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 1-3 of 3 results.

A116415 a(n) = 5*a(n-1) - 3*a(n-2).

Original entry on oeis.org

1, 5, 22, 95, 409, 1760, 7573, 32585, 140206, 603275, 2595757, 11168960, 48057529, 206780765, 889731238, 3828313895, 16472375761, 70876937120, 304967558317, 1312206980225, 5646132226174, 24294040190195, 104531804272453
Offset: 0

Views

Author

Paul Barry, Feb 13 2006

Keywords

Comments

Row sums of A116414.
Partial sums of A018902. - Greg Dresden and Mulong Xu, Aug 31 2024
Binomial transform of the sequence A006190. - Sergio Falcon, Nov 23 2007
a(n+1) equals the number of words of length n over {0,1,2,3,4} avoiding 01, 02 and 03. - Milan Janjic, Dec 17 2015

Crossrefs

Programs

Formula

G.f.: 1/(1 - 5*x + 3*x^2).
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n-j,k)*C(k+j,j)*3^j.
a(n) = (1/sqrt(13))*(((5+sqrt(13))/2)^n - ((5-sqrt(13))/2)^n). - Sergio Falcon, Nov 23 2007
If p[i] = (3^i-1)/2, and if A is the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010
a(n) = 4*a(n-1) + a(n-2) + a(n-3) + ... + a(0) + 1. These expansions with the partial sums on one side can be generated en masse by taking the g.f. of the partial sum and its partial fraction, 1/(1-x)/(1 - 5*x + 3*x^2) = -1/(1-x)+(2-3*x)/(1 - 5*x + 3*x^2) and reading this as a(0) + a(1) + ... + a(n) = -1 + 2*a(n)- 3*a(n-1). - Gary W. Adamson, Feb 18 2011

A110441 Triangular array formed by the Mersenne numbers.

Original entry on oeis.org

1, 3, 1, 7, 6, 1, 15, 23, 9, 1, 31, 72, 48, 12, 1, 63, 201, 198, 82, 15, 1, 127, 522, 699, 420, 125, 18, 1, 255, 1291, 2223, 1795, 765, 177, 21, 1, 511, 3084, 6562, 6768, 3840, 1260, 238, 24, 1, 1023, 7181, 18324, 23276, 16758, 7266, 1932, 308, 27, 1
Offset: 0

Views

Author

Asamoah Nkwanta (nkwanta(AT)jewel.morgan.edu), Aug 08 2005

Keywords

Comments

This sequence factors A038255 into a product of Riordan arrays.
Subtriangle of the triangle given by (0, 3, -2/3, 2/3, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 19 2012
From Peter Bala, Jul 22 2014: (Start)
Let M denote the lower unit triangular array A130330 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)
For 1<=k<=n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01 and 02. - Milan Janjic, Dec 20 2016
The convolution triangle of the Mersenne numbers. - Peter Luschny, Oct 09 2022

Examples

			Triangle starts:
   1;
   3,  1;
   7,  6,  1;
  15, 23,  9,  1;
  31, 72, 48, 12,  1;
(0, 3, -2/3, 2/3, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, ...) begins:
  1
  0,  1
  0,  3,  1
  0,  7,  6,  1
  0, 15, 23,  9,  1
  0, 31, 72, 48, 12, 1. - _Philippe Deléham_, Mar 19 2012
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
/ 1          \/1         \/1        \      / 1       \
| 3  1       ||0  1      ||0 1      |      | 3  1    |
| 7  3 1     ||0  3 1    ||0 0 1    |... = | 7  6 1  |
|15  7 3 1   ||0  7 3 1  ||0 0 3 1  |      |15 23 9 1|
|31 15 7 3 1 ||0 15 7 3 1||0 0 7 3 1|      |...      |
|...         ||...       ||...      |      |...      | - _Peter Bala_, Jul 22 2014
		

Crossrefs

Programs

  • Maple
    # Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
    PMatrix(10, n -> 2^n - 1); # Peter Luschny, Oct 09 2022
  • Mathematica
    With[{n = 9}, DeleteCases[#, 0] & /@ CoefficientList[Series[1/(1 - (3 + y) x + 2 x^2), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)

Formula

Riordan array M(n, k): (1/(1-3z+2z^2), z/(1-3z+2z^2)). Leftmost column M(n, 0) is the Mersenne numbers A000225, first column is A045618, second column is A055582, row sum is A007070 and diagonal sum is even-indexed Fibonacci numbers A001906.
T(n,k) = Sum_{j=0..n} C(j+k,k)C(n-j,k)2^(n-j-k). - Paul Barry, Feb 13 2006
From Philippe Deléham, Mar 19 2012: (Start)
G.f.: 1/(1-(3+y)*x+2*x^2).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) -2*T(n-2,k), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
Sum_{k, 0<=k<=n} T(n,k)*x^k = A000225(n+1), A007070(n), A107839(n), A154244(n), A186446(n), A190975(n+1), A190979(n+1), A190869(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7 respectively. (End)
Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (2^(i+1) - 1)*T(n-i,k). - Peter Bala, Jul 22 2014
From Peter Bala, Oct 07 2019: (Start)
Recurrence for row polynomials: R(n,x) = (3 + x)*R(n-1,x) - 2*R(n-2,x) with R(0,x) = 1 and R(1,x) = 3 + x.
The row reverse polynomial x^n*R(n,1/x) is equal to the numerator polynomial of the finite continued fraction 1 + x/(1 + 2*x/(1 + ... + x/(1 + 2*x/(1)))) (with 2*n partial numerators). Cf. A116414. (End)

A116412 Riordan array ((1+x)/(1-2x),x(1+x)/(1-2x)).

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 12, 21, 9, 1, 24, 60, 45, 12, 1, 48, 156, 171, 78, 15, 1, 96, 384, 558, 372, 120, 18, 1, 192, 912, 1656, 1473, 690, 171, 21, 1, 384, 2112, 4608, 5160, 3225, 1152, 231, 24, 1, 768, 4800, 12240, 16584, 13083, 6219, 1785, 300, 27, 1, 1536, 10752
Offset: 0

Views

Author

Paul Barry, Feb 13 2006

Keywords

Comments

Row sums are A003688. Diagonal sums are A116413. Product of A007318 and A116413 is A116414. Product of A007318 and A105475.
Subtriangle of triangle given by (0, 3, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 18 2012

Examples

			Triangle begins
1,
3, 1,
6, 6, 1,
12, 21, 9, 1,
24, 60, 45, 12, 1,
48, 156, 171, 78, 15, 1
Triangle T(n,k), 0<=k<=n, given by (0, 3, -1, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, ...) begins :
1
0, 1
0, 3, 1
0, 6, 6, 1
0, 12, 21, 9, 1
0, 24, 60, 45, 12, 1
0, 48, 156, 171, 78, 15, 1
... - _Philippe Deléham_, Jan 18 2012
		

Crossrefs

Programs

  • Mathematica
    With[{n = 10}, DeleteCases[#, 0] & /@ CoefficientList[Series[(1 + x)/(1 - (y + 2) x - y x^2), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)

Formula

Number triangle T(n,k)=sum{j=0..n, C(k+1,j)*C(n-j,k)2^(n-k-j)}
From Vladimir Kruchinin, Mar 17 2011: (Start)
T((m+1)*n+r-1, m*n+r-1) * r/(m*n+r) = sum(k=1..n, k/n * T((m+1)*n-k-1, m*n-1) * T(r+k-1,r-1)), n>=m>1.
T(n-1,m-1) = m/n * sum(k=1..n-m+1, k*A003945(k-1)*T(n-k-1,m-2)), n>=m>1. (End)
G.f.: (1+x)/(1-(y+2)*x -y*x^2). - Philippe Deléham, Jan 18 2012
Sum_{k, 0<=k<=n} T(n,k)*x^k = A104537(n), A110523(n), (-2)^floor(n/2), A057079(n), A003945(n), A003688(n+1), A123347(n), A180035(n) for x = -4, -3, -2, -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Jan 18 2012
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(2,0) = T(2,1) = 6, T(2,2) = 1, T(n,k) = 0 if k>n or if k<0. - Philippe Deléham, Oct 31 2013
Showing 1-3 of 3 results.