A081138 -id:A081138 - 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)

A053107 Expansion of 1/(1-8*x)^8.

Original entry on oeis.org

1, 64, 2304, 61440, 1351680, 25952256, 449839104, 7197425664, 107961384960, 1535450808320, 20882130993152, 273366078455808, 3462636993773568, 42617070692597760, 511404848311173120, 6000483553517764608, 69005560865454292992, 779356922715719073792

Offset: 0

Wolfdieter Lang

With a different offset, number of n-permutations (n>=7) of 9 objects: p, r, s, t, u, v, z, x, y with repetition allowed, containing exactly 7 u's. - Zerinvary Lajos, Feb 11 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (64, -1792, 28672, -286720, 1835008, -7340032, 16777216, -16777216).

Cf. A036226.

Cf. A081138, A140802, A175210, A140406, A053107, A141054, A173155. - Zerinvary Lajos, Feb 11 2010

[8^n* Binomial(n+7, 7): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011

Table[Binomial[n + 7, 7]*8^n, {n, 0, 20}] (* Zerinvary Lajos, Feb 11 2010 *)
CoefficientList[Series[1/(1-8x)^8,{x,0,20}],x] (* or *) LinearRecurrence[ {64,-1792,28672,-286720,1835008,-7340032,16777216,-16777216},{1,64,2304,61440,1351680,25952256,449839104,7197425664},20] (* Harvey P. Dale, Jul 19 2018 *)

vector(30, n, n--; 8^n*binomial(n+7,7)) \\ G. C. Greubel, Aug 16 2018

[lucas_number2(n, 8, 0)*binomial(n,7)/8^7 for n in range(7, 22)] # Zerinvary Lajos, Mar 13 2009

a(n) = 8^n*binomial(n+7, 7).

G.f.: 1/(1-8*x)^8.

More terms from Harvey P. Dale, Jul 19 2018

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

A173155 a(n) = binomial(n + 5, 5) * 8^n.

Original entry on oeis.org

1, 48, 1344, 28672, 516096, 8257536, 121110528, 1660944384, 21592276992, 268703891456, 3224446697472, 37520834297856, 425236122042368, 4710307813392384, 51140484831117312, 545498504865251328, 5727734301085138944, 59298896293587320832, 606166495445559279616

Offset: 0

Zerinvary Lajos, Feb 11 2010

With a different offset, number of n-permutations (n>=5) of 9 objects: p, r, s, t, u, v, z, x, y 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 (48,-960,10240,-61440,196608,-262144).

Cf. A081138, A140802, A175210, A140406, A053107, A141054.

[8^n* Binomial(n+5, 5): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011

Table[Binomial[n + 5, 5]*8^n, {n, 0, 20}]

a(n) = C(n + 5, 5)*8^n, n>=0.
G.f.: 1/(1-8*x)^6. - Vincenzo Librandi, Oct 16 2011
From Amiram Eldar, Aug 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 96040*log(8/7) - 38470/3.
Sum_{n>=0} (-1)^n/a(n) = 262440*log(9/8) - 30910. (End)

A172510 a(n) = binomial(n + 4, 4) * 8^n.

Original entry on oeis.org

1, 40, 960, 17920, 286720, 4128768, 55050240, 692060160, 8304721920, 95965675520, 1074815565824, 11725260718080, 125069447659520, 1308418837053440, 13458022323978240, 136374626216312832, 1363746262163128320, 13477021884906209280, 131775325096860712960

Offset: 0

Zerinvary Lajos, Feb 05 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..138
Index entries for linear recurrences with constant coefficients, signature (40,-640,5120,-20480,32768).

Cf. A081138, A140802, A140406, A053107, A141054.

[Binomial(n + 4, 4)*8^n: n in [0..30]]; // Vincenzo Librandi, Jun 06 2011

Table[Binomial[n + 4, 4]*8^n, {n, 0, 25}]

Vec(1 / (1-8*x)^5 + O(x^30)) \\ Colin Barker, Jul 24 2017

G.f.: 1 / (1-8*x)^5. - R. J. Mathar, Feb 11 2010
a(n) = (8^(-1 + n)*(1 + n)*(2 + n)*(3 + n)*(4 + n)) / 3. - Colin Barker, Jul 24 2017
From Amiram Eldar, Aug 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 4400/3 - 10976*log(8/7).
Sum_{n>=0} (-1)^n/a(n) = 23328*log(9/8) - 8240/3. (End)

A317028 Triangle read by rows: T(0,0) = 1; T(n,k) = 8 * 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, 8, 64, 1, 512, 16, 4096, 192, 1, 32768, 2048, 24, 262144, 20480, 384, 1, 2097152, 196608, 5120, 32, 16777216, 1835008, 61440, 640, 1, 134217728, 16777216, 688128, 10240, 40, 1073741824, 150994944, 7340032, 143360, 960, 1, 8589934592, 1342177280, 75497472, 1835008, 17920, 48

Offset: 0

Zagros Lalo, Jul 19 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013615 ((1+8*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038279 ((8+x)^n).
The coefficients in the expansion of 1/(1-8x-x^2) are given by the sequence generated by the row sums.
The row sums are Denominators of continued fraction convergents to sqrt(17), see A041025.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 8.12310562561766054982... (a metallic mean), when n approaches infinity (see A176458: (4+sqrt(17))).

			Triangle begins:
1;
8;
64, 1;
512, 16;
4096, 192, 1;
32768, 2048, 24;
262144, 20480, 384, 1;
2097152, 196608, 5120, 32;
16777216, 1835008, 61440, 640, 1;
134217728, 16777216, 688128, 10240, 40;
1073741824, 150994944, 7340032, 143360, 960, 1;
8589934592, 1342177280, 75497472, 1835008, 17920, 48;
68719476736, 11811160064, 754974720, 22020096, 286720, 1344, 1;
549755813888, 103079215104, 7381975040, 251658240, 4128768, 28672, 56;
4398046511104, 893353197568, 70866960384, 2768240640, 55050240, 516096, 1792, 1;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, Pages 70, 98

Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 8x)^n
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (8 + x)^n

Row sums give A041025.
Cf. A013615, A038279, A176458.
Cf. A001018 (column 0), A053539 (column 1), A081138 (column 2), A140802 (column 3), A172510 (column 4).

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 8 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, 8*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

A116166 a(n) = 8^n * n*(n+1).

Original entry on oeis.org

0, 16, 384, 6144, 81920, 983040, 11010048, 117440512, 1207959552, 12079595520, 118111600640, 1133871366144, 10720238370816, 100055558127616, 923589767331840, 8444249301319680, 76561193665298432, 689050742987685888

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 (24,-192,512).

Cf. A007758, A036289, A081138, A128796.

List([0..30], n-> 8^n*n*(n+1)); # G. C. Greubel, May 11 2019

[(n^2+n)*8^n: n in [0..30]]; // Vincenzo Librandi, Feb 28 2013

Table[(n^2 + n) 8^n, {n, 0, 30}]  (* Harvey P. Dale, Mar 09 2011 *)
CoefficientList[Series[16 x/(1 - 8 x)^3, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 28 2013 *)

a(n)=(n^2+n)*8^n \\ Charles R Greathouse IV, Feb 28 2013

[8^n*n*(n+1) for n in (0..30)] # G. C. Greubel, May 11 2019

G.f.: 16*x/(1-8*x)^3. - Vincenzo Librandi, Feb 28 2013
a(n) = 24*a(n-1) - 192*a(n-2) + 512*a(n-3). - Vincenzo Librandi, Feb 28 2013
a(n) = 16*A081138(n+1). - Bruno Berselli, Feb 28 2013
E.g.f.: 16*x*(1 + 4*x)*exp(8*x). - G. C. Greubel, May 11 2019
From Amiram Eldar, Jul 20 2020: (Start)
Sum_{n>=1} 1/a(n) = 1 - 7*log(8/7).
Sum_{n>=1} (-1)^(n+1)/a(n) = 9*log(9/8) - 1. (End)

A128802 a(n) = n(n-1)8^n.

Original entry on oeis.org

0, 0, 128, 3072, 49152, 655360, 7864320, 88080384, 939524096, 9663676416, 96636764160, 944892805120, 9070970929152, 85761906966528, 800444465020928, 7388718138654720, 67553994410557440, 612489549322387456

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 (24,-192,512).

Cf. A036289, A007758.

[(n^2-n)*8^n: n in [0..20]]; // Vincenzo Librandi, Nov 16 2011

Table[(n^2-n)8^n,{n,0,20}] (* or *) LinearRecurrence[{24,-192,512},{0,0,128},20] (* Harvey P. Dale, Nov 05 2011 *)

a(n) = 24*a(n-1) - 192*a(n-2) + 512*a(n-3); a(0)=0, a(1)=0, a(2)=128. - Harvey P. Dale, Nov 05 2011
G.f.: -128*x^2/(8*x-1)^3. - Harvey P. Dale, Nov 05 2011
a(n) = 128*A081138(n). - R. J. Mathar, Apr 26 2015

cp's OEIS Frontend

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

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A053107 Expansion of 1/(1-8*x)^8.

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

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A173155 a(n) = binomial(n + 5, 5) * 8^n.

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A172510 a(n) = binomial(n + 4, 4) * 8^n.

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A317028 Triangle read by rows: T(0,0) = 1; T(n,k) = 8 * 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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A116166 a(n) = 8^n * n*(n+1).

A128802 a(n) = n(n-1)8^n.