cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-18 of 18 results.

A027476 Third column of A027467.

Original entry on oeis.org

1, 45, 1350, 33750, 759375, 15946875, 318937500, 6150937500, 115330078125, 2114384765625, 38058925781250, 674680957031250, 11806916748046875, 204350482177734375, 3503151123046875000, 59553569091796875000
Offset: 3

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Author

Keywords

Crossrefs

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), A081142 (q=12), this sequence (q=15).

Programs

  • Magma
    [(n-1)*(n-2)/2 * 15^(n-3): n in [3..20]]; // Vincenzo Librandi, Dec 29 2012
    
  • Maple
    seq((15)^(n-3)*binomial(n-1, 2), n=3..30) # G. C. Greubel, May 13 2021
  • Mathematica
    Table[(n-1)*(n-2)/2 * 15^(n-3), {n, 3, 30}] (* Vincenzo Librandi, Dec 29 2012 *)
  • Sage
    [(15)^(n-3)*binomial(n-1,2) for n in (3..30)] # G. C. Greubel, May 13 2021

Formula

Numerators of sequence a[3,n] in (a[i,j])^4 where a[i,j] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i.
a(n) = 15^(n-3)*binomial(n-1, 2).
From G. C. Greubel, May 13 2021: (Start)
a(n) = 45*a(n-1) - 675*a(n-2) + 3375*a(n-3).
G.f.: x^3/(1 - 15*x)^3.
E.g.f.: (-2 + (2 - 30*x + 225*x^2)*exp(15*x))/6750. (End)
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=3} 1/a(n) = 30 - 420*log(15/14).
Sum_{n>=3} (-1)^(n+1)/a(n) = 480*log(16/15) - 30. (End)

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

Original entry on oeis.org

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

Views

Author

Paul Barry, Mar 08 2003

Keywords

Comments

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

Crossrefs

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).

Programs

  • GAP
    List([0..20],n->12^(n-2)*Binomial(n,2)); # Muniru A Asiru, Nov 24 2018
  • Magma
    [12^(n-2)* Binomial(n, 2): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011
    
  • Maple
    seq(coeff(series(x^2/(1-12*x)^3,x,n+1), x, n), n = 0 .. 20); # Muniru A Asiru, Nov 24 2018
  • Mathematica
    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 *)
  • PARI
    vector(20, n, n--; 2^(2*n-5)*3^(n-2)*n*(n-1)) \\ G. C. Greubel, Nov 23 2018
    
  • Sage
    [2^(2*n-5)*3^(n-2)*n*(n-1) for n in range(20)] # G. C. Greubel, Nov 23 2018
    

Formula

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

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Author

Paul Barry, Mar 08 2003

Keywords

Comments

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

Examples

			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;
		

Crossrefs

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).

Programs

  • Magma
    [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
    
  • Mathematica
    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 *)
  • PARI
    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
  • Sage
    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
    

Formula

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)

Extensions

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

A218017 Triangle, read by rows, where T(n,k) = k!*C(n, k)*7^(n-k) for n>=0, k=0..n.

Original entry on oeis.org

1, 7, 1, 49, 14, 2, 343, 147, 42, 6, 2401, 1372, 588, 168, 24, 16807, 12005, 6860, 2940, 840, 120, 117649, 100842, 72030, 41160, 17640, 5040, 720, 823543, 823543, 705894, 504210, 288120, 123480, 35280, 5040, 5764801, 6588344, 6588344, 5647152, 4033680, 2304960, 987840, 282240, 40320
Offset: 0

Views

Author

Vincenzo Librandi, Nov 10 2012

Keywords

Comments

Triangle formed by the derivatives of x^n evaluated at x=7. Also:
first column: A000420;
second column: A027473;
third column: 2*A027474;
fourth column: 6*A140107.

Examples

			Triangle begins:
1;
7,       1;
49,      14,      2;
343,     147,     42,      6;
2401,    1372,    588,     168,     24;
16807,   12005,   6860,    2940,    840,     120;
117649,  100842,  72030,   41160,   17640,   5040,    720;
823543,  823543,  705894,  504210,  288120,  123480,  35280,  5040; etc.
		

Crossrefs

Programs

  • Magma
    [Factorial(n)/Factorial(n-k)*7^(n-k): k in [0..n], n in [0..10]];
  • Mathematica
    Flatten[Table[n!/(n-k)!*7^(n-k), {n, 0, 10}, {k, 0, n}]]

Formula

T(n,k) = 7^(n-k)*n!/(n-k)! for n>=0, k=0..n.
E.g.f. (by columns): exp(7x)*x^k.

A170932 a(n) = binomial(n + 8, 8)*7^n .

Original entry on oeis.org

1, 63, 2205, 56595, 1188495, 21630609, 353299947, 5299499205, 74192988870, 980996186170, 12360551945742, 149450309889426, 1743586948709970, 19715944727720430, 216875392004924730, 2327795874186192102, 24441856678955017071, 251607348165713411025
Offset: 0

Views

Author

Zerinvary Lajos, Feb 08 2010

Keywords

Comments

With a different offset, number of n-permutations of 8 objects: r, s, t, u, v, z, x, y with repetition allowed, containing exactly eight, (8) u's.

Crossrefs

Programs

  • Magma
    [Binomial(n + 8, 8)*7^n: n in [0..20]]; // Vincenzo Librandi, Oct 12 2011
  • Mathematica
    Table[Binomial[n + 8, 8]*7^n, {n, 0, 20}]

Formula

a(n) = C(n + 8, 8)*7^n.
From Amiram Eldar, Aug 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 12082656/5 - 15676416*log(7/6).
Sum_{n>=0} (-1)^n/a(n) = 117440512*log(8/7) - 235229912/15. (End)

A317014 Triangle read by rows: T(0,0) = 1; T(n,k) = 7 * 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, 7, 49, 1, 343, 14, 2401, 147, 1, 16807, 1372, 21, 117649, 12005, 294, 1, 823543, 100842, 3430, 28, 5764801, 823543, 36015, 490, 1, 40353607, 6588344, 352947, 6860, 35, 282475249, 51883209, 3294172, 84035, 735, 1, 1977326743, 403536070, 29647548, 941192, 12005, 42
Offset: 0

Views

Author

Zagros Lalo, Jul 19 2018

Keywords

Comments

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013614 ((1+7*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A027466 ((7+x)^n).
The coefficients in the expansion of 1/(1-7x-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 7.14005494464025913554... ((7+sqrt(53))/2), a metallic mean (see A176439), when n approaches infinity.

Examples

			Triangle begins:
1;
7;
49, 1;
343, 14;
2401, 147, 1;
16807, 1372, 21;
117649, 12005, 294, 1;
823543, 100842, 3430, 28;
5764801, 823543, 36015, 490, 1;
40353607, 6588344, 352947, 6860, 35;
282475249, 51883209, 3294172, 84035, 735, 1;
1977326743, 403536070, 29647548, 941192, 12005, 42;
13841287201, 3107227739, 259416045, 9882516, 168070, 1029, 1;
96889010407, 23727920916, 2219448385, 98825160, 2117682, 19208, 49;
678223072849, 179936733613, 18643366434, 951192165, 24706290, 302526, 1372, 1;
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 96.

Crossrefs

Row sums give A054413.
Cf. A000420 (column 0), A027473 (column 1), A027474 (column 2), A140107 (column 3), A139641 (column 4).

Programs

  • Mathematica
    t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 7 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten
  • PARI
    T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, 7*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

A116165 a(n) = 7^n * n*(n+1).

Original entry on oeis.org

0, 14, 294, 4116, 48020, 504210, 4941258, 46118408, 415065672, 3631824630, 31072277390, 261007130076, 2159240803356, 17633799894074, 142426845298290, 1139414762386320, 9039357114931472, 71184937280085342, 556917450485373558
Offset: 0

Views

Author

Mohammad K. Azarian, Apr 08 2007

Keywords

Crossrefs

Programs

Formula

G.f.: 14*x/(1-7*x)^3. - Vincenzo Librandi, Feb 28 2013
a(n) = 21*a(n-1) - 147*a(n-2) + 343*a(n-3). - Vincenzo Librandi, Feb 28 2013
a(n+1) = 14*A027474(n+2). - Bruno Berselli, Feb 28 2013
E.g.f.: 7*x*(2 + 7*x)*exp(7*x). - G. C. Greubel, May 11 2019
From Amiram Eldar, Jul 20 2020: (Start)
Sum_{n>=1} 1/a(n) = 1 - 6*log(7/6).
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(8/7) - 1. (End)

A128801 a(n) = n*(n-1)*7^n.

Original entry on oeis.org

0, 98, 2058, 28812, 336140, 3529470, 34588806, 322828856, 2905459704, 25422772410, 217505941730, 1827049910532, 15114685623492, 123436599258518, 996987917088030, 7975903336704240, 63275499804520304, 498294560960597394
Offset: 1

Views

Author

Mohammad K. Azarian, Apr 07 2007

Keywords

Crossrefs

Programs

  • Magma
    [(n^2-n)*7^n: n in [1..25]] // Vincenzo Librandi, Feb 12 2013
  • Maple
    g:=98*x^2/(1-7*x)^3: gser:=series(g,x=0,22): seq(coeff(gser,x,n),n=1..18); # Emeric Deutsch, May 04 2007
  • Mathematica
    LinearRecurrence[{21, -147, 343}, {0, 98, 2058}, 30] (* Vincenzo Librandi, Feb 12 2013 *)

Formula

G.f.: 98*x^2/(1-7*x)^3. - Emeric Deutsch, May 04 2007
a(n) = 21*a(n-1) - 147*a(n-2) + 343*a(n-3). - Vincenzo Librandi, Feb 12 2013
a(n) = 98*A027474(n). - R. J. Mathar, Apr 26 2015
Previous Showing 11-18 of 18 results.