A004525 -id:A004525 - cp's OEIS frontend

A187480 Rank transform of the sequence round(n/2); complement of A187481.

1, 2, 4, 5, 6, 8, 9, 10, 11, 13, 15, 16, 17, 19, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 46, 48, 49, 50, 51, 53, 54, 55, 57, 59, 60, 61, 63, 64, 65, 66, 68, 69, 70, 71, 73, 75, 76, 77, 79, 80, 81, 82, 84, 85, 86, 87, 89, 91, 92, 93, 95, 97, 98, 99, 100, 102, 103, 104, 106, 108, 109, 110, 112, 113, 114, 115, 117, 119, 120, 121, 123, 124, 125

Offset: 1

Clark Kimberling, Mar 10 2011

nonn

See A187224. A187480 is the rank transform of (A004525 with initial two zeros removed).

Cf. A187224, A187481, A004524.

seqA = Table[Round[n/2], {n, 1, 180}]  (* A004524 *)
seqB = Table[n, {n, 1, 80}];           (* A000027 *)
jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA,
{#1, 2} & /@ seqB}, 1]];
limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                     (* A187480 *)
Complement[Range[Length[seqA]], limseqU]  (* A187481 *)
(* by Peter J. C. Moses, Mar 10 2011 *)

A380286 Number of distinct values of the number of regions between the free polyominoes with n cells and their bounding boxes.

Original entry on oeis.org

1, 1, 2, 3, 5, 5, 5, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 11, 12, 13, 13, 13, 14, 15, 15, 15, 16, 17, 17, 17, 18, 19, 19, 19, 20, 21, 21, 21, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 27, 28, 29, 29, 29, 30, 31, 31, 31, 32, 33, 33, 33, 34, 35, 35, 35, 36, 37, 37, 37

Offset: 1

Omar E. Pol, Jan 24 2025

The regions include any holes in the polyominoes.
From Andrew Howroyd, Mar 01 2025: (Start)
Consider the following sequence of polyominoes for n >= 5:
    O       O         O           O   O       O   O       O   O
  O O O   O O O O   O O O O O   O O O O O   O O O O O   O O O O O O
    O       O         O           O           O   O       O   O
This construction shows how the number of regions between the polyomino and its bounding box can be increased by 2 with the addition of 4 cells. It is also easy to see that any number of fewer holes can also be realized. Moreover, this construction gives the greatest number of regions since except for four corner regions every other region must be bounded on 3 sides by at least one cell separating it from a neighboring region. This leads to a formula for a(n). (End)

			Illustration for n = 4:
The free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
    _
   |_|     _       _       _
   |_|    |_|     |_|_    |_|_     _ _
   |_|    |_|_    |_|_|   |_|_|   |_|_|
   |_|    |_|_|     |_|   |_|     |_|_|
.
The bounding boxes are respectively as shown below:
    _
   | |     _ _     _ _     _ _
   | |    |   |   |   |   |   |    _ _
   | |    |   |   |   |   |   |   |   |
   |_|    |_ _|   |_ _|   |_ _|   |_ _|
.
From left to right the number of regions between the free tetrominoes and their bounding boxes are respectively [0, 1, 2, 2, 0], hence there are three distinct values of the number of regions, they are [0, 1, 2], so a(4) = 3.
.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for sequences related to polyominoes.
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Row lengths of A380282.

Cf. A000105, A379627, A379628, A380283, A380284, A380285.

LinearRecurrence[{2, -2, 2, -1}, {1, 1, 2, 3, 5, 5, 5, 6}, 100] (* Paolo Xausa, Mar 02 2025 *)

From Andrew Howroyd, Mar 01 2025: (Start)
a(n) = A004525(n + 4) for n >= 5.
G.f.: x*(1 - x + 2*x^2 - x^3 + 2*x^4 - 2*x^5 + x^6 - x^7)/((1 - x)^2*(1 + x^2)). (End)
E.g.f.: (exp(x)*(4 + x) + sin(x))/2 - 2 - 2*x - x^2 - x^3/6 - x^4/24. - Stefano Spezia, Mar 03 2025

a(8)-a(16) from Pontus von Brömssen, Jan 24 2025
a(17)-a(18) from John Mason, Feb 14 2025
a(19) onwards from Andrew Howroyd, Feb 17 2025

A049206 Maximum mean distance between cards during perfect faro shuffles, with cut, to return to original order in A024222.

Original entry on oeis.org

0, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 11, 12, 13, 13, 13, 14, 15, 15, 15, 16, 17, 17, 17, 18, 19, 19, 19, 20, 21, 21, 21, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 27, 28, 29, 29, 29, 30, 31, 31, 31, 32, 33, 33, 33, 34, 35, 35, 35, 36, 37, 37, 37

Offset: 1

Enoch Haga

			Consider n=6. There are 4 shuffles to return to original order in a 6-card deck. The maximum mean distance between cards during these 4 shuffles and cuts, s1-s4, is 3, computed as follows: s1, 415263, cut, 263415; s2, 421653, cut 653421; s3, 462513, cut 513462; s4, 456123, cut, 123456. Mean distances: s1 15/5=3, maximum; s2 7/5=1.4; s3 13/5=2.6; s4 5/5; mean cumulative distance: 40/20=2.

Cf. A024222, A024542.

Take difference between successive cards after each shuffle. Compute mean (if necessary, round to nearest integer). Retain until replaced by a higher mean in a succeeding shuffle.
(1/4) {2n + 2 - (-1)^[n/2] + (-1)^[(n-1)/2] }. - Ralf Stephan, Jun 10 2005
a(n)=A004525(n), n>1. [From R. J. Mathar, Oct 15 2008]

A104563 A floretion-generated sequence relating to centered square numbers.

Original entry on oeis.org

0, 1, 3, 5, 8, 13, 19, 25, 32, 41, 51, 61, 72, 85, 99, 113, 128, 145, 163, 181, 200, 221, 243, 265, 288, 313, 339, 365, 392, 421, 451, 481, 512, 545, 579, 613, 648, 685, 723, 761, 800, 841, 883, 925, 968, 1013, 1059, 1105, 1152, 1201, 1251

Offset: 0

Creighton Dement, Mar 15 2005

Floretion Algebra Multiplication Program, FAMP Code: a(n) = 1vesrokseq[A*B] with A = - .5'i - .5i' + .5'ii' + .5e, B = + .5'ii' - .5'jj' + .5'kk' + .5e. RokType: Y[sqa.Findk()] = Y[sqa.Findk()] + Math.signum(Y[sqa.Findk()])*p (internal program code). Note: many slight variations of the "RokType" already exist, such that it has become difficult to assign them all names.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A001844, A004525, A104564, A098180, A059100, A010000, A047599, A007877.

LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 3, 5, 8}, 60] (* Amiram Eldar, Dec 14 2024 *)

concat(0, Vec(x*(1 + x)*(1 - x + x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Apr 29 2019

G.f.: x*(1 + x^3)/((1 + x^2)*(1 - x)^3).
FAMP result: 2*a(n) + 2*A004525(n+1) = A104564(n) + a(n+1).
Superseeker results:
a(2*n+1) = A001844(n) = 2*n*(n+1) + 1 (Centered square numbers);
a(n+1) - a(n) = A098180(n) (Odd numbers with two times the odd numbers repeated in order between them);
a(n) + a(n+2) = A059100(n+1) = A010000(n+1);
a(n+2) - a(n) = A047599(n+1) (Numbers that are congruent to {0, 3, 4, 5} mod 8);
a(n+2) - 2*a(n+1) + a(n) = A007877(n+3) (Period 4 sequence with initial period (0, 1, 2, 1));
Coefficients of g.f.*(1-x)/(1+x) = convolution of this with A280560 gives A004525;
Coefficients of g.f./(1+x) = convolution of this with A033999 gives A054925.
a(n) = (1/2)*(n^2 + 1 - cos(n*Pi/2)). - Ralf Stephan, May 20 2007
From Colin Barker, Apr 29 2019: (Start)
a(n) = (2 - (-i)^n - i^n + 2*n^2) / 4 where i=sqrt(-1).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4. (End)
a(n) = A011848(n-1)+A011848(n+2). - R. J. Mathar, Sep 11 2019
Sum_{n>=1} 1/a(n) = Pi^2/48 + (Pi/2) * tanh(Pi/2) + (Pi/(4*sqrt(2)) * tanh(Pi/(2*sqrt(2)))). - Amiram Eldar, Dec 14 2024

Stephan's formula corrected by Bruno Berselli, Apr 29 2019

A215415 a(2n) = n, a(4n+1) = 2n-1, a(4n+3) = 2*n+3.

Original entry on oeis.org

0, -1, 1, 3, 2, 1, 3, 5, 4, 3, 5, 7, 6, 5, 7, 9, 8, 7, 9, 11, 10, 9, 11, 13, 12, 11, 13, 15, 14, 13, 15, 17, 16, 15, 17, 19, 18, 17, 19, 21, 20, 19, 21, 23, 22, 21, 23, 25, 24, 23, 25, 27, 26, 25, 27, 29, 28, 27, 29, 31, 30, 29, 31, 33, 32, 31, 33, 35, 34, 33, 35, 37

Offset: 0

Paul Curtz, Aug 09 2012

a(n) and higher order differences in further rows:
0,  -1,  1,  3,  2,  1,
-1,  2,  2, -1, -1, -2,    A134430(n).
3,   0, -3,  0,  3,  0,
-3, -3,  3,  3, -3, -3,
0,   6,  0, -6,  0,  6,
6,  -6, -6,  6,  6, -6.
a(n) is the binomial transform of 0, -1, 3, -3, 0, 6, -12, 12, 0, -24, 48, -48, 0, 96..., essentially negated A134813.
By definition, all differences a(n+k)-a(n) are periodic sequences with period length 4. For k=1, 3 and 4 these are A134430, A021307 and A007395, for example.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Quadrisections: A005843, A060747, A005408, A144396.

Flatten[Table[{2n, 2n - 1, 2n + 1, 2n + 3}, {n, 0, 19}]] (* Alonso del Arte, Aug 09 2012 *)

a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4 \\ Colin Barker, Oct 19 2015

concat(0, Vec(-x*(1-3*x+x^2)/((x^2+1)*(x-1)^2) + O(x^100))) \\ Colin Barker, Oct 19 2015

a(2*n) = n, a(2*n+1) = A097062(n+1).
a(n) = (A214297(n+1) - A214297(n-1))/2.
a(3*n) =3*A004525(n).
a(n) = +2*a(n-1) -2*a(n-2) +2*a(n-3) -a(n-4).
G.f. -x*(1-3*x+x^2) / ( (x^2+1)*(x-1)^2 ). - R. J. Mathar, Aug 11 2012
a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4. - Colin Barker, Oct 19 2015

cp's OEIS Frontend

A187480 Rank transform of the sequence round(n/2); complement of A187481.

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A380286 Number of distinct values of the number of regions between the free polyominoes with n cells and their bounding boxes.

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A049206 Maximum mean distance between cards during perfect faro shuffles, with cut, to return to original order in A024222.

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A104563 A floretion-generated sequence relating to centered square numbers.

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A215415 a(2*n) = n, a(4*n+1) = 2*n-1, a(4*n+3) = 2*n+3.

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A215415 a(2n) = n, a(4n+1) = 2n-1, a(4n+3) = 2*n+3.