)^(n+1)*A081144(n), 7^4*(-1)^n*A139641(n-4), 2^15*(-1)^(n+1)*A173155(n-5), 3^12*(-1)^n*A173191(n-6), respectively.
The e.g.f. of the triangle (see below) needs the exponential convolution (LambertW(-z)/(-z))^2 = Sum_{n>=0} c(2; n)*z^n/n!, where c(2; n) = Sum_{m=0..n} |A137352(n+1, m)|*2^m = A007334(n+2).
The row sums give n! = A000142(n).

			The triangle T begins:
n\k    0       1        2         3         4          5          6         7
0:     1
1:    -3       4
2:     9     -32       25
3:   -27     192     -375       216
4:    81   -1024     3750     -5184      2401
5:  -243    5120   -31250     77760    -84035      32768
6:   729  -24576   234375   -933120   1764735   -1572864     531441
7: -2187  114688 -1640625   9797760 -28824005   44040192  -33480783  10000000
...
n = 8:  6561 -524288 10937500 -94058496 403536070 -939524096 1205308188 -800000000 2143588,
n = 9: -19683 2359296 -70312500 846526464 -5084554482 16911433728 -32543321076 36000000000 -21221529219 5159780352.

Paolo Xausa, Table of n, a(n) for n = 0..5049 (rows 0..100 of the triangle, flattened)
Wolfdieter Lang, On a Certain Family of Sidi Polynomials, May 2023.
Avram Sidi, Numerical Quadrature and Nonlinear Sequence Transformations; Unified Rules for Efficient Computation of Integrals with Algebraic and Logarithmic Endpoint Singularities, Math. Comp., 35 (1980), 851-874.

Cf. A000142 (row sums), A075513, A154715, A258773.

Columns k = 0..6 involve (see above): A002697, A007334, A018215, A081135, A081144, A128964, A137352, A139641, A141413, A173155, A173191.

A362353row[n_]:=Table[(-1)^(n-k)Binomial[n,k](k+3)^n,{k,0,n}];Array[A362353row,10,0] (* Paolo Xausa, Jul 30 2023 *)

T(n, k) = (-1)^(n-k)*binomial(n, k)*(k + 3)^n, for n >= 0, k = 0, 1, ..., n.
O.g.f. of column k: (x*(k + 3))^k/(1 - (k + 3)*x)^(k+1), for k >= 0.
E.g.f. of column k: exp(-(k + 3)*x)*((k + 3)*x)^k/k!, for k >= 0.
E.g.f. of the triangle, that is, the e.g.f. of its row polynomials {PS(2;n,y)}_{n>=0}): ES(2;y,x) = exp(-3*x)*(1/3)*(d/dz)(W(-z)/(-z))^2, after replacing z by x*y*exp(-x), where W is the Lambert W-function for the principal branch. This becomes ES(2;y,x) = exp(-3*x)*exp(3*(-W(-z)))/(1 - (-W(-z)), with z = x*y*exp(-x).

a(41)-a(44) from Paolo Xausa, Jul 31 2023

cp's OEIS Frontend

A081142 12th binomial transform of (0,0,1,0,0,0,...).

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A081143 5th binomial transform of (0,0,0,1,0,0,0,0,......).

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

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A081130 Square array of binomial transforms of (0,0,1,0,0,0,...), read by antidiagonals.

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A084902 a(n) = 5^(n-1)*n*(n+1)/2.

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A082151 A transform of C(n,2).

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A084902 a(n) = 5^(n-1)n(n+1)/2.

A128799 a(n) = n(n-1)5^n.

A362353 Triangle read by rows: T(n,k) = (-1)^(n-k)binomial(n, k)(k+3)^n, for n >= 0, and k = 0,1, ..., n. Coefficients of certain Sidi polynomials.