A016116 -id:A016116 - cp's OEIS frontend

A087503 a(n) = 3*(a(n-2) + 1), with a(0) = 1, a(1) = 3.

1, 3, 6, 12, 21, 39, 66, 120, 201, 363, 606, 1092, 1821, 3279, 5466, 9840, 16401, 29523, 49206, 88572, 147621, 265719, 442866, 797160, 1328601, 2391483, 3985806, 7174452, 11957421, 21523359, 35872266, 64570080, 107616801, 193710243, 322850406, 581130732

Offset: 0

Reinhard Zumkeller, Sep 11 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3).

Sequences with similar recurrence rules: A027383 (p=2), A133628 (p=4), A133629 (p=5).
Other related sequences for different p: A016116 (p=2), A038754 (p=3), A084221 (p=4), A133632 (p=5).
See A133629 for general formulas with respect to the recurrence rule parameter p.
Related sequences: A132666, A132667, A132668, A132669.

[(3/2)*(3^Floor((n+1)/2)+3^Floor(n/2)-3^Floor((n-1)/2)-1): n in [0..40]]; // Vincenzo Librandi, Aug 16 2011

A087503 := proc(n)
    option remember;
    if n <=1 then
        op(n+1,[1,3]) ;
    else
        3*procname(n-2)+3 ;
    end if;
end proc:
seq(A087503(n),n=0..20) ; # R. J. Mathar, Sep 10 2021

RecurrenceTable[{a[0]==1,a[1]==3,a[n]==3(a[n-2]+1)},a,{n,40}] (* or *) LinearRecurrence[{1,3,-3},{1,3,6},40] (* Harvey P. Dale, Jan 01 2015 *)

def A087503(n): return (3+((n+1&1)<<1))*3**(n+1>>1)-3>>1 # Chai Wah Wu, Sep 02 2025

a(n) = a(n-1) + A038754(n). (i.e., partial sums of A038754).
From Hieronymus Fischer, Sep 19 2007, formulas adjusted to offset, Dec 29 2012: (Start)
G.f.: (1+2*x)/((1-3*x^2)*(1-x)).
a(n) = (3/2)*(3^((n+1)/2)-1) if n is odd, else a(n) = (3/2)*(5*3^((n-2)/2)-1).
a(n) = (3/2)*(3^floor((n+1)/2) + 3^floor(n/2) - 3^floor((n-1)/2)-1).
a(n) = 3^floor((n+1)/2) + 3^floor((n+2)/2)/2 - 3/2.
a(n) = A132667(a(n+1)) - 1.
a(n) = A132667(a(n-1) + 1) for n > 0.
A132667(a(n)) = a(n-1) + 1 for n > 0.
Also numbers such that: a(0)=1, a(n) = a(n-1) + (p-1)*p^((n+1)/2 - 1) if n is odd, else a(n) = a(n-1) + p^(n/2), where p=3. (End)
a(n) = A052993(n)+2*A052993(n-1). - R. J. Mathar, Sep 10 2021

Additional comments from Hieronymus Fischer, Sep 19 2007
Edited by N. J. A. Sloane, May 04 2010. I merged two essentially identical entries with different offsets, so some of the formulas may need to be adjusted.
Formulas and MAGMA prog adjusted to offset 0 by Hieronymus Fischer, Dec 29 2012

A112030 a(n) = (2 + (-1)^n) * (-1)^floor(n/2).

Original entry on oeis.org

3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1

Offset: 0

Reinhard Zumkeller, Aug 27 2005

The fractions A112031(n)/A112032(n) give the partial sums of a(n)/floor((n+4)/2).

Sum of the two Cartesian coordinates from the image of the point (2,1) after n 90-degree counterclockwise rotations about the origin. - Wesley Ivan Hurt, Jul 06 2013

Index entries for linear recurrences with constant coefficients, signature (0,-1).

Cf. A010684, A016116, A112031, A112032, A112033.

A112030 := proc(n)
        (2 + (-1)^n) * (-1)^floor(n/2) ;
end proc: # R. J. Mathar, Jul 09 2013

LinearRecurrence[{0, -1}, {3, 1}, 100] (* Jean-François Alcover, Nov 24 2020 *)

a(n)=[3,1,-3,-1][n%4+1] \\ Charles R Greathouse IV, Aug 21 2011

def A112030(n): return (3, 1, -3, -1)[n&3] # Chai Wah Wu, Jan 31 2023

a(n) = A010684(n+1) * (-1)^floor(n/2).

O.g.f.: (3+x)/(1+x^2). - R. J. Mathar, Jan 09 2008

A228405 Pellian Array, A(n, k) with numbers m such that 2*m^2 +- 2^k is a square, and their corresponding square roots, read downward by diagonals.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 2, 2, 3, 5, 0, 2, 4, 7, 12, 4, 4, 6, 10, 17, 29, 0, 4, 8, 14, 24, 41, 70, 8, 8, 12, 20, 34, 58, 99, 169, 0, 8, 16, 28, 48, 82, 140, 239, 408, 16, 16, 24, 40, 68, 116, 198, 338, 577, 985, 0, 16, 32, 56, 96, 164, 280, 478, 816, 1393, 2378

Offset: 0

Richard R. Forberg, Aug 21 2013

The left column, A(n,0), is A000129(n), Pell Numbers.
The top row, A(0,k), is A077957(k) plus an initial 0, which is the inverse binomial transform of A000129.
These may be considered initializing values, or results, depending the perspective taken, since there are several ways to generate the array. See Formula section for details.
The columns of the array hold all values, in sequential order, of numbers m such that 2m^2 + 2^k or  2m^2 - 2^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for k+1.
For example A(n,0) are numbers such that 2m^2 +- 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 2m^2 +- 2 are squares, with those square roots in A(n,2), etc.
A(n, k)/A(n,k-2) = 2; A(n,k)/A(n,k-1) converges to sqrt(2) for large n.
A(n,k)/A(n-1,k) converges to 1 + sqrt(2) for large n.
The other columns of this array hold current OEIS sequences as follows:
  A(n,1) = A001333(n);   A(n,2) = A163271(n);  A(n,3) = A002203(n);
Bisections of these column-oriented sequences also appear in the OEIS, corresponding to the even and odd rows of the array, which in turn correspond to the two different recursive square root equations in the formula section below.
Farey fraction denominators interleave columns 0 and 1, and the corresponding numerators interleave columns 1 and 2, for approximating sqrt(2). See A002965 and A119016, respectively.
The other rows of this array hold current OEIS sequences as follows:
A(1,k) = A016116(k);   A(2,k) = A029744(k) less the initial 1;
A(3,k) = A070875(k);    A(4,k) = A091523(k) less the initial 8.
The Pell Numbers (A000219) are the only initializing set of numbers where the two recursive square root equations (see below) produce exclusively integer values, for all iterations of k.  For any other initial values only even iterations (at k = 2, 4, ...) produce integers.
The numbers in this array, especially the first three columns, are also integer square roots of these expressions: floor(m^2/2), floor(m^2/2 + 1), floor (m^2/2 - 1). See A227972 for specific arrangements and relationships. Also: ceiling(m^2/2), ceiling(m^2/2 + 1), ceiling (m^2/2 -1), m^2+1, m^2-1, m^2*(m^2-1)/2, m^2*(m^2-1)/2, in various different arrangements. Many of these involve: A000129(2n)/2 = A001109(n).
A001109 also appears when multiplying adjacent columns:   A(n,k) * A(n,k+1) = (k+1) * A001109(n), for all k.

			With row # as n. and column # as k, and n, k =>0, the array begins:
0,     1,     0,     2,     0,     4,     0,     8, ...
1,     1,     2,     2,     4,     4,     8,     8, ...
2,     3,     4,     6,     8,    12,    16,    24, ...
5,     7,    10,    14,    20,    28,    40,    56, ...
12,   17,    24,    34,    48,    68,    96,   136, ...
29,   41,    58,    82,   116,   164,   232,   328, ...
70,   99,   140,   198,   280,   396,   560,   792, ...
169,  239,  338,   478,   676,   956,  1352,  1912, ...
408,  577,  816,  1154,  1632,  2308,  3264,  4616, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
MacTutor, D'Arcy Thompson on Greek irrationals
D'Arcy Thompson, Excess and Defect: Or the Little More and the Little Less, Mind, New Series, Vol. 38, No. 149 (Jan., 1929), pp. 43-55 (13 pages). See page 50.

Cf. A000219, A077957, A001333, A163271, A002203, A001109, A227972.

Main diagonal is A007070.

If using the left column and top row to initialize: A(n,k) = A(n,k-1) + A(n-1,k-1).
If using only the top row to initialize, then each column for k = i is the binomial transform of A(0,k) restricted to k=> i as input to the transform with an appropriate down shift of index. The inverse binomial transform with a similar condition can produce each row from A000129.
If using only the first two rows to initialize then the Pell equation produces each column, as: A(n,k) = 2*A(n-1, k) + A(n-2, k).
If using only the left column (A000219(n) = Pell Numbers) to initialize then the following two equations will produce each row:
     A(n,k) = sqrt(2*A(n,k-1) + (-2)^(k-1)) for even rows
     A(n,k) = sqrt(2*A(n,k-1) - (-2)^(k-1)) for odd rows.
Interestingly, any portion of the array can also be filled "backwards" given the top row and any column k, using only:  A(n,k-1) = A(n-1,k-1) + A(n-1, k), or if given any column and its column number by rearranging the sqrt recursions above.

A239030 T(n,k)=Number of nXk 0..2 arrays with no element equal to the sum of elements to its left or one plus the sum of the elements above it, modulo 3.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 4, 4, 4, 1, 5, 7, 11, 4, 1, 6, 11, 28, 16, 8, 1, 7, 16, 59, 54, 43, 8, 1, 8, 22, 110, 149, 212, 64, 16, 1, 9, 29, 189, 354, 806, 428, 171, 16, 1, 10, 37, 306, 757, 2592, 2195, 1652, 256, 32, 1, 11, 46, 473, 1495, 7265, 9319, 11768, 3410, 683, 32, 1, 12, 56

Offset: 1

R. H. Hardin, Mar 09 2014

Table starts
..1...1.....1......1.......1........1.........1.........1..........1
..2...3.....4......5.......6........7.........8.........9.........10
..2...4.....7.....11......16.......22........29........37.........46
..4..11....28.....59.....110......189.......306.......473........704
..4..16....54....149.....354......757......1495......2773.......4888
..8..43...212....806....2592.....7265.....18362.....42809......93464
..8..64...428...2195....9319....33699....107611....311585.....833304
.16.171..1652..11768...69288...339315...1435014...5388959...18371174
.16.256..3410..33417..265247..1719471...9453266..45358859..194626082
.32.683.13004.177087.1965398.17562449.131139508.838702960.4711005062

			Some solutions for n=5 k=4
..2..0..0..0....2..0..0..0....2..0..0..0....2..0..0..0....2..0..0..0
..2..0..0..0....1..2..2..0....2..0..0..0....1..0..2..2....1..2..2..0
..1..0..2..2....2..1..2..0....1..0..2..2....2..0..1..2....2..1..2..0
..2..0..1..1....2..0..1..2....1..0..2..1....2..0..0..1....2..0..1..2
..1..0..2..2....1..0..2..2....2..0..0..0....1..0..2..1....1..2..2..1

R. H. Hardin, Table of n, a(n) for n = 1..480

Column 1 is A016116
Row 2 is A000027(n+1)
Row 3 is A000124

Empirical for column k:
k=1: a(n) = 2*a(n-2)
k=2: a(n) = 5*a(n-2) -4*a(n-4)
k=3: a(n) = 17*a(n-2) -96*a(n-4) +210*a(n-6) -152*a(n-8)
k=4: [order 18]
k=5: [order 38]
k=6: [order 90]
Empirical for row n:
n=1: a(n) = 1
n=2: a(n) = n + 1
n=3: a(n) = (1/2)*n^2 + (1/2)*n + 1
n=4: a(n) = (1/12)*n^4 - (1/6)*n^3 + (47/12)*n^2 - (29/6)*n + 5
n=5: [polynomial of degree 6] for n>1
n=6: [polynomial of degree 9] for n>2
n=7: [polynomial of degree 12] for n>3

A056451 Number of palindromes using a maximum of five different symbols.

Original entry on oeis.org

1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125, 30517578125, 152587890625, 152587890625

Offset: 0

Marks R. Nester

Number of achiral rows of n colors using up to five colors. For a(3) = 25, the rows are AAA, ABA, ACA, ADA, AEA, BAB, BBB, BCB, BDB, BEB, CAC, CBC, CCC, CDC, CEC, DAD, DBD, DCD, DDD, DED, EAE, EBE, ECE, EDE, and EEE. - Robert A. Russell, Nov 09 2018

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0,5).

Column k=5 of A321391.
Cf. A000007, A016116, A056391, A056453, A056454, A056455, A056456, A057427,
Cf. A000351 (oriented), A032122 (unoriented), A032088(n>1) (chiral).

[5^Floor((n+1)/2): n in [0..40]]; // Vincenzo Librandi, Aug 16 2011

LinearRecurrence[{0,5},{1,5},30] (* or *) Riffle[5^Range[0, 20], 5^Range[20]] (* Harvey P. Dale, Jul 28 2018 *)
Table[5^Ceiling[n/2], {n,0,40}] (* Robert A. Russell, Nov 07 2018 *)

vector(40, n, n--; 5^floor((n+1)/2)) \\ G. C. Greubel, Nov 07 2018

a(n) = 5^floor((n+1)/2).
a(n) = 5*a(n-2). - Colin Barker, May 06 2012
G.f.: (1+5*x) / (1-5*x^2). - Colin Barker, May 06 2012 [Adapted to offset 0 by Robert A. Russell, Nov 07 2018]
a(n) = C(5,0)*A000007(n) + C(5,1)*A057427(n) + C(5,2)*A056453(n) + C(5,3)*A056454(n) + C(5,4)*A056455(n) + C(5,5)*A056456(n). - Robert A. Russell, Nov 08 2018
E.g.f.: cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x). - Stefano Spezia, Jun 06 2023

a(0)=1 prepended by Robert A. Russell, Nov 07 2018

A059053 Number of chiral pairs of necklaces with n beads and two colors (color complements being equivalent); i.e., turning the necklace over neither leaves it unchanged nor simply swaps the colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 2, 7, 12, 31, 58, 126, 234, 484, 906, 1800, 3402, 6643, 12624, 24458, 46686, 90157, 172810, 333498, 641340, 1238671, 2388852, 4620006, 8932032, 17302033, 33522698, 65042526, 126258960, 245361172, 477091232

Offset: 0

Henry Bottomley, Dec 21 2000

nonn

Number of chiral pairs of set partitions of a cycle of n elements using exactly two different elements. - Robert A. Russell, Oct 02 2018

			For a(7) = 1, the chiral pair is AAABABB-AAABBAB.
For a(8) = 2, the chiral pairs are AAAABABB-AAAABBAB and AAABAABB-AAABBAAB.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Illustration of initial terms
Index entries for sequences related to necklaces

Column 2 of A320647 and A320742.

Cf. A056295 (oriented), A056357 (unoriented), A052551(n-2) (achiral).

Prepend[Table[DivisorSum[n, EulerPhi[#] StirlingS2[n/# + If[Divisible[#,2],1,0], 2] &] / (2n) - StirlingS2[1+Floor[n/2],2] / 2, {n, 1, 40}],0] (* Robert A. Russell, Oct 02 2018 *)

a(n) = {if(n<1, 0, (sumdiv(n, k, eulerphi(2*k) * 2^(n/k)) / (2*n) - 2^(n\2))/2)}; \\ Andrew Howroyd, Nov 03 2019

a(n) = A000013(n) - A000011(n) = A000011(n) - A016116(n) = (A000013(n) - A016116(n))/2.
From Robert A. Russell, Oct 02 2018: (Start)
a(n) = (A056295(n)-A052551(n-2)) / 2 = A056295(n) - A056357(n) = A056357(n) - A052551(n-2).
a(n) = -S2(1+floor(n/2),2) + (1/2n) * Sum_{d|n} phi(d) * S2(n/d+[2|d],2), where S2 is a Stirling subset number A008277.
G.f.: -x(1+2x)/(2-4x^2) - Sum_{d>0} phi(d) * log(1-2x^d) / (2d*(2-[2|d])).
(End)

Name clarified by Robert A. Russell, Oct 02 2018

A074872 Inverse BinomialMean transform of the Fibonacci sequence A000045 (with the initial 0 omitted).

Original entry on oeis.org

1, 1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125, 30517578125, 152587890625

Offset: 1

John W. Layman, Sep 12 2002

See A075271 for the definition of the BinomialMean transform.

The inverse binomial transform of 2^n*c(n+1), where c(n) is the solution to c(n) = c(n-1) + k*c(n-2), a(0)=0, a(1)=1 is 1, 1, 4k+1, 4k+1, (4k+1)^2, ... - Paul Barry, Feb 12 2004

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (0,5).

Cf. A000045, A075271, A056451, A063727, A016116, A108411.

[5^Floor((n-1)/2): n in [1..40]]; // Vincenzo Librandi, Aug 16 2011

a[1] := 1; a[2] := 1; a[n_] := 5a[n - 2]; Table[a[n], {n, 30}] (* Alonso del Arte, Mar 04 2011 *)

a(n)=5^((n-1)\2) \\ Charles R Greathouse IV, Oct 03 2016

a(n) = 5^floor((n-1)/2).
a(1)=1, a(2)=1 and, for n > 2, a(n) = 5*a(n-2).
From Paul Barry, Feb 12 2004: (Start)
G.f.: x*(1+x)/(1-5*x^2);
a(n) = (1/(2*sqrt(5))*((1+sqrt(5))*(sqrt(5))^n - (1-sqrt(5))*(-sqrt(5))^n)).
Inverse binomial transform of A063727 (2^n*Fibonacci(n+1)). (End)
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
E.g.f.: (cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x) - 1)/5. - Stefano Spezia, May 24 2024

A135318 The Kentucky-2 sequence: a(n) = a(n-2) + 2*a(n-4), with a[0..3] = [1, 1, 1, 2].

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 5, 8, 11, 16, 21, 32, 43, 64, 85, 128, 171, 256, 341, 512, 683, 1024, 1365, 2048, 2731, 4096, 5461, 8192, 10923, 16384, 21845, 32768, 43691, 65536, 87381, 131072, 174763, 262144, 349525, 524288, 699051, 1048576, 1398101, 2097152, 2796203

Offset: 0

Paul Curtz, Feb 16 2008

Shifted Jacobsthal recurrence.
From L. Edson Jeffery, Apr 21 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U=U_(6,2)=
(0 0 1)
(0 2 0)
(2 0 1),
let i in {0,1}, m>=0 an integer and n=2*m+i. Then a(n)=a(2*m+i)=Sum_{j=0..2} (U^m)_(i,j). (End)
a(n) is also the pebbling number of the cycle graph C_{n+1} for n > 1. - Eric W. Weisstein, Jan 07 2021
From Greg Dresden and Ziyi Xie, Aug 25 2023: (Start)
a(n) is the number of ways to tile a zig-zag strip of n cells using squares (of 1 cell) and triangles (of 3 cells). Here is the zig-zag strip corresponding to n=11, with 11 cells:
       _     _
   _|   |_|   |_
  |   |_|   |_|   |_
  |_|   |_|   |_|   |
  |   |_|   |_|   |_|
  |_|   |_|   |_|,
and here are the two types of triangles (where one is just a reflection of the other):
   _               _
  |   |_       _|   |
  |       |     |       |
  |    _| and |_    |
  |_|             |_|.
As an example, here is one of the a(11) = 32 ways to tile the zig-zag strip of 11 cells:
       _     _
   _|   |_|   |_
  |   |_|       |   |_
  |       |_    |       |
  |    _|   |_|    _|
  |_|   |_|   |_|.   (End)

			Let i=0 and m=3. Then U^3 = (2,0,3;0,8,0;6,0,5), and the first-row sum (corresponding to i=0) is 2 + 0 + 3 = 5. Hence a(n) = a(2*m+i) = a(2*3+0) = a(6) = 2 + 3 = 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Minerva Catral et al., Generalizing Zeckendorf's Theorem: The Kentucky Sequence, arXiv:1409.0488 [math.NT], 2014. See 1.3 p. 2, same sequence without the first 2 terms.
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See pp. 2, 17.
L. E. Jeffery, Unit-primitive matrices.
Eric Weisstein's World of Mathematics, Cycle Graph
Eric Weisstein's World of Mathematics, Pebbling Number
Index entries for linear recurrences with constant coefficients, signature (0,1,0,2).

Cf. A000079, A001045, A112387.

Cf. A048573.

[(2^Floor(n/2)*(5-(-1)^n)+(-1)^Floor(n/2)*(1+(-1)^n))/6: n in [0..50]]; // Vincenzo Librandi, Aug 10 2011

a:= n-> (<<0|1>, <2|1>>^(iquo(n, 2, 'm')). <<1, 1+m>>)[1,1]:
seq(a(n), n=0..50);  # Alois P. Heinz, May 30 2022

LinearRecurrence[{0,1,0,2},{1,1,1,2},40] (* Harvey P. Dale, Oct 14 2015 *)

From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: (1/(1+x^2)+(-2-3*x)/(2*x^2-1))/3.
a(2n) = A001045(n+1).
a(2n+1) = A000079(n). (End)
From L. Edson Jeffery, Apr 21 2011: (Start)
G.f.: (1+x+x^3)/((1+x^2)*(1-2*x^2)).
a(n) = (((-i)^(n+1)-i^(n+1))*2*i*sqrt(2)+3*(1+(-1)^(n+1))*2^((n+2)/2)+(1-(-1)^(n+1))*2^((n+5)/2))/(12*sqrt(2)), where i=sqrt(-1). (End)
a(n) = (2^floor(n/2)*(5-(-1)^n)+(-1)^floor(n/2)*(1+(-1)^n))/6 = (A016116(n)*A010711(n)+2*A056594(n))/6. - Bruno Berselli, Apr 21 2011
a(2n) = 2*a(2n-1) - a(2n-2); a(2n+1) = a(2n) + a(2n-2). - Richard R. Forberg, Aug 19 2013
a(n) = A112387(n + (-1)^n). - Alois P. Heinz, Sep 28 2023
E.g.f.: (2*cos(x) + 4*cosh(sqrt(2)*x) + 3*sqrt(2)*sinh(sqrt(2)*x))/6. - Stefano Spezia, Nov 09 2024
a(2*n) + a(2*n+1) = A048573(n) for n >= 0. - Paul Curtz, May 18 2025

More terms from R. J. Mathar, Feb 19 2008

A215870 T(n,k) = Number of permutations of 0..floor((n*k-2)/2) on odd squares of an n X k array such that each row, column, diagonal and (downwards) antidiagonal of odd squares is increasing.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 4, 4, 1, 1, 1, 1, 5, 12, 10, 4, 1, 1, 1, 1, 14, 29, 78, 20, 8, 1, 1, 1, 1, 14, 110, 262, 189, 50, 8, 1, 1, 1, 1, 42, 290, 3001, 1642, 1233, 100, 16, 1, 1, 1, 1, 42, 1274, 11694, 26451, 15485, 2988, 250, 16, 1, 1, 1, 1, 132

Offset: 1

R. H. Hardin, Aug 25 2012

Table starts
.1.1.1..1....1......1.......1.........1.........1..........1........1
.1.1.1..2....2......5.......5........14........14.........42.......42
.1.1.1..2....4.....12......29.......110.......290.......1274.....3532
.1.1.1..4...10.....78.....262......3001.....11694.....170594...727846
.1.1.1..4...20....189....1642.....26451....307874....7027942.98057806
.1.1.1..8...50...1233...15485....767560..14296434.1124811332
.1.1.1..8..100...2988...97289...6812794.386699176
.1.1.1.16..250..19494..918637.198409297
.1.1.1.16..500..47241.5772013
.1.1.1.32.1250.308205
.1.1.1.32.2500
.1.1.1.64

			Some solutions for n=6, k=4:
..x..0..x..1....x..0..x..2....x..0..x..2....x..0..x..1....x..0..x..1
..2..x..3..x....1..x..3..x....1..x..3..x....2..x..3..x....2..x..3..x
..x..4..x..5....x..4..x..6....x..4..x..5....x..4..x..6....x..4..x..6
..6..x..7..x....5..x..7..x....6..x..7..x....5..x..7..x....5..x..7..x
..x..8..x.10....x..8..x.10....x..8..x.10....x..8..x.10....x..8..x..9
..9..x.11..x....9..x.11..x....9..x.11..x....9..x.11..x...10..x.11..x

R. H. Hardin, Table of n, a(n) for n = 1..125

Column 5 is A026395(n-1).
Row 2 is A000108(floor(n/2)).
Even squares: A215788.

Empirical for column k:
k=4: a(n) = 2*a(n-2), A016116.
k=5: a(n) = 5*a(n-2) for n>3, A026395.
k=6: a(n) = 16*a(n-2) -3*a(n-4), A215866.
k=7: a(n) = 61*a(n-2) -99*a(n-4) -2*a(n-6), A215867.
k=8: a(n) = 272*a(n-2) -3439*a(n-4) -3336*a(n-6) +140*a(n-8).
k=9: a(n) = 1385*a(n-2) -131648*a(n-4) -318070*a(n-6) -4160916*a(n-8) -1097892*a(n-10) +648*a(n-12).

A216218 Square array T, read by antidiagonals: T(n,k) = 0 if n-k>=2 or if k-n>=2, T(1,0) = T(0,0) = T(0,1) = 1, T(n,k) = T(n-1,k) + T(n,k-1).

Original entry on oeis.org

1, 1, 1, 0, 2, 0, 0, 2, 2, 0, 0, 0, 4, 0, 0, 0, 0, 4, 4, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 8, 8, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0, 0, 0, 0, 16, 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Mar 13 2013

With zeros omitted, this is A173862.

			Square array begins:
1, 1, 0, 0,  0,  0,  0, 0, ... row n=0
1, 2, 2, 0,  0,  0,  0, 0, ... row n=1
0, 2, 4, 4,  0,  0,  0, 0, ... row n=2
0, 0, 4, 8,  8,  0,  0, 0, ... row n=3
0, 0, 0, 8, 16, 16,  0, 0, ... row n=4
0, 0, 0, 0, 16, 32, 32, 0, ... row n=5
...

Cf. A000079, A016116, A216216, A216210, A068913

T(n,n) = T(n+1,n) = T(n,n+1) = 2^n = A000079(n).

Sum_{k, 0<=k<=n} T(n-k,k) = A016116(n+1) = A163403(n+1).

cp's OEIS Frontend

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