A081139 -id:A081139 - 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)

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

A173187 a(n) = binomial(n + 3, 3)*9^n.

Original entry on oeis.org

1, 36, 810, 14580, 229635, 3306744, 44641044, 573956280, 7102708965, 85232507580, 997220338686, 11422705697676, 128505439098855, 1423444863864240, 15556218869373480, 168007163789233584

Offset: 0

Zerinvary Lajos, Feb 12 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (36,-486,2916,-6561).

Cf. A081139, A173000.

[Binomial(n+3, 3)*9^n: n in [0..20]]; // Vincenzo Librandi, Oct 13 2011

A173187:=n->binomial(n+3,3)*9^n: seq(A173187(n), n=0..30); # Wesley Ivan Hurt, Jul 24 2017

Table[Binomial[n + 3, 3]*9^n, {n, 0, 20}]
LinearRecurrence[{36,-486,2916,-6561},{1,36,810,14580},30] (* Harvey P. Dale, May 19 2011 *)

a(n)=binomial(n+3,3)*9^n \\ Charles R Greathouse IV, Oct 07 2015

From Harvey P. Dale, May 19 2011: (Start)
a(n) = 36*a(n-1)-486*a(n-2)+ 2916*a(n-3)-6561*a(n-4).
G.f.: 1/(1-9*x)^4. (End)
From Amiram Eldar, Aug 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 1728*log(9/8) - 405/2.
Sum_{n>=0} (-1)^n/a(n) = 2700*log(10/9) - 567/2. (End)

A173188 a(n) = binomial(n + 5, 5)*9^n.

Original entry on oeis.org

1, 54, 1701, 40824, 826686, 14880348, 245525742, 3788111448, 55401129927, 775615818978, 10470813556203, 137072468372112, 1747673971744428, 21778706417122872, 266011342666286508, 3192136111995438096, 37707107822946112509, 439176902879019428046

Offset: 0

Zerinvary Lajos, Feb 12 2010

Number of n-permutations (n>=5) of 10 objects p, r, q, u, v, w, z, x, y, z with repetition allowed, containing exactly five (5) u's.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (54,-1215,14580,-98415,354294,-531441).

Cf. A081139, A173000, A173187.

[Binomial(n+5, 5)*9^n: n in [0..20]]; // Vincenzo Librandi, Oct 13 2011

Table[Binomial[n + 5, 5]*9^n, {n, 0, 20}]

a(n) = C(n + 5, 5)*9^n.
From Amiram Eldar, Aug 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 184320*log(9/8) - 86835/4.
Sum_{n>=0} (-1)^n/a(n) = 450000*log(10/9) - 189645/4. (End)

A173191 a(n) = binomial(n + 6, 6)*9^n.

Original entry on oeis.org

1, 63, 2268, 61236, 1377810, 27280638, 491051484, 8207574804, 129269303163, 1939039547445, 27922169483208, 388371993720984, 5243021915233284, 68965903654222428, 886704475554288360, 11172476391984033336, 138259395350802412533, 1683511461036241140843

Offset: 0

Zerinvary Lajos, Feb 12 2010

Number of n-permutations (n>=6) of 10 objects p, r, q, u, v, w, z, x, y, z with repetition allowed, containing exactly six (6) u's.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (63,-1701,25515,-229635,1240029,-3720087,4782969).

Cf. A081139, A173000, A173187, A173188.

[Binomial(n+6, 6)*9^n: n in [0..20]]; // Vincenzo Librandi, Oct 13 2011

A173191:=n->binomial(n+6,6)*9^n: seq(A173191(n), n=0..25); # Wesley Ivan Hurt, Jul 24 2017

Table[Binomial[n + 6, 6]*9^n, {n, 0, 20}]

a(n) = C(n + 6, 6)*9^n.
From Amiram Eldar, Aug 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 1042074/5 - 1769472*log(9/8).
Sum_{n>=0} (-1)^n/a(n) = 5400000*log(10/9) - 2844729/5. (End)

A173192 a(n) = binomial(n + 7, 7)*9^n.

Original entry on oeis.org

1, 72, 2916, 87480, 2165130, 46766808, 911952756, 16415149608, 277005649635, 4432090394160, 67810983030648, 998670840996816, 14231059484204628, 197045439012064080, 2660113426662865080, 35113497231949819056, 454280870438350784037, 5772039294981398197176

Offset: 0

Zerinvary Lajos, Feb 12 2010

Number of n-permutations (n>=7) of 10 objects p, r, q, u, v, w, z, x, y, z with repetition allowed, containing exactly 7 u's.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (72,-2268,40824,-459270,3306744,-14880348,38263752,-43046721).

Cf. A081139, A173000, A173187, A173188, A173191.

[Binomial(n+7, 7)*9^n: n in [0..20]]; // Vincenzo Librandi, Oct 13 2011

A173192:=n->binomial(n+7,7)*9^n: seq(A173192(n), n=0..25); # Wesley Ivan Hurt, Jul 24 2017

Table[Binomial[n + 7, 7]*9^n, {n, 0, 20}]

a(n) = C(n + 7, 7)*9^n.
From Amiram Eldar, Aug 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 16515072*log(9/8) - 19451943/10.
Sum_{n>=0} (-1)^n/a(n) = 63000000*log(10/9) - 13275423/2. (End)

A317052 Triangle read by rows: T(0,0) = 1; T(n,k) = 9*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, 9, 81, 1, 729, 18, 6561, 243, 1, 59049, 2916, 27, 531441, 32805, 486, 1, 4782969, 354294, 7290, 36, 43046721, 3720087, 98415, 810, 1, 387420489, 38263752, 1240029, 14580, 45, 3486784401, 387420489, 14880348, 229635, 1215, 1, 31381059609, 3874204890, 172186884, 3306744, 25515, 54

Offset: 0

Zagros Lalo, Jul 20 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013616 ((1+9*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038291 ((9+x)^n).
The coefficients in the expansion of 1/(1-9*x-x^2) are given by the sequence generated by the row sums (see A099371).
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 9.109772228646443655... (a metallic mean), when n approaches infinity; (see A176522: ((9+sqrt(85))/2)).

			Triangle begins:
  1;
  9;
  81, 1;
  729, 18;
  6561, 243, 1;
  59049, 2916, 27;
  531441, 32805, 486, 1;
  4782969, 354294, 7290, 36;
  43046721, 3720087, 98415, 810, 1;
  387420489, 38263752, 1240029, 14580, 45;
  3486784401, 387420489, 14880348, 229635, 1215, 1;
  31381059609, 3874204890, 172186884, 3306744, 25515, 54;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 100.

Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 9x)^n
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (9 + x)^n

Row sums give A099371.
Cf. A013616, A038291, A176522.
Cf. A001019 (column 0), A053540 (column 1), A081139 (column 2), A173187 (column 3), A173000 (column 4).

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 9 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten

T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, 9*T(n-1, k)+T(n-2, k-1)));
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018

A116176 a(n) = 9^n * n*(n+1).

Original entry on oeis.org

0, 18, 486, 8748, 131220, 1771470, 22320522, 267846264, 3099363912, 34867844010, 383546284110, 4142299868388, 44059007691036, 462619580755878, 4804126415541810, 49413871702715760, 504021491367700752

Offset: 0

Mohammad K. Azarian, Apr 08 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (27,-243,729).

Cf. A007758, A036289, A081139, A128796.

List([0..20], n-> 9^n*n*(n+1)); # G. C. Greubel, May 11 2019

[(n^2+n)*9^n: n in [0..20]]; // Vincenzo Librandi, Feb 28 2013

Table[(n^2 + n) 9^n, {n, 0, 20}]  (* Vincenzo Librandi, Feb 28 2013 *)

{a(n) = 9^n*n*(n+1)}; \\ G. C. Greubel, May 11 2019

[9^n*n*(n+1) for n in (0..20)] # G. C. Greubel, May 11 2019

G.f.: 18*x/(1-9*x)^3. - Vincenzo Librandi, Feb 28 2013
a(n) = 27*a(n-1) - 243*a(n-2) + 729*a(n-3). - Vincenzo Librandi, Feb 28 2013
a(n) = 18*A081139(n+1). - Bruno Berselli, Mar 01 2013
E.g.f.: 9*x*(2 + 9*x)*exp(9*x). - G. C. Greubel, May 11 2019
From Amiram Eldar, Jul 20 2020: (Start)
Sum_{n>=1} 1/a(n) = 1 - 8*log(9/8).
Sum_{n>=1} (-1)^(n+1)/a(n) = 10*log(10/9) - 1. (End)

A128803 a(n) = n(n-1)9^n.

Original entry on oeis.org

0, 0, 162, 4374, 78732, 1180980, 15943230, 200884698, 2410616376, 27894275208, 313810596090, 3451916556990, 37280698815492, 396531069219324, 4163576226802902, 43237137739876290, 444724845324441840

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 (27,-243,729).

Cf. A036289, A007758, A081139.

[(n^2-n)*9^n: n in [0..25]]; // Vincenzo Librandi, Feb 12 2013

I:=[0,0,162]; [n le 3 select I[n] else 27*Self(n-1)-243*Self(n-2)+729*Self(n-3): n in [1..25]]; // Vincenzo Librandi, Feb 12 2013

LinearRecurrence[{27, -243, 729}, {0, 0, 162}, 30] (* or *) CoefficientList[Series[162 x^2/(1 - 9 x)^3, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 12 2013 *)

From Vincenzo Librandi, Feb 12 2013: (Start)
G.f.: 162*x^2/(1-9*x)^3.
a(n) = 27*a(n-1)-243*a(n-2)+729*a(n-3). (End)
a(n) = 162*A081139(n). - R. J. Mathar, Apr 26 2015
From Amiram Eldar, Jun 26 2025: (Start)
Sum_{n>=2} 1/a(n) = 1/9 - (8/9)*log(9/8).
Sum_{n>=2} (-1)^n/a(n) = (10/9)*log(10/9) - 1/9. (End)

A196221 Binomial(n+10, 10)*9^n.

Original entry on oeis.org

1, 99, 5346, 208494, 6567561, 177324147, 4255779528, 93019181112, 1883638417518, 35789129932842, 644204338791156, 11068601821048044, 182631930047292726, 2908062270753045714, 44867246463046991016, 673008696945704865240, 9842752192830933654135, 140693457815171581056165, 1969708409412402134786310

Offset: 0

Vincenzo Librandi, Oct 13 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..400

Cf. A053108, A081139, A173000, A173187, A173188, A173191, A173192.

Magma
```
[Binomial(n+10, 10)*9^n: n in [0..20]];
```

a(n) = C(n+10, 10)*9^n.

G.f. -1 / (9*x-1)^11 . - R. J. Mathar, Oct 13 2011

cp's OEIS Frontend

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

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

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A173187 a(n) = binomial(n + 3, 3)*9^n.

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A173188 a(n) = binomial(n + 5, 5)*9^n.

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A173191 a(n) = binomial(n + 6, 6)*9^n.

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A173192 a(n) = binomial(n + 7, 7)*9^n.

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A317052 Triangle read by rows: T(0,0) = 1; T(n,k) = 9*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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A128803 a(n) = n(n-1)9^n.