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.

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A034182 Number of not-necessarily-symmetric n X 2 crossword puzzle grids.

Original entry on oeis.org

1, 5, 15, 39, 97, 237, 575, 1391, 3361, 8117, 19599, 47319, 114241, 275805, 665855, 1607519, 3880897, 9369317, 22619535, 54608391, 131836321, 318281037, 768398399, 1855077839, 4478554081, 10812186005, 26102926095, 63018038199, 152139002497, 367296043197
Offset: 1

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Author

Keywords

Comments

n X 2 binary arrays with a path of adjacent 1's and no path of adjacent 0's from top row to bottom row. - R. H. Hardin, Mar 21 2002
Define a triangle with T(n,1) = T(n,n) = n*(n-1) + 1, n>=1, and its interior terms via T(r,c) = T(r-1,c) + T(r-1,c-1)+ T(r-2,c-1), 2<=cJ. M. Bergot, Mar 16 2013

Crossrefs

Row 2 of A292357.
Column sums of A059678.
Cf. A001333, A034184, A034187, A052542 (first differences).

Programs

  • Haskell
    a034182 n = a034182_list !! (n-1)
    a034182_list = 1 : 5 : (map (+ 4) $
       zipWith (+) a034182_list (map (* 2) $ tail a034182_list))
    -- Reinhard Zumkeller, May 23 2013
  • Mathematica
    {1}~Join~NestList[{#2, 2 #2 + #1 + 4} & @@ # &, {1, 5}, 28][[All, -1]] (* Michael De Vlieger, Oct 02 2017 *)

Formula

a(n) = 2a(n-1) + a(n-2) + 4.
(1 + 5x + 15x^2 + ...) = (1 + 2x + 2x^2 + ...) * (1 + 3x + 7x^2 + ...), convolution of A040000 and left-shifted A001333.
a(n) = (-4 + (1-sqrt(2))^(1+n) + (1+sqrt(2))^(1+n))/2. G.f.: x*(1+x)^2/((1-x)*(1 - 2*x - x^2)). - Colin Barker, May 22 2012
a(n) = A001333(n+1)-2. - R. J. Mathar, Mar 28 2013
a(n) = A048739(n-3) +2*A048739(n-2) +A048739(n-1). - R. J. Mathar, Jun 15 2020

A059020 Number of 2 X n checkerboards (with at least one red square) in which the set of red squares is edge connected.

Original entry on oeis.org

0, 3, 13, 40, 108, 275, 681, 1664, 4040, 9779, 23637, 57096, 137876, 332899, 803729, 1940416, 4684624, 11309731, 27304157, 65918120, 159140476, 384199155, 927538873, 2239276992, 5406092952, 13051462995, 31509019045, 76069501192, 183648021540, 443365544387
Offset: 0

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Author

John W. Layman, Dec 14 2000

Keywords

Comments

In other words, the number of connected (non-null) induced subgraphs in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, May 02 2017
Also, the number of cycles in the grid graph P_3 X P_{n+1}. - Andrew Howroyd, Jun 12 2017

Crossrefs

Row 2 of A287151 and row 2 of A231829.
See also A059021, A059524.
Cf. A000129. - Jaume Oliver Lafont, Sep 28 2009
Other sequences counting connected induced subgraphs: A020873, A059525, A286139, A286182, A286183, A286184, A286185, A286186, A286187, A286188, A286189, A286191, A285765, A285934, A286304.

Programs

  • Magma
    I:=[0, 3, 13, 40];[n le 4 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-4):n in [1..30]]; // Marius A. Burtea, Aug 25 2019
  • Mathematica
    Join[{0},LinearRecurrence[{4, -4, 0, 1}, {3, 13, 40, 108}, 20]] (* Eric W. Weisstein, May 02 2017 *) (* adapted by Vincenzo Librandi, May 09 2017 *)
    Table[(LucasL[n + 3, 2] - 8 n - 14)/4, {n, 0, 20}] (* Eric W. Weisstein, May 02 2017 *)

Formula

a(n) = 2*a(n-1) + a(n-2) + 4*n - 1.
From Jaume Oliver Lafont, Nov 23 2008: (Start)
a(n) = 3*a(n-1) - a(n-2) - a(n-3) + 4;
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4). (End)
G.f.: x*(3+x)/((1-2*x-x^2)*(1-x)^2). - Jaume Oliver Lafont, Sep 28 2009
Empirical observations (from Superseeker):
(1) if b(n) = a(n) + n then {b(n)} is A048777;
(2) if b(n) = a(n+3) - 3*a(n+2) - 3*a(n+1) + a(n) then {b(n)} is A052542;
(3) if b(n) = a(n+2) - 2*a(n+1) + a(n) then {b(n)} is A001333.
4*a(n) = A002203(n+3) - 8*n - 14. - Eric W. Weisstein, May 02 2017
a(n) = 3*A048776(n-1) + A048776(n-2). - R. J. Mathar, May 12 2019
E.g.f.: (1/2)*exp(x)*(-7-4*x+7*cosh(sqrt(2)*x)+5*sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, Aug 25 2019

A083100 a(n) = 2*a(n-1) + 7*a(n-2).

Original entry on oeis.org

1, 9, 25, 113, 401, 1593, 5993, 23137, 88225, 338409, 1294393, 4957649, 18976049, 72655641, 278143625, 1064876737, 4076758849, 15607654857, 59752621657, 228758827313, 875786006225, 3352883803641, 12836269650857, 49142725927201
Offset: 0

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Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 23 2003

Keywords

Comments

a(n) = a(n-1) + 8*A015519(n). a(n)/A015519(n+1) converges to sqrt(8).
a(n-1) is the number of compositions of n when there is 1 type of 1 and 8 types of other natural numbers. - Milan Janjic, Aug 13 2010

Crossrefs

Essentially a duplicate of A084058.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) + 7*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Mathematica
    CoefficientList[Series[(1 + 7 x)/(1 - 2 x - 7 x^2), {x, 0, 25}], x] (* Or *) a[n_] := Simplify[((1 + Sqrt[8])^n + (1 - Sqrt[8])^n)/2]; Array[a, 25, 0] (* Or *) LinearRecurrence[{2, 7}, {1, 1}, 28] (* Or *) Table[ MatrixPower[{{1, 2}, {4, 1}}, n][[1, 1]], {n, 0, 25}] (* Robert G. Wilson v, Sep 18 2013 *)
  • PARI
    a(n)=([0,1; 7,2]^n*[1;9])[1,1] \\ Charles R Greathouse IV, Apr 06 2016
    
  • PARI
    x='x+O('x^30); Vec((1+7*x)/(1-2*x-7*x^2)) \\ G. C. Greubel, Jan 08 2018
    

Formula

G.f.: (1+7*x)/(1-2*x-7*x^2).
a(n) = binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
If p[1]=1, and p[i]=8,(i>1), and if A is 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, Apr 29 2010
G.f.: G(0)/(2*x) - 1/x, where G(k)= 1 + 1/(1 - x*(8*k-1)/(x*(8*k+7) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013

A266213 Square array A(n,r), the number of neighbors at a sharp Manhattan distance r in a finite n-hypercube lattice, read by upwards antidiagonals; A(n,r) = Sum_{k=0..min(n,r)} binomial(r-1,k-1)*binomial(n,k)* 2^k.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 2, 0, 1, 6, 8, 2, 0, 1, 8, 18, 12, 2, 0, 1, 10, 32, 38, 16, 2, 0, 1, 12, 50, 88, 66, 20, 2, 0, 1, 14, 72, 170, 192, 102, 24, 2, 0, 1, 16, 98, 292, 450, 360, 146, 28, 2, 0, 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2, 0
Offset: 0

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Author

Dmitry Zaitsev, Dec 24 2015

Keywords

Comments

In an n-dimensional hypercube lattice, the array A(n,r) gives the number of nodes situated at a Manhattan distance equal to r, counting the current node. When counting coordinate offsets for neighboring nodes, binomial(n,k) chooses k nonzero coordinates from n coordinates, binomial(r-1,k-1) partitions the number r as the sum of exactly k nonzero numbers, and 2^k counts combinations of signs for coordinate offsets; starting indexing from 0 adds 1, which counts the current node.
In cellular automata theory, the cell surrounding with Manhattan distance less than or equal to r is called the von Neumann neighborhood of radius r or the diamond-shaped neighborhood to distinguish it from other generalizations of the von Neumann neighborhood for radius r>1, for instance, as a neighborhood having a difference in the range from -r to r in exactly one coordinate (the "narrow" von Neumann neighborhood of radius r).
The square array of partial sums of A(n,r) on rows gives us the Delannoy numbers A008288, which correspond to the number of nodes in the diamond-shaped neighborhood of radius r. - Dmitry Zaitsev, Dec 24 2015
For n >= 2, the term A(n,r) gives the number of polyominoes of bounding box 2 x (r+n-1) of area (r + 2(n-1)). Let A'(n,k) be the table A(n,k) without the first two rows. The sum of the terms in the i-th anti-diagonal of A'(n,k) gives the i-th term of A034182. - Louis Marin, Dec 11 2024

Examples

			The array A(n, k) begins:
n \ k  0  1   2   3    4     5     6      7      8      9
---------------------------------------------------------
0:     1  0   0   0    0     0     0      0      0      0
1:     1  2   2   2    2     2     2      2      2      2
2:     1  4   8  12   16    20    24     28     32     36
3:     1  6  18  38   66   102   146    198    258    326
4:     1  8  32  88  192   360   608    952   1408   1992
5:     1 10  50 170  450  1002  1970   3530   5890   9290
6:     1 12  72 292  912  2364  5336  10836  20256  35436
7:     1 14  98 462 1666  4942 12642  28814  59906 115598
8:     1 16 128 688 2816  9424 27008  68464 157184 332688
9:     1 18 162 978 4482 16722 53154 148626 374274 864146
...
For instance, in a 5-hypercube lattice there are 170 nodes situated at a Manhattan distance of 3 for a chosen node.
The triangle T(m, r) begins:
m\r 0  1   2   3   4    5   6   7  8 9 10 ...
0:  1
1:  1  0
2:  1  2   0
3:  1  4   2   0
4:  1  6   8   2   0
5:  1  8  18  12   2    0
6:  1 10  32  38  16    2   0
7:  1 12  50  88  66   20   2   0
8:  1 14  72 170 192  102  24   2  0
9:  1 16  98 292 450  360 146  28  2 0
10: 1 18 128 462 912 1002 608 198 32 2  0
... Formatted by _Wolfdieter Lang_, Jan 31 2016
		

Crossrefs

Other versions: A035607, A113413, A119800, A122542.
Partial sums on rows of A give A008288.
Cf. A001333 (row sums of T). A057077 (alternating row sums of T). - Wolfdieter Lang, Jan 31 2016

Programs

  • Maple
    # Prints the array by rows.
    gf := n -> ((1 + x)/(1 - x))^n: ser := n -> series(gf(n), x, 40):
    seq(lprint(seq(coeff(ser(n), x, k), k=0..6)), n=0..9); # Peter Luschny, Mar 20 2020
  • Mathematica
    Table[Sum[Binomial[r - 1, k - 1] Binomial[n - r, k] 2^k, {k, 0, Min[n - r, r]}], {n, 0, 10}, {r, 0, n}] // Flatten (* Michael De Vlieger, Dec 24 2015 *)
  • Python
    from sympy import binomial
    def T(n, r):
        if r==0: return 1
        return sum(binomial(r - 1, k - 1) * binomial(n - r, k) * 2**k for k in range(min(n - r, r) + 1))
    for n in range(11): print([T(n, r) for r in range(n + 1)]) # Indranil Ghosh, May 23 2017

Formula

A(n, 0)=1, n>=0, A(0, r)=0, r>0.
A(n, r) = A(n, r-1) + A(n-1, r-1) + A(n-1, r).
A(n, r) = Sum_{k=0..min(n,r)} binomial(r-1,k-1)*binomial(n,k)*2^k.
Triangle T(m, r) = A(m-r, r), n >= 0, 0 <= r <= n, otherwise 0. - Wolfdieter Lang, Jan 31 2016
A(n, k) = [x^k] ((1 + x)/(1 - x))^n. - Ilya Gutkovskiy, May 23 2017

A214992 Power ceiling-floor sequence of (golden ratio)^4.

Original entry on oeis.org

7, 47, 323, 2213, 15169, 103969, 712615, 4884335, 33477731, 229459781, 1572740737, 10779725377, 73885336903, 506417632943, 3471038093699, 23790849022949, 163064905066945, 1117663486445665, 7660579500052711
Offset: 0

Views

Author

Clark Kimberling, Nov 08 2012, Jan 24 2013

Keywords

Comments

Let f = floor and c = ceiling. For x > 1, define four sequences as functions of x, as follows:
p1(0) = f(x), p1(n) = f(x*p1(n-1));
p2(0) = f(x), p2(n) = c(x*p2(n-1)) if n is odd and p2(n) = f(x*p1(n-1)) if n is even;
p3(0) = c(x), p3(n) = f(x*p3(n-1)) if n is odd and p3(n) = c(x*p3(n-1)) if n is even;
p4(0) = c(x), p4(n) = c(x*p4(n-1)).
The present sequence is given by a(n) = p3(n).
Following the terminology at A214986, call the four sequences power floor, power floor-ceiling, power ceiling-floor, and power ceiling sequences. In the table below, a sequence is identified with an A-numbered sequence if they appear to agree except possibly for initial terms. Notation: S(t)=sqrt(t), r = (1+S(5))/2 = golden ratio, and Limit = limit of p3(n)/p2(n).
x ......p1..... p2..... p3..... p4.......Limit
r^2.....A001519 A001654 A061646 A001906..-1+S(5)
r^3.....A024551 A001076 A015448 A049652..-1+S(5)
r^4.....A049685 A157335 A214992 A004187..-19+9*S(5)
r^5.....A214993 A049666 A015457 A214994...(-9+5*S(5))/2
r^6.....A007805 A156085 A214995 A049660..-151+68*S(5)
2+S(2)..A007052 A214996 A214997 A007070..(1+S(2))/2
1+S(3)..A057960 A002605 A028859 A077846..(1+S(3))/2
2+S(3)..A001835 A109437 A214998 A001353..-4+3*S(3)
S(5)....A214999 A215091 A218982 A218983..1.26879683...
2+S(5)..A024551 A001076 A015448 A049652..-1+S(5)
2+S(6)..A218984 A090017 A123347 A218985..S(3/2)
2+S(7)..A218986 A015530 A126473 A218987..(1+S(7))/3
2+S(8)..A218988 A057087 A086347 A218989..(1+S(2))/2
3+S(8)..A001653 A084158 A218990 A001109..-13+10*S(2)
3+S(10).A218991 A005668 A015451 A218992..-2+S(10)
...
Properties of p1, p2, p3, p4:
(1) If x > 2, the terms of p2 and p3 interlace: p2(0) < p3(0) < p2(1) < p3(1) < p2(2) < p3(2)... Also, p1(n) <= p2(n) <= p3(n) <= p4(n) <= p1(n+1) for all x>0 and n>=0.
(2) If x > 2, the limits L(x) = limit(p/x^n) exist for the four functions p(x), and L1(x) <= L2(x) <= L3(x) <= L4 (x). See the Mathematica programs for plots of the four functions; one of them also occurs in the Odlyzko and Wilf article, along with a discussion of the special case x = 3/2.
(3) Suppose that x = u + sqrt(v) where v is a nonsquare positive integer. If u = f(x) or u = c(x), then p1, p2, p3, p4 are linear recurrence sequences. Is this true for sequences p1, p2, p3, p4 obtained from x = (u + sqrt(v))^q for every positive integer q?
(4) Suppose that x is a Pisot-Vijayaraghavan number. Must p1, p2, p3, p4 then be linearly recurrent? If x is also a quadratic irrational b + c*sqrt(d), must the four limits L(x) be in the field Q(sqrt(d))?
(5) The Odlyzko and Wilf article (page 239) raises three interesting questions about the power ceiling function; it appears that they remain open.

Examples

			a(0) = ceiling(r) = 7, where r = ((1+sqrt(5))/2)^4 = 6.8...; a(1) = floor(7*r) = 47; a(2) = ceiling(47) = 323.
		

Crossrefs

Programs

  • Mathematica
    (* Program 1.  A214992 and related sequences *)
    x = GoldenRatio^4; z = 30; (* z = # terms in sequences *)
    z1 = 100; (* z1 = # digits in approximations *)
    f[x_] := Floor[x]; c[x_] := Ceiling[x];
    p1[0] = f[x]; p2[0] = f[x]; p3[0] = c[x]; p4[0] = c[x];
    p1[n_] := f[x*p1[n - 1]]
    p2[n_] := If[Mod[n, 2] == 1, c[x*p2[n - 1]], f[x*p2[n - 1]]]
    p3[n_] := If[Mod[n, 2] == 1, f[x*p3[n - 1]], c[x*p3[n - 1]]]
    p4[n_] := c[x*p4[n - 1]]
    Table[p1[n], {n, 0, z}]  (* A049685 *)
    Table[p2[n], {n, 0, z}]  (* A157335 *)
    Table[p3[n], {n, 0, z}]  (* A214992 *)
    Table[p4[n], {n, 0, z}]  (* A004187 *)
    Table[p4[n] - p1[n], {n, 0, z}]  (* A004187 *)
    Table[p3[n] - p2[n], {n, 0, z}]  (* A098305 *)
    (* Program 2.  Plot of power floor and power ceiling functions, p1(x) and p4(x) *)
    f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
    p1[x_, 0] := f[x]; p1[x_, n_] := f[x*p1[x, n - 1]];
    p4[x_, 0] := c[x]; p4[x_, n_] := c[x*p4[x, n - 1]];
    Plot[Evaluate[{p1[x, 10]/x^10, p4[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]
    (* Program 3. Plot of power floor-ceiling and power ceiling-floor functions, p2(x) and p3(x) *)
    f[x_] := f[x] = Floor[x]; c[x_] := c[x] = Ceiling[x];
    p2[x_, 0] := f[x]; p3[x_, 0] := c[x];
    p2[x_, n_] := If[Mod[n, 2] == 1, c[x*p2[x, n - 1]], f[x*p2[x, n - 1]]]
    p3[x_, n_] := If[Mod[n, 2] == 1, f[x*p3[x, n - 1]], c[x*p3[x, n - 1]]]
    Plot[Evaluate[{p2[x, 10]/x^10, p3[x, 10]/x^10}], {x, 2, 3}, PlotRange -> {0, 4}]

Formula

a(n) = floor(r*a(n-1)) if n is odd and a(n) = ceiling(r*a(n-1)) if n is even, where a(0) = ceiling(r), r = (golden ratio)^4 = (7 + sqrt(5))/2.
a(n) = 6*a(n-1) + 6*a(n-2) - a(n-3).
G.f.: (7 + 5*x - x^2)/((1 + x)*(1 - 7*x + x^2)).
a(n) = (10*(-2)^n+(10+3*sqrt(5))*(7-3*sqrt(5))^(n+2)+(10-3*sqrt(5))*(7+3*sqrt(5))^(n+2))/(90*2^n). - Bruno Berselli, Nov 14 2012
a(n) = 7*A157335(n) + 5*A157335(n-1) - A157335(n-2). - R. J. Mathar, Feb 05 2020
E.g.f.: exp(-x)*(5 + 2*exp(9*x/2)*(155*cosh(3*sqrt(5)*x/2) + 69*sqrt(5)*sinh(3*sqrt(5)*x/2)))/45. - Stefano Spezia, Oct 28 2024

A122950 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 0, 1, -1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 3, 5, 0, 0, 0, 1, 7, 8, 0, 0, 0, 0, 4, 15, 13, 0, 0, 0, 0, 1, 12, 30, 21, 0, 0, 0, 0, 0, 5, 31, 58, 34, 0, 0, 0, 0, 0, 1, 18, 73, 109, 55, 0, 0, 0, 0, 0, 0, 6, 54, 162, 201, 89, 0, 0, 0, 0, 0, 0, 1, 25, 145, 344, 365, 144, 0, 0, 0, 0, 0, 0, 0, 7, 85, 361, 707
Offset: 0

Views

Author

Philippe Deléham, Oct 25 2006

Keywords

Comments

Skew triangle associated with the Fibonacci numbers.

Examples

			Triangle begins:
  1;
  0, 1;
  0, 0, 2;
  0, 0, 1, 3;
  0, 0, 0, 3, 5;
  0, 0, 0, 1, 7, 8;
  0, 0, 0, 0, 4, 15, 13;
  0, 0, 0, 0, 1, 12, 30, 21;
  0, 0, 0, 0, 0,  5, 31, 58,  34;
  0, 0, 0, 0, 0,  1, 18, 73, 109,  55;
  0, 0, 0, 0, 0,  0,  6, 54, 162, 201,  89;
  0, 0, 0, 0, 0,  0,  1, 25, 145, 344, 365, 144;
  0, 0, 0, 0, 0,  0,  0,  7,  85, 361, 707, 655, 233;
		

Crossrefs

Cf. A055830 (another version).

Programs

  • Mathematica
    T[0, 0] = T[1, 1] = 1; T[, 0] = T[, 1] = 0; T[n_, n_] := Fibonacci[n+1]; T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-2, k-1] + T[n-2, k-2]; T[, ] = 0;
    Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 29 2018 *)

Formula

Sum_{k=0..n} T(n,k) = A011782(n).
Sum_{n>=k} T(n,k) = A001333(k).
T(n,k) = 0 if k < 0 or if k > n, T(0,0) = 1, T(2,1) = 0, T(n,k) = T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2).
T(n,n) = Fibonacci(n+1) = A000045(n+1).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A133592(n), A133594(n), A133642(n), A133646(n), A133678(n), A133679(n), A133680(n), A133681(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Jan 03 2008
G.f.: (1-y*x^2)/(1-y*x-y*(y+1)*x^2). - Philippe Deléham, Nov 26 2011

A084068 a(1) = 1, a(2) = 2; a(2*k) = 2*a(2*k-1) - a(2*k-2), a(2*k+1) = 4*a(2*k) - a(2*k-1).

Original entry on oeis.org

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350
Offset: 1

Views

Author

Benoit Cloitre, May 10 2003

Keywords

Comments

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have
a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ).
It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)
Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Crossrefs

Programs

  • Maple
    a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
    end proc:
    seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018
  • Mathematica
    a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
    Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Formula

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
From Paul Barry, Jun 06 2006: (Start)
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007
Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007
From Peter Bala, Mar 23 2018: (Start)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
sqrt(1 + 2*a(2*n)^2) = A001541(n).
1 + 2*a(2*n)^2 = A055792(n+1).
a(2*n) - a(2*n-1) = A001653(n).
(1 + a(2*n+1)^2)/2 = A008844(n). (End)
a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021

A002538 Second-order Eulerian numbers <>.

Original entry on oeis.org

1, 8, 58, 444, 3708, 33984, 341136, 3733920, 44339040, 568356480, 7827719040, 115336085760, 1810992556800, 30196376985600, 532953524275200, 9927928075161600, 194677319705702400, 4008789120817152000, 86495828444928000000, 1951566265951948800000, 45958933902500720640000
Offset: 1

Views

Author

Keywords

Comments

Second-order Eulerian numbers <> count permutations of the multiset {1,1,2,2,...,n,n} with k ascents and the restriction that for any m <= n, all numbers between the two copies of m are less than m.
a(n) = number of edges in the Hasse diagram for the Bruhat order on permutations of [n+1]. - David Callan, Sep 03 2005
Proof. As explained on page 1 of the Stanley link, edges in the Hasse diagram of the (strong) Bruhat order on S_n are associated with pairs (pi,(i,j)) with pi in S_n and 1 <= i < j <= n, such that pi_i < pi_j and each entry of pi lying between pi_i and pi_j in POSITION does not lie between pi_i and pi_j in VALUE.
For example, pi = (3, 5, 1, 2, 4) gives edges for the (i,j) pairs (1,2), (1,5), (3,4), (4,5) but not, e.g., for (i,j) = (3,5) because 2 lies between pi_3=1 and pi_5=4 both in position and in value.
Let us count edges for a given pair (i,j). Consider the j-i+1 entries pi_i, pi_(i+1),...,pi_j. There are (j-i+1)! possible orderings for their values and (i,j) contributes an edge <=> the values of pi_i, pi_j are adjacent in this ordering with pi_i < pi_j.
There are (j-i)! such orderings (just coalesce the items pi_i, pi_j into a single item). The net result is that (i,j) contributes an edge 1/(j-i+1) of the time. So the total number of edges in the Hasse diagram is Sum_{1 <= i < j <= n} n!/(j-i+1) = (n+1)!(H_(n+1) - 2) + n! where H_n = 1 + 1/2 + 1/3 + ... + 1/n is the harmonic sum. QED - David Callan, Mar 07 2006
Number of reentrant corners along the lower contours of all deco polyominoes of height n+2. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. a(n) = Sum_{k>=1} k*A121579(n+2,k). - Emeric Deutsch, Aug 16 2006
a(n) is the total sum of the cycle maxima minus the cycle minima over all permutations of [n+1]. a(2) = 8 = 2+2+1+2+1+0: (123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3). - Alois P. Heinz, Dec 22 2023
a(n-1) is the number of parking functions of order n with exactly n-1 lucky cars, where a lucky car is a car which parks in the spot it prefers. - Kimberly P. Hadaway, Jun 20 2024

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 83.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Second diagonal of A008517 and second column of A112007.
Cf. A121579.

Programs

  • Magma
    [n le 1 select n else (n+2)*Self(n-1) + n*Factorial(n): n in [1..30]]; // Vincenzo Librandi, Aug 11 2018
  • Maple
    egf:= (x+2*log(1-x))/(x-1)^3:
    a:= n-> n!*coeff(series(egf, x, n+1), x, n):
    seq(a(n), n=1..21);  # Peter Luschny, Feb 12 2021
    # Alternative:
    a := n -> (n + 1)! * ((n + 2)*harmonic(n + 2) - 2*n - 3);
    seq(a(n), n = 1..22);  # Peter Luschny, Apr 09 2024
  • Mathematica
    Table[(-1)^(n + 1)* Sum[(-1)^(n - k) k (-1)^(n - k) StirlingS1[n + 3, k + 3], {k, 0, n}], {n, 1, 16}] (* Zerinvary Lajos, Jul 08 2009 *)
    a[n_]:=(-1)*((2*n+3)*(n+1)!-Abs[StirlingS1[n+3,2]]);Flatten[Table[a[n],{n,1,21}]] (* Detlef Meya, Apr 09 2024 *)
  • PARI
    N=66; x='x+O('x^66); Vec(serlaplace((x+2*log(1-x))/(x-1)^3)) \\ Joerg Arndt, Apr 09 2016
    

Formula

From Vladeta Jovovic, Sep 15 2003: (Start)
a(n) = Sum_{k=1..n} k * |Stirling1(n+2, k+2)|.
E.g.f.: (x+2*log(1-x))/(x-1)^3. (End)
With alternating signs: Ramanujan polynomials psi_2(n, x) evaluated at -1. - Ralf Stephan, Apr 16 2004
a(n) = (n+2)*a(n-1) + n*n!, n>=1, a(0):=0.
a(n) = (n+2)!*HarmonicSum(n+2) + (n+1)! - 2(n+2)! where HarmonicSum(n) = 1 + 1/2 + 1/3 + ... + 1/n. - David Callan, Mar 07 2006
a(n) = (n+1)!*((n+2)*h(n+2)-2*n-3) where h(n) = Sum_{k=1..n} 1/k. - Gary Detlefs, Mar 25 2011
Conjecture: a(n) + 2*(-n-2)*a(n-1) + (n^2+4*n+1)*a(n-2) - n*(n-1)*a(n-3) = 0. - R. J. Mathar, Oct 27 2014
a(n) = (-1)*((2*n + 3)*(n + 1)! - abs(Stirling1(n + 3, 2))). - Detlef Meya, Apr 09 2024

Extensions

More terms from Joerg Arndt, Apr 09 2016

A055997 Numbers k such that k*(k - 1)/2 is a square.

Original entry on oeis.org

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522
Offset: 1

Views

Author

Barry E. Williams, Jun 14 2000

Keywords

Comments

Numbers k such that (k-th triangular number - k) is a square.
Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002
Number of closed walks of length 2k on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - Bradley Klee & Bill Gosper, Jul 22 2015
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
  • P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.

Crossrefs

A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
a(n) = A001108(n-1)+1.
A001110(n-1)=a(n)*(a(n)-1)/2.
Identical to A115599, but with additional leading term.

Programs

  • Magma
    I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015
  • Maple
    A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
    map(A,[$1..100]); # Robert Israel, Jul 22 2015
  • Mathematica
    Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
    CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
    (1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
    a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
    LinearRecurrence[{7,-7,1},{1,2,9},30] (* Harvey P. Dale, Dec 06 2015 *)
  • PARI
    Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */
    
  • PARI
    t(n)=(1+sqrt(2))^(n-1);
    for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ Hugo Pfoertner, Nov 29 2019
    
  • PARI
    a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020
    

Formula

a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - Charlie Marion, Jul 21 2003; corrected by Michel Marcus, Apr 20 2020
a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004; corrected by Bernard Schott, Apr 20 2020
A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004
a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Feb 21 2006; corrected by Michel Marcus, Apr 20 2020
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with mA001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020

A090390 Repeatedly multiply (1,0,0) by ([1,2,2],[2,1,2],[2,2,3]); sequence gives leading entry.

Original entry on oeis.org

1, 1, 9, 49, 289, 1681, 9801, 57121, 332929, 1940449, 11309769, 65918161, 384199201, 2239277041, 13051463049, 76069501249, 443365544449, 2584123765441, 15061377048201, 87784138523761, 511643454094369, 2982076586042449, 17380816062160329, 101302819786919521, 590436102659356801
Offset: 0

Views

Author

Vim Wenders, Jan 30 2004

Keywords

Comments

The values of a and b in (a,b,c)*A give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1; the values of c are A000129(2n)
Binomial transform of A086348. - Johannes W. Meijer, Aug 01 2010
All values of a(n) are squares. sqrt(a(n+1)) = A001333(n). The ratio a(n+1)/a(n) converges to 3 + 2*sqrt(2). - Richard R. Forberg, Aug 14 2013

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 60 at p. 123.

Crossrefs

Programs

  • Haskell
    a090390 n = a090390_list !! n
    a090390_list = 1 : 1 : 9 : zipWith (-) (map (* 5) $
       tail $ zipWith (+) (tail a090390_list) a090390_list) a090390_list
    -- Reinhard Zumkeller, Aug 17 2013
    
  • Magma
    [Evaluate(DicksonFirst(n,-1),2)^2/4: n in [0..40]]; // G. C. Greubel, Aug 21 2022
    
  • Maple
    a:= n-> (<<1|0|0>>. <<1|2|2>, <2|1|2>, <2|2|3>>^n)[1, 1]:
    seq(a(n), n=0..30);  # Alois P. Heinz, Aug 17 2013
  • Mathematica
    CoefficientList[Series[(1-4x-x^2)/((1+x)(1-6x+x^2)),{x, 0, 30}], x] (* Harvey P. Dale, May 20 2012 *)
    LinearRecurrence[{5,5,-1}, {1,1,9}, 30] (* Harvey P. Dale, May 20 2012 *)
    Table[(ChebyshevT[n,3]+(-1)^n)/2, {n,0,30}] (* Eric W. Weisstein, Apr 17 2018 *)
    (LucasL[Range[0, 40], 2]/2)^2 (* G. C. Greubel, Aug 21 2022 *)
  • PARI
    a(n)=polcoeff((1-4*x-x^2)/((1+x)*(1-6*x+x^2))+x*O(x^n),n)
    
  • PARI
    a(n)=if(n<0,0,([1,2,2;2,1,2;2,2,3]^n)[1,1])
    
  • PARI
    Vec( (1-4*x-x^2)/((1+x)*(1-6*x+x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 16 2013
    
  • Perl
    use Math::Matrix; use Math::BigInt; $a = new Math::Matrix ([ 1, 2, 2], [ 2, 1, 2], [ 2, 2, 3]); $p = new Math::Matrix ([1, 0, 0]); $p->print(); for ($i=1; $i<20;$i++) { $p = $p->multiply($a); $p->print(); }
    
  • SageMath
    [lucas_number2(n,2,-1)^2/4 for n in (0..40)] # G. C. Greubel, Aug 21 2022

Formula

G.f.: (1-4*x-x^2)/((1+x)*(1-6*x+x^2)).
a(n) = A001333(n)^2
(a, b, c) = (1, 0, 0). Recursively multiply (a, b, c)*( [1, 2, 2], [2, 1, 2], [2, 2, 3] ).
M^n * [ 1 1 1] = [a(n+1) q a(n)], where M = the 3 X 3 matrix [4 4 1 / 2 1 0 / 1 0 0]. E.g. M^5 * [1 1 1] = [9801 4059 1681] where 9801 = a(6), 1681 = a(5). Similarly, M^n * [1 0 0] generates A079291 (Pell number squares). - Gary W. Adamson, Oct 31 2004
a(n) = (((1+sqrt(2))^(2*n) + (1-sqrt(2))^(2*n)) + 2*(-1)^n)/4 - Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 09 2005
a(n) = (A001541(n) + (-1)^n)/2. - R. J. Mathar, Nov 20 2009
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3), with a(0)=1, a(1)=1, a(2)=9. - Harvey P. Dale, May 20 2012
(a(n)) = tesseq(- .5'j + .5'k - .5j' + .5k' - 2'ii' + 'jj' - 'kk' + .5'ij' + .5'ik' + .5'ji' + 'jk' + .5'ki' + 'kj' + e), apart from initial term. - Creighton Dement, Nov 16 2004
a(n) = A302946(n)/4. - Eric W. Weisstein, Apr 17 2018
E.g.f.: exp(-x)*(1 + exp(4*x)*cosh(2*sqrt(2)*x))/2. - Stefano Spezia, Aug 03 2024
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