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.

A006684 Convolve Fibonacci and Pell numbers.

Original entry on oeis.org

0, 0, 1, 3, 9, 24, 62, 156, 387, 951, 2323, 5652, 13716, 33228, 80405, 194415, 469845, 1135092, 2741626, 6620928, 15987663, 38603019, 93204647, 225030024, 543293352, 1311663096, 3166694569, 7645173627, 18457238241, 44559967920, 107577688310, 259716176580
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Define a triangle T(r,c) by T(n,0) = T(n,n) = A000045(n) and T(r,c) = T(r-1,c) + T(r-1,c-1) + T(r-2,c-1). The sum of the terms in the first n rows is 2*a(n+1). - J. M. Bergot, Apr 07 2013

Links

Crossrefs

Cf. A000045, A000129, A106515 (first differences).

Programs

  • Magma
    Pell:= func< n | Round(((1+Sqrt(2))^n - (1-Sqrt(2))^n)/(2*Sqrt(2))) >;
[Pell(n) - Fibonacci(n): n in [0..30]]; // G. C. Greubel, Aug 05 2021
  • Maple
    with(combinat):seq(fibonacci(i,2)-fibonacci(i, 1),i=0..27); # Zerinvary Lajos, Mar 20 2008
  • Mathematica
    LinearRecurrence[{3,0,-3,-1}, {0,0,1,3}, 50] (* T. D. Noe, Apr 16 2013 *)
Table[Fibonacci[n, 2] - Fibonacci[n], {n, 0, 30}] (* Vladimir Reshetnikov, Sep 27 2016 *)
  • Sage
    [lucas_number1(n, 2, -1) - lucas_number1(n, 1, -1) for n in (0..30)] # G. C. Greubel, Aug 05 2021

Formula

a(n) = Pell(n) - Fibonacci(n).
G.f.: x^2/( (1-x-x^2)*(1-2*x-x^2) ). - Joerg Arndt, Apr 17 2013
a(n) = 3*a(n-1) - 3*a(n-3) - a(n-4) with a(0) = a(1) = 0, a(2) = 1, a(3) = 3. - Taras Goy, Mar 12 2019

A106513 A Pell-Pascal matrix.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 12, 8, 4, 1, 29, 20, 12, 5, 1, 70, 49, 32, 17, 6, 1, 169, 119, 81, 49, 23, 7, 1, 408, 288, 200, 130, 72, 30, 8, 1, 985, 696, 488, 330, 202, 102, 38, 9, 1, 2378, 1681, 1184, 818, 532, 304, 140, 47, 10, 1, 5741, 4059, 2865, 2002, 1350, 836, 444, 187, 57, 11, 1
Offset: 0

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Author

Paul Barry, May 05 2005

Keywords

Comments

This triangle gives the iterated partial sums of the Pell sequence A000129(n+1), n>=0. - Wolfdieter Lang, Oct 05 2014

Examples

			The triangle T(n,k) begins:
n\k    0    1    2    3    4   5   6   7  8  9 10 ...
0:     1
1:     2    1
2:     5    3    1
3:    12    8    4    1
4:    29   20   12    5    1
5:    70   49   32   17    6   1
6:   169  119   81   49   23   7   1
7:   408  288  200  130   72  30   8   1
8:   985  696  488  330  202 102  38   9  1
9:  2378 1681 1184  818  532 304 140  47 10  1
10: 5741 4059 2865 2002 1350 836 444 187 57 11  1
... Reformatted and extended. - _Wolfdieter Lang_, Oct 05 2014
-----------------------------------------------------
Recurrence from the Z-sequence (see the formula above) for T(0,n) in terms of the entries of row n-1. For example, 29 = T(4,0) = 2*12 + 1*8 + (-1)*4 + 1*1 = 29. - _Wolfdieter Lang_, Oct 05 2014

Links

Crossrefs

Cf. A000129, A001333, A106514 (row sums), A106515 (antidiagonal sums), A248156.

Programs

  • Magma
    [ (&+[Binomial(n+1, 2*j+k+1)*2^j: j in [0..Floor((n+1)/2)]]) : k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 05 2021
  • Mathematica
    T[n_, k_]= Sum[Binomial[n+1, 2*j+k+1]*2^j, {j, 0, Floor[(n+1)/2]}];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 05 2021 *)
  • Sage
    @CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return lucas_number1(n+1, 2, -1)
    else: return T(n-1,k-1) + T(n-1,k)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 05 2021

Formula

Riordan array (1/(1-2*x-x^2), x/(1-x)).
Number triangle T(n,0) = A000129(n+1), T(n,k) = T(n-1,k-1) + T(n-1,k).
T(n,k) = Sum_{j=0..floor((n+1)/2)} binomial(n+1, 2*j+k+1)*2^j.
Sum_{k=0..n} T(n, k) = A106514(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A106515(n).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-3,k) - T(n-3,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 14 2014
From Wolfdieter Lang, Oct 05 2014: (Start)
O.g.f. for row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k: (1 - z)/((1 - 2*z - z^2)*(1 - (1+x)*z)).
O.g.f. column m: (1/(1 - 2*z - z^2))*(z/(1 - z))^m, m >= 0. (Riordan property).
The alternating row sums are shown in A001333.
A-sequence: [1, 1] (see the three term recurrence given above). Z-sequence has o.g.f. (2 + 3*x)/(1 + x), [2, 1, repeat(-1,1)] (unsigned A054977). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
The inverse Riordan triangle is shown in A248156. (End)

A212362 Triangle by rows, binomial transform of the beheaded Pascal's triangle A074909.

Original entry on oeis.org

1, 2, 2, 4, 7, 3, 8, 19, 15, 4, 16, 47, 52, 26, 5, 32, 111, 155, 110, 40, 6, 64, 255, 426, 385, 200, 57, 7, 128, 575, 1113, 1211, 805, 329, 77, 8, 256, 1279, 2808, 3556, 2856, 1498, 504, 100, 9, 512, 2815, 6903, 9948, 9324, 5922, 2562, 732, 126, 10
Offset: 0

Views

Author

Gary W. Adamson, Jun 29 2012

Keywords

Comments

Row sums of the triangle inverse = A027641/A027642, the Bernoulli numbers; (1, -1/2, 1/6, 0, -1/30,...)

Examples

			First few rows of the triangle are:
    1;
    2,    2;
    4,    7,    3;
    8,   19,   15,    4
   16,   47,   52,   26,    5;
   32,  111,  155,  110,   40,    6;
   64,  255,  426,  385,  200,   57,   7;
  128,  575, 1113, 1211,  805,  329,  77,   8;
  256, 1279, 2808, 3556, 2856, 1498, 504, 100, 9;
  ...

Links

Crossrefs

Cf. A074909, A027641/A027642, A027649 (row sums), A006589 (2nd column), A106515.

Programs

  • Magma
    A074909:= func< n,k | k lt 0 or k gt n select 0 else Binomial(n+1, k) >;
A212362:= func< n,k | (&+[ Binomial(n,j)*A074909(j, k) : j in [0..n]]) >;
[A212362(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 05 2021
  • Maple
    A212362 := proc(n,k)
        add( binomial(n,i)*A074909(i,k),i=0..n) ;
end proc: # R. J. Mathar, Aug 03 2015
  • Mathematica
    T[n_, k_]= 2^(n-k)*Binomial[n+1, k] + (2^(n-k) -1)*Binomial[n, k-1];
Table[T[n, k] , {n,0,12}, {k,0,n}] //Flatten (* G. C. Greubel, Aug 05 2021 *)
  • Sage
    def T(n, k): return 2^(n-k)*binomial(n+1, k) + (2^(n-k) - 1)*binomial(n, k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 05 2021

Formula

Binomial transform of the beheaded Pascal's triangle (A074909) as a matrix. (The beheaded Pascal matrix deletes the rightmost border of 1's.)
From G. C. Greubel, Aug 05 2021: (Start)
T(n, k) = Sum_{j=0..n} binomial(n, j)*binomial(j+1, k) - binomial(n, k-1), with T(n, 0) = 2^n.
T(n, k) = 2^(n-k)*binomial(n+1, k) + (2^(n-k) - 1)*binomial(n, k-1).
Sum_{k=0..n} T(n, k) = A027649(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A106515(n). (End)

Extensions

a(22) corrected by G. C. Greubel, Aug 05 2021
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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