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.

Showing 1-10 of 15 results. Next

A027465 Cube of lower triangular normalized binomial matrix.

Original entry on oeis.org

1, 3, 1, 9, 6, 1, 27, 27, 9, 1, 81, 108, 54, 12, 1, 243, 405, 270, 90, 15, 1, 729, 1458, 1215, 540, 135, 18, 1, 2187, 5103, 5103, 2835, 945, 189, 21, 1, 6561, 17496, 20412, 13608, 5670, 1512, 252, 24, 1, 19683, 59049, 78732, 61236, 30618, 10206, 2268
Offset: 0

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Author

Keywords

Comments

Rows of A013610 reversed. - Michael Somos, Feb 14 2002
Row sums are powers of 4 (A000302), antidiagonal sums are A006190 (a(n) = 3*a(n-1) + a(n-2)). - Gerald McGarvey, May 17 2005
Triangle of coefficients in expansion of (3+x)^n.
Also: Pure Galton board of scheme (3,1). Also: Multiplicity (number) of pairs of n-dimensional binary vectors with dot product (overlap) k. There are 2^n = A000079(n) binary vectors of length n and 2^(2n) = 4^n = A000302(n) different pairs to form dot products k = Sum_{i=1..n} v[i]*u[i] between these, 0 <= k <= n. (Since dot products are symmetric, there are only 2^n*(2^n-1)/2 different non-ordered pairs, actually.) - R. J. Mathar, Mar 17 2006
Mirror image of A013610. - Zerinvary Lajos, Nov 25 2007
T(i,j) is the number of i-permutations of 4 objects a,b,c,d, with repetition allowed, containing j a's. - Zerinvary Lajos, Dec 21 2007
The antidiagonals of the sequence formatted as a square array (see Examples section) and summed with alternating signs gives a bisection of Fibonacci sequence, A001906. Example: 81-(27-1)=55. Similar rule applied to rows gives A000079. - Mark Dols, Sep 01 2009
Triangle T(n,k), read by rows, given by (3,0,0,0,0,0,0,0,...)DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 09 2011
T(n,k) = binomial(n,k)*3^(n-k), the number of subsets of [2n] with exactly k symmetric pairs, where elements i and j of [2n] form a symmetric pair if i+j=2n+1. Equivalently, if n couples attend a (ticketed) event that offers door prizes, then the number of possible prize distributions that have exactly k couples as dual winners is T(n,k). - Dennis P. Walsh, Feb 02 2012
T(n,k) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that the intersection of A and B contains exactly k elements. For example, T(2,1) = 6 because we have ({1},{1}); ({1},{1,2}); ({2},{2}); ({2},{1,2}); ({1,2},{1}); ({1,2},{2}). Sum_{k=0..n} T(n,k)*k = A002697(n) (see comment there by Ross La Haye). - Geoffrey Critzer, Sep 04 2013
Also the convolution triangle of A000244. - Peter Luschny, Oct 09 2022

Examples

			Example: n = 3 offers 2^3 = 8 different binary vectors (0,0,0), (0,0,1), ..., (1,1,0), (1,1,1). a(3,2) = 9 of the 2^4 = 64 pairs have overlap k = 2: (0,1,1)*(0,1,1) = (1,0,1)*(1,0,1) = (1,1,0)*(1,1,0) = (1,1,1)*(1,1,0) = (1,1,1)*(1,0,1) = (1,1,1)*(0,1,1) = (0,1,1)*(1,1,1) = (1,0,1)*(1,1,1) = (1,1,0)*(1,1,1) = 2.
For example, T(2,1)=6 since there are 6 subsets of {1,2,3,4} that have exactly 1 symmetric pair, namely, {1,4}, {2,3}, {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}.
The present sequence formatted as a triangular array:
     1
     3     1
     9     6     1
    27    27     9     1
    81   108    54    12    1
   243   405   270    90   15    1
   729  1458  1215   540  135   18   1
  2187  5103  5103  2835  945  189  21  1
  6561 17496 20412 13608 5670 1512 252 24 1
  ...
A013610 formatted as a triangular array:
  1
  1  3
  1  6   9
  1  9  27   27
  1 12  54  108   81
  1 15  90  270  405   243
  1 18 135  540 1215  1458   729
  1 21 189  945 2835  5103  5103  2187
  1 24 252 1512 5670 13608 20412 17496 6561
   ...
A099097 formatted as a square array:
      1     0     0    0   0 0 0 0 0 0 0 ...
      3     1     0    0   0 0 0 0 0 0 ...
      9     6     1    0   0 0 0 0 0 ...
     27    27     9    1   0 0 0 0 ...
     81   108    54   12   1 0 0 ...
    243   405   270   90  15 1 ...
    729  1458  1215  540 135 ...
   2187  5103  5103 2835 ...
   6561 17496 20412 ...
  19683 59049 ...
  59049 ...
		

Crossrefs

Programs

  • Haskell
    a027465 n k = a027465_tabl !! n !! k
    a027465_row n = a027465_tabl !! n
    a027465_tabl = iterate (\row ->
       zipWith (+) (map (* 3) (row ++ [0])) (map (* 1) ([0] ++ row))) [1]
    -- Reinhard Zumkeller, May 26 2013
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*3^(i-j), j = 0 .. i) od; # Zerinvary Lajos, Nov 25 2007
    # Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
    PMatrix(10, n -> 3^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    t[n_, k_] := Binomial[n, k]*3^(n-k); Table[t[n, n-k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 19 2012 *)
  • PARI
    {T(n, k) = polcoeff( (3 + x)^n, k)}; /* Michael Somos, Feb 14 2002 */
    

Formula

Numerators of lower triangle of (b^2)[ i, j ] where b[ i, j ] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
Triangle whose (i, j)-th entry is binomial(i, j)*3^(i-j).
a(n, m) = 4^(n-1)*Sum_{j=m..n} b(n, j)*b(j, m) = 3^(n-m)*binomial(n-1, m-1), n >= m >= 1; a(n, m) := 0, n < m. G.f. for m-th column: (x/(1-3*x))^m (m-fold convolution of A000244, powers of 3). - Wolfdieter Lang, Feb 2006
G.f.: 1 / (1 - x(3+y)).
a(n,k) = 3*a(n-1,k) + a(n-1,k-1) - R. J. Mathar, Mar 17 2006
From the formalism of A133314, the e.g.f. for the row polynomials of A027465 is exp(x*t)*exp(3x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-3x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*3x). The results generalize for 3 replaced by any number. - Tom Copeland, Aug 18 2008
T(n,k) = A164942(n,k)*(-1)^k. - Philippe Deléham, Oct 09 2011
Let P and P^T be the Pascal matrix and its transpose and H = P^3 = A027465. Then from the formalism of A132440 and A218272,
exp[x*z/(1-3z)]/(1-3z) = exp(3z D_z z) e^(x*z)= exp(3D_x x D_x) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T = Sum_{n>=0} (3z)^n L_n(-x/3), where D is the derivative operator and L_n(x) are the regular (not normalized) Laguerre polynomials. - Tom Copeland, Oct 26 2012
E.g.f. for column k: x^k/k! * exp(3x). - Geoffrey Critzer, Sep 04 2013

A036216 Expansion of 1/(1 - 3*x)^4; 4-fold convolution of A000244 (powers of 3).

Original entry on oeis.org

1, 12, 90, 540, 2835, 13608, 61236, 262440, 1082565, 4330260, 16888014, 64481508, 241805655, 892820880, 3252418920, 11708708112, 41712272649, 147219785820, 515269250370, 1789882659180, 6175095174171, 21171754882872
Offset: 0

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Comments

With three leading zeros, 3rd binomial transform of (0,0,0,1,0,0,0,0,...). - Paul Barry, Mar 07 2003
Number of n-permutations (n=4) of 4 objects u, v, w, z, with repetition allowed, containing exactly three u's. - Zerinvary Lajos, May 23 2008

Crossrefs

Cf. A027465.
Sequences of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), this sequence (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Programs

  • Magma
    [3^n* Binomial(n+3, 3): n in [0..30]]; // Vincenzo Librandi, Oct 14 2011
    
  • Maple
    seq(3^n*binomial(n+3, 3), n=0..30)]; # Zerinvary Lajos, Dec 21 2006
  • Mathematica
    CoefficientList[Series[1/(1-3x)^4,{x,0,30}],x] (* or *) LinearRecurrence[ {12,-54,108,-81},{1,12,90,540},30] (* Harvey P. Dale, Jul 27 2017 *)
  • PARI
    a(n) = 3^n*binomial(n+3, 3) \\ Charles R Greathouse IV, Oct 03 2016
  • Sage
    [3^n*binomial(n+3,3) for n in range(30)] # Zerinvary Lajos, Mar 10 2009
    

Formula

a(n) = 3^n*binomial(n+3, 3).
a(n) = A027465(n+4, 4).
G.f.: 1/(1 - 3*x)^4.
With three leading zeros, a(n) = 12*a(n-1) - 54*a(n-2) + 108*a(n-3) - 81*a(n-4), a(0) = a(1) = a(2) = 0, a(3) = 1. - Paul Barry, Mar 07 2003
With three leading zeros, C(n, 3)*3^(n-3) is the second binomial transform of C(n, 3). - Paul Barry, Jul 24 2003
E.g.f.: (1/2)*(2 + 18*x + 27*x^2 + 9*x^3)*exp(3*x). - Franck Maminirina Ramaharo, Nov 23 2018
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=0} 1/a(n) = 36*log(3/2) - 27/2.
Sum_{n>=0} (-1)^n/a(n) = 144*log(4/3) - 81/2. (End)

A013610 Triangle of coefficients in expansion of (1+3*x)^n.

Original entry on oeis.org

1, 1, 3, 1, 6, 9, 1, 9, 27, 27, 1, 12, 54, 108, 81, 1, 15, 90, 270, 405, 243, 1, 18, 135, 540, 1215, 1458, 729, 1, 21, 189, 945, 2835, 5103, 5103, 2187, 1, 24, 252, 1512, 5670, 13608, 20412, 17496, 6561, 1, 27, 324, 2268, 10206, 30618, 61236, 78732, 59049, 19683
Offset: 0

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Comments

T(n,k) is the number of lattice paths from (0,0) to (n,k) with steps (1,0) and three kinds of steps (1,1). The number of paths with steps (1,0) and s kinds of steps (1,1) corresponds to the expansion of (1+s*x)^n. - Joerg Arndt, Jul 01 2011
Rows of A027465 reversed. - Michael Somos, Feb 14 2002
T(n,k) equals the number of n-length words on {0,1,2,3} having n-k zeros. - Milan Janjic, Jul 24 2015
T(n-1,k-1) is the number of 3-compositions of n with zeros having k positive parts; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020

Examples

			Triangle begins
  1;
  1,    3;
  1,    6,    9;
  1,    9,   27,   27;
  1,   12,   54,  108,   81;
  1,   15,   90,  270,  405,  243;
  1,   18,  135,  540, 1215, 1458,  729;
  1,   21,  189,  945, 2835, 5103, 5103, 2187;
		

Crossrefs

Cf. A007318, A013609, A027465, etc.
Diagonals of the triangle: A000244 (k=n), A027471 (k=n-1), A027472 (k=n-2), A036216 (k=n-3), A036217 (k=n-4), A036219 (k=n-5), A036220 (k=n-6), A036221 (k=n-7), A036222 (k=n-8), A036223 (k=n-9), A172362 (k=n-10).

Programs

  • Haskell
    a013610 n k = a013610_tabl !! n !! k
    a013610_row n = a013610_tabl !! n
    a013610_tabl = iterate (\row ->
       zipWith (+) (map (* 1) (row ++ [0])) (map (* 3) ([0] ++ row))) [1]
    -- Reinhard Zumkeller, May 26 2013
    
  • Magma
    [3^k*Binomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 19 2021
    
  • Maple
    T:= n-> (p-> seq(coeff(p, x, k), k=0..n))((1+3*x)^n):
    seq(T(n), n=0..10);  # Alois P. Heinz, Jul 25 2015
  • Mathematica
    t[n_, k_] := Binomial[n, k]*3^(n-k); Table[t[n, n-k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2013 *)
    BinomialROW[n_, k_, t_] := Sum[Binomial[n, k]*Binomial[k, j]*(-1)^(k - j)*t^j, {j, 0, k}]; Column[Table[BinomialROW[n, k, 4], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jan 28 2019 *)
    T[0, 0] := 1; T[n_, k_]/;0<=k<=n := T[n, k] = 3T[n-1, k-1]+T[n-1, k]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 0, 7}, {k, 0, n}] (* Oliver Seipel, Jan 26 2025 *)
  • PARI
    {T(n, k) = polcoeff((1 + 3*x)^n, k)}; /* Michael Somos, Feb 14 2002 */
    
  • PARI
    /* same as in A092566 but use */
    steps=[[1,0], [1,1], [1,1], [1,1]]; /* note triple [1,1] */
    /* Joerg Arndt, Jul 01 2011 */
    
  • Sage
    flatten([[3^k*binomial(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 19 2021

Formula

G.f.: 1 / (1 - x*(1+3*y)).
Row sums are 4^n. - Joerg Arndt, Jul 01 2011
T(n,k) = 3^k*C(n,k) = Sum_{i=n-k..n} C(i,n-k)*C(n,i)*2^(n-i). - Mircea Merca, Apr 28 2012
From Peter Bala, Dec 22 2014: (Start)
Riordan array ( 1/(1 - x), 3*x/(1 - x) ).
exp(3*x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(3*x)*(1 + 9*x + 27*x^2/2! + 27*x^3/3!) = 1 + 12*x + 90*x^2/2! + 540*x^3/3! + 2835*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), 3*x/(1 - x) ). (End)
T(n,k) = Sum_{j=0..k} (-1)^(k-j) * binomial(n,k) * binomial(k,j) * 4^j. - Kolosov Petro, Jan 28 2019
T(0,0)=1, T(n,k)=3*T(n-1,k-1)+T(n-1,k) for 0<=k<=n, T(n,k)=0 for k<0 or k>n. - Oliver Seipel, Feb 10 2025

A036219 Expansion of 1/(1-3*x)^6; 6-fold convolution of A000244 (powers of 3).

Original entry on oeis.org

1, 18, 189, 1512, 10206, 61236, 336798, 1732104, 8444007, 39405366, 177324147, 773778096, 3288556908, 13660159464, 55616363532, 222465454128, 875957725629, 3400777052442, 13036312034361, 49400761393368, 185252855225130, 688082033693340, 2533392942234570
Offset: 0

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Crossrefs

Cf. A027465.
Sequences of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), this sequence (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Programs

  • Magma
    [3^n*Binomial(n+5, 5): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
  • Maple
    seq(3^n*binomial(n+5,5), n=0..30); # Zerinvary Lajos, Jun 13 2008
  • Mathematica
    Table[3^n*Binomial[n+5, 5], {n, 0, 30}] (* G. C. Greubel, May 19 2021 *)
    CoefficientList[Series[1/(1-3x)^6,{x,0,30}],x] (* or *) LinearRecurrence[ {18,-135,540,-1215,1458,-729},{1,18,189,1512,10206,61236},30] (* Harvey P. Dale, Jan 02 2022 *)
  • Sage
    [3^n*binomial(n+5,5) for n in range(30)] # Zerinvary Lajos, Mar 10 2009
    

Formula

a(n) = 3^n*binomial(n+5, 5).
a(n) = A027465(n+6, 6).
G.f.: 1/(1-3*x)^6.
E.g.f.: (1/40)*(40 + 600*x + 1800*x^2 + 1800*x^3 + 675*x^4 + 81*x^5)*exp(3*x). - G. C. Greubel, May 19 2021
From Amiram Eldar, Sep 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 240*log(3/2) - 385/4.
Sum_{n>=0} (-1)^n/a(n) = 3840*log(4/3) - 4415/4. (End)

A036220 Expansion of 1/(1-3*x)^7; 7-fold convolution of A000244 (powers of 3).

Original entry on oeis.org

1, 21, 252, 2268, 17010, 112266, 673596, 3752892, 19702683, 98513415, 472864392, 2192371272, 9865670724, 43257171636, 185387878440, 778629089448, 3211844993973, 13036312034361, 52145248137444, 205836505805700, 802762372642230, 3096369151620030
Offset: 0

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Crossrefs

Cf. A027465.
Sequences 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), this sequence (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Programs

  • Magma
    [3^n*Binomial(n+6, 6): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
  • Maple
    seq(3^n*binomial(n+6,6), n=0..20); # Zerinvary Lajos, Jun 16 2008
  • Mathematica
    Table[3^n*Binomial[n+6, 6], {n,0,30}] (* G. C. Greubel, May 19 2021 *)
  • Sage
    [3^n*binomial(n+6,6) for n in range(30)] # Zerinvary Lajos, Mar 10 2009
    

Formula

a(n) = 3^n*binomial(n+6, 6).
a(n) = A027465(n+7,7).
G.f.: 1/(1-3*x)^7.
E.g.f.: (1/80)*(80 + 1440*x + 5400*x^2 + 7200*x^3 + 4050*x^4 + 972*x^5 + 81*x^6)*exp(3*x). - G. C. Greubel, May 19 2021
From Amiram Eldar, Sep 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 1173/5 - 576*log(3/2).
Sum_{n>=0} (-1)^n/a(n) = 18432*log(4/3) - 26508/5. (End)

A036221 Expansion of 1/(1-3*x)^8; 8-fold convolution of A000244 (powers of 3).

Original entry on oeis.org

1, 24, 324, 3240, 26730, 192456, 1250964, 7505784, 42220035, 225173520, 1148384952, 5637526128, 26778249108, 123591918960, 556163635320, 2447119995408, 10553204980197, 44695926974952, 186233029062300
Offset: 0

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Comments

With a different offset, number of n-permutations (n>=7) of 4 objects: u, v, z, x with repetition allowed, containing exactly seven (7) u's. Example: a(1)=24 because we have uuuuuuuv, uuuuuuuz, uuuuuuux, uuuuuuvu, uuuuuuzu, uuuuuuxu, uuuuuvuu, uuuuuzuu, uuuuuxuu, uuuuvuuu, uuuuzuuu, uuuuxuuu, uuuvuuuu, uuuzuuuu, uuuxuuuu, uuvuuuuu, uuzuuuuu, uuxuuuuu, uvuuuuuu, uzuuuuuu, uxuuuuuu, vuuuuuuu, zuuuuuuu, xuuuuuuu. - Zerinvary Lajos, Jun 23 2008

Crossrefs

Cf. A027465.
Sequences 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), this sequence (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Programs

  • Magma
    [3^n*Binomial(n+7, 7): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
  • Maple
    seq(3^n*binomial(n+7,7), n=0..30); # Zerinvary Lajos, Jun 23 2008
  • Mathematica
    Table[3^n*Binomial[n+7,7], {n,0,30}] (* G. C. Greubel, May 19 2021 *)
  • Sage
    [3^n*binomial(n+7, 7) for n in range(30)] # Zerinvary Lajos, Mar 13 2009
    

Formula

a(n) = 3^n*binomial(n+7, 7).
a(n) = A027465(n+8, 8.)
G.f.: 1/(1-3*x)^8.
E.g.f.: (1/560)*(560 +11760*x +52920*x^2 +88200*x^3 +66150*x^4 +23814*x^5 +3969*x^6 +243*x^7)*exp(3*x). - G. C. Greubel, May 19 2021
From Amiram Eldar, Sep 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 1344*log(3/2) - 5439/10.
Sum_{n>=0} (-1)^n/a(n) = 86016*log(4/3) - 247443/10. (End)

A036222 Expansion of 1/(1-3*x)^9; 9-fold convolution of A000244 (powers of 3).

Original entry on oeis.org

1, 27, 405, 4455, 40095, 312741, 2189187, 14073345, 84440070, 478493730, 2583866142, 13389124554, 66945622770, 324428787270, 1529449997130, 7035469986798, 31659614940591, 139674771796725, 605257344452475
Offset: 0

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Keywords

Comments

With a different offset, number of n-permutations (n>=8) of 4 objects: u, v, z, x with repetition allowed, containing exactly eight (8) u's. Example: a(1)=27 because we have uuuuuuuuv, uuuuuuuuz, uuuuuuuux, uuuuuuuvu, uuuuuuuzu, uuuuuuuxu, uuuuuuvuu, uuuuuuzuu, uuuuuuxuu, uuuuuvuuu, uuuuuzuuu, uuuuuxuuu, uuuuvuuuu, uuuuzuuuu, uuuuxuuuu, uuuvuuuuu, uuuzuuuuu, uuuxuuuuu, uuvuuuuuu, uuzuuuuuu, uuxuuuuuu, uvuuuuuuu, uzuuuuuuu, uxuuuuuuu, vuuuuuuuu, zuuuuuuuu, xuuuuuuuu. - Zerinvary Lajos, Jun 23 2008

Crossrefs

Cf. A027465.
Sequences 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), this sequence (m=8), A036223 (m=9), A172362 (m=10).

Programs

  • Magma
    [3^n*Binomial(n+8, 8): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
  • Maple
    seq(3^n*binomial(n+8,8), n=0..18); # Zerinvary Lajos, Jun 23 2008
  • Mathematica
    Table[3^n*Binomial[n+8, 8], {n, 0, 20}] (* Zerinvary Lajos, Jan 31 2010 *)
    CoefficientList[Series[1/(1-3x)^9,{x,0,30}],x] (* or *) LinearRecurrence[{27,-324, 2268,-10206,30618,-61236,78732,-59049,19683}, {1,27,405,4455,40095,312741, 2189187,14073345,84440070}, 30] (* Harvey P. Dale, Jan 07 2016 *)
  • Sage
    [3^n*binomial(n+8, 8) for n in range(30)] # Zerinvary Lajos, Mar 13 2009
    

Formula

a(n) = 3^n*binomial(n+8, 8).
a(n) = A027465(n+9, 9).
G.f.: 1/(1-3*x)^9.
a(0)=1, a(1)=27, a(2)=405, a(3)=4455, a(4)=40095, a(5)=312741, a(6)=2189187, a(7)=14073345, a(8)=84440070, a(n) = 27*a(n-1) - 324*a(n-2) + 2268*a(n-3) - 10206*a(n-4) + 30618*a(n-5) - 61236*a(n-6) + 78732*a(n-7) - 59049*a(n-8) + 19683*a(n-9). - Harvey P. Dale, Jan 07 2016
From Amiram Eldar, Sep 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 43632/35 - 3072*log(3/2).
Sum_{n>=0} (-1)^n/a(n) = 393216*log(4/3) - 3959208/35. (End)

A036223 Expansion of 1/(1-3*x)^10; 10-fold convolution of A000244 (powers of 3).

Original entry on oeis.org

1, 30, 495, 5940, 57915, 486486, 3648645, 25019280, 159497910, 956987460, 5454828522, 29753610120, 156206453130, 793048146660, 3908594437110, 18761253298128, 87943374834975, 403504896301650, 1815772033357425
Offset: 0

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Author

Keywords

Comments

With a different offset, number of n-permutations (n >= 9) of 4 objects: u, v, z, x with repetition allowed, containing exactly nine (9) u's. - Zerinvary Lajos, Jul 02 2008

Crossrefs

Cf. A027465.
Sequences 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), this sequence (m=9), A172362 (m=10).

Programs

  • Magma
    [3^n*Binomial(n+9, 9): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
  • Maple
    seq(3^n*binomial(n+9, 9), n=0..20); # Zerinvary Lajos, Jul 02 2008
  • Mathematica
    Table[3^n*Binomial[n+9,9], {n,0,30}] (* G. C. Greubel, May 18 2021 *)
    CoefficientList[Series[1/(1-3x)^10,{x,0,30}],x] (* or *) LinearRecurrence[ {30,-405,3240,-17010,61236,-153090,262440,-295245,196830,-59049},{1,30,495,5940,57915,486486,3648645,25019280,159497910,956987460},30] (* Harvey P. Dale, Jan 16 2022 *)
  • Sage
    [3^n*binomial(n+9,9) for n in range(30)] # Zerinvary Lajos, Mar 13 2009
    

Formula

a(n) = 3^n*binomial(n+9, 9).
a(n) = A027465(n+10, 10).
G.f.: 1/(1-3*x)^10.
E.g.f.: (4480 + 120960*x + 725760*x^2 + 1693440*x^3 + 1905120*x^4 + 1143072*x^5 + 381024*x^6 + 69984*x^7 + 6561*x^8 + 243*x^9)*exp(3*x)/4480. - G. C. Greubel, May 18 2021
From Amiram Eldar, Sep 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 6912*log(3/2) - 784431/280.
Sum_{n>=0} (-1)^n/a(n) = 1769472*log(4/3) - 142532433/280. (End)

A038221 Triangle whose (i,j)-th entry is binomial(i,j)*3^(i-j)*3^j.

Original entry on oeis.org

1, 3, 3, 9, 18, 9, 27, 81, 81, 27, 81, 324, 486, 324, 81, 243, 1215, 2430, 2430, 1215, 243, 729, 4374, 10935, 14580, 10935, 4374, 729, 2187, 15309, 45927, 76545, 76545, 45927, 15309, 2187, 6561, 52488, 183708, 367416, 459270, 367416, 183708, 52488, 6561
Offset: 0

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Author

Keywords

Comments

Triangle of coefficients in expansion of (3 + 3x)^n = 3^n (1 +x)^n, where n is a nonnegative integer. (Coefficients in expansion of (1 +x)^n are given in A007318: Pascal's triangle). - Zagros Lalo, Jul 23 2018

Examples

			Triangle begins as:
     1;
     3,     3;
     9,    18,      9;
    27,    81,     81,     27;
    81,   324,    486,    324,     81;
   243,  1215,   2430,   2430,   1215,    243;
   729,  4374,  10935,  14580,  10935,   4374,    729;
  2187, 15309,  45927,  76545,  76545,  45927,  15309,  2187;
  6561, 52488, 183708, 367416, 459270, 367416, 183708, 52488, 6561;
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 44, 48

Crossrefs

Columns k: A000244 (k=0), 3*A027471 (k=1), 3^2*A027472 (k=2), 3^3*A036216 (k=3), 3^4*A036217 (k=4), 3^5*A036219 (k=5), 3^6*A036220 (k=6), 3^7*A036221 (k=7), 3^8*A036222 (k=8), 3^9*A036223 (k=9), 3^10*A172362 (k=10).

Programs

  • GAP
    Flat(List([0..8],i->List([0..i],j->Binomial(i,j)*3^(i-j)*3^j))); # Muniru A Asiru, Jul 23 2018
    
  • Haskell
    a038221 n = a038221_list !! n
    a038221_list = concat $ iterate ([3,3] *) [1]
    instance Num a => Num [a] where
       fromInteger k = [fromInteger k]
       (p:ps) + (q:qs) = p + q : ps + qs
       ps + qs         = ps ++ qs
       (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
        *                = []
    -- Reinhard Zumkeller, Apr 02 2011
    
  • Magma
    [3^n*Binomial(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Oct 17 2022
    
  • Mathematica
    (* programs from Zagros Lalo, Jul 23 2018 *)
    t[0, 0]=1; t[n_, k_]:= t[n, k]= If[n<0 || k<0, 0, 3 t[n-1, k] + 3 t[n-1, k-1]]; Table[t[n, k], {n,0,10}, {k,0,n}]//Flatten
    Table[CoefficientList[Expand[3^n *(1+x)^n], x], {n,0,10}]//Flatten
    Table[3^n Binomial[n, k], {n,0,10}, {k,0,n}]//Flatten  (* End *)
  • SageMath
    def A038221(n,k): return 3^n*binomial(n,k)
    flatten([[A038221(n,k) for k in range(n+1)] for n in range(10)]) # G. C. Greubel, Oct 17 2022

Formula

G.f.: 1/(1 - 3*x - 3*x*y). - Ilya Gutkovskiy, Apr 21 2017
T(0,0) = 1; T(n,k) = 3 T(n-1,k) + 3 T(n-1,k-1) for k = 0...n; T(n,k)=0 for n or k < 0. - Zagros Lalo, Jul 23 2018
From G. C. Greubel, Oct 17 2022: (Start)
T(n, k) = T(n, n-k).
T(n, n) = A000244(n).
T(n, n-1) = 3*A027471(n).
T(n, n-2) = 9*A027472(n+1).
T(n, n-3) = 27*A036216(n-3).
T(n, n-4) = 81*A036217(n-4).
T(n, n-5) = 243*A036219(n-5).
Sum_{k=0..n} T(n, k) = A000400(n).
Sum_{k=0..n} (-1)^k * T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A030195(n+1), n >= 0.
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = A057083(n).
T(n, k) = 3^k * A027465(n, k). (End)

A172362 a(n) = binomial(n+10, 10)*3^n.

Original entry on oeis.org

1, 33, 594, 7722, 81081, 729729, 5837832, 42532776, 287096238, 1818276174, 10909657044, 62482581252, 343654196886, 1824010737318, 9380626649064, 46903133245320, 228652774570935, 1089463220014455, 5084161693400790
Offset: 0

Views

Author

Zerinvary Lajos, Feb 01 2010

Keywords

Comments

With a different offset, number of n-permutations (n>=10) of 4 objects: u, v, z, x with repetition allowed, containing exactly ten, (10) u's.

Crossrefs

Programs

  • Magma
    [3^n*Binomial(n+10, 10): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
  • Maple
    seq(binomial(n+10, 10)*3^n, n=0..30);
  • Mathematica
    Table[Binomial[n + 10, 10]*3^n, {n, 0, 20}]

Formula

G.f.: 1/(1-3*x)^11. - Vincenzo Librandi, Oct 15 2011
From Amiram Eldar, Aug 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 261617/42 - 15360*log(3/2).
Sum_{n>=0} (-1)^n/a(n) = 7864320*log(4/3) - 47510881/21. (End)
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