A326118 -id:A326118 - cp's OEIS frontend

A338329 First differences of A326118.

1, 1, 3, 1, 3, 5, 7, 3, 5, 7, 9, 5, 7, 9, 11, 7, 9, 11, 13, 9, 11, 13, 15, 11, 13, 15, 17, 13, 15, 17, 19, 15, 17, 19, 21, 17, 19, 21, 23, 19, 21, 23, 25, 21, 23, 25, 27, 23, 25, 27, 29, 25, 27, 29, 31, 27, 29, 31, 33, 29, 31, 33, 35, 31, 33, 35, 37, 33, 35, 37, 39

Offset: 0

Stefano Spezia, Oct 23 2020

It includes exclusively all the odd numbers (A005408). Except for 1, 3 and 5 that appear three times, each other odd number appears four times.

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

Cf. A004277 (averages of the increasing runs), A005408, A326118.

LinearRecurrence[{1,0,0,1,-1},{1,1,3,1,3,5,7,3},71]

O.g.f.: (1 + 2*x^2 - 2*x^3 + x^4 + 2*x^5 - 2*x^7)/((1 - x)^2*(1 + x + x^2 + x^3)).
E.g.f.: (3*exp(-x) + exp(x)*(7 + 2*x) - 6*cos(x) + 6*sin(x))/4 - 2*x - x^2.
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 7.
a(n) = (7 + 2*n - 6*cos(n*Pi/2) + 3*(-1)^n + 6*sin(n*Pi/2))/4 for n > 2.

A309038 Irregular triangle T read by rows: given a square made of n^2 squares of unit area, T(n, k) is the longest perimeter that can be obtained by removing k of n^2 squares such that the modified figure remains connected and without holes (n >= 0 and 0 <= k <= n^2).

Original entry on oeis.org

0, 4, 0, 8, 8, 8, 4, 0, 12, 14, 16, 18, 20, 16, 12, 8, 4, 0, 16, 18, 20, 22, 24, 26, 28, 28, 28, 26, 24, 20, 16, 12, 8, 4, 0, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 42, 40, 38, 36, 32, 28, 24, 20, 16, 12, 8, 4, 0, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 56, 56, 56, 56, 56, 56, 52, 48, 44, 40, 36, 32, 28, 24, 20, 16, 12, 8, 4, 0

Offset: 0

Stefano Spezia, Jul 08 2019

All the terms of this sequence are even numbers (A005843).
In the figure, two unit area squares can be connected in a corner or sideways.
Every n-th row of the triangle is made of almost four successive finite arithmetic progressions characterized respectively by the following common differences: 2, 0, -2, -4. If we let h_i(n) be the number of first differences of i-th progression (i = 1,2,3,4), we have that 4*n + 2*h_1(n) - 2*h_3(n) - 4*h_4(n) = 0 and h_1(n) + h_2(n) + h_3(n) + h_4(n) = n^2.

			The triangle T(n, k) begins:
---+-------------------------------------------------------------------
n\k|  0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16
---+-------------------------------------------------------------------
0  |  0
1  |  4   0
2  |  8   8   8   4   0
3  | 12  14  16  18  20  16  12   8   4   0
4  | 16  18  20  22  24  26  28  28  28  26  24  20  16  12   8   4   0
...
Here are the values of h_i's for the first seven rows of the triangle T:
n    h_1(n)   h_2(n)   h_3(n)   h_4(n)
--------------------------------------
0         0        0        0        0
1         0        0        0        1
2         0        2        0        2
3         4        0        0        5
4         6        2        2        6
5        12        0        4        9
6        16        6        0       14
...
Illustrations for n = 4, k=0..15 by _Andrew Howroyd_, Sep 01 2019: (Start)
   __.__.__.__    __.__.__.__    __.__.__.__    __.__.__.__
  |           |  |           |  |           |  |           |
  |           |  |__         |  |__         |  |__       __|
  |           |   __|        |   __|__      |   __|__   |__
  |__.__.__.__|  |__.__.__.__|  |__|  |__.__|  |__|  |__.__|
       (16)           (18)          (20)           (22)
   __.__.__.__    __.  .__.__    __    __.__    __    __.__
  |           |  |  |__|     |  |  |  |     |  |  |  |   __|
  |__    __.__|  |__    __.__|  |__|__|__.__|  |__|__|__|
   __|__|__.__    __|__|__.__    __|__|__.__    __|__|__.__
  |__|  |__.__|  |__|  |__.__|  |__|  |__.__|  |__|  |__.__|
      (24)            (26)          (28)           (28)
   __       __             __             __
  |  |   __|__|   __    __|__|   __    __|__|   __    __
  |__|__|__|     |__|__|__|     |__|__|__|     |__|__|__|
   __|__|__.__    __|__|__.__    __|__|__       __|__|__
  |__|  |__.__|  |__|  |__.__|  |__|  |__|     |__|  |__|
      (28)            (26)         (24)           (20)
   __    __             __
  |__|__|__|         __|__|         __             __
   __|__|         __|__|         __|__|           |__|
  |__|           |__|           |__|
      (16)         (12)            (8)             (4)
(End)

Cf. A000290, A005843, A008586, A048759, A326118 (h_4).

h4[n_]:=If[n>2,(1/8)(-29+12n+2n^2-3*Cos[n*Pi]-12*Sin[n*Pi/2]),n]; h3[n_]:=1-Cos[n*Pi]-4*KroneckerDelta[n,1]+2*KroneckerDelta[n,4]+2*Sin[n*Pi/2]; h2[n_]:=If[n>4,(1/8)(71-20n+2n^2+25Cos[n*Pi]+4Sin[n*Pi/2]),2*(KroneckerDelta[n,2]+KroneckerDelta[n,4])]; h1[n_]:=n^2-(h2[n]+h3[n]+h4[n]); T[n_,k_]:=If[0<=k<=h1[n],2(2n+k),If[h1[n]

T(n, 0) = A008586(n).

T(n, k) = 2*(2*n + k) for 0 <= k <= h_1(n), T(n, k) = 2*(2*n + h_1(n)) for h_1(n) <= k <= h_1(n) + h_2(n), T(n, k) = 2*(2*(n + h_1(n)) + h_2(n) - k) for h_1(n) + h_2(n) <= k <= h_1(n) + h_2(n) + h_3(n), T(n, k) = 2*(2*(n + h_2(n) - k) + 3*h_1(n) + h_3(n)), for h_1(n) + h_2(n) + h_3(n) <= k <= n^2, where h_4(n) = n for 0 <= n <= 2 and h_4(n) = (1/8)*(-29 + 12*n + 2*n^2 - 3*cos(n*Pi) - 12*sin(n*Pi/2)) for n > 2, h_3(n) = 2*delta(n, 4) - 4*delta(n, 1) + 1 - cos(n*Pi) + 2*sin(n*Pi/2) and delta(i, j) is the Kronecker delta, h_2(n) = 2*(delta(n, 2) + delta(n, 4)) for 0 <= n <= 4 and h_2(n) = (1/8)*(71 - 20*n + 2*n^2 + 25*cos(n*Pi) + 4*sin(n*Pi/2)) for n > 4, h_1(n) = n^2 - (h_1(n) + h_2(n) + h_3(n)).

A327480 a(n) is the maximum number of squares of unit area that can be removed from an n X n square while still obtaining a connected figure without holes and of the longest perimeter.

Original entry on oeis.org

0, 0, 2, 4, 8, 12, 22, 28, 40, 48, 64, 76, 94, 108, 130, 148, 172, 192, 220, 244, 274, 300, 334, 364, 400, 432, 472, 508, 550, 588, 634, 676, 724, 768, 820, 868, 922, 972, 1030, 1084, 1144, 1200, 1264, 1324, 1390, 1452, 1522, 1588, 1660, 1728, 1804, 1876, 1954

Offset: 0

Stefano Spezia, Sep 16 2019

a(n) is equal to h_1(n) + h_2(n) as defined in A309038.

			Illustrations for n = 2..7:
      __                      __    __                __    __
     |__|__                  |__|__|__|              |__|__|__|
        |__|                  __|__|__                __|__|__ __
                             |__|  |__|              |__|  |     |
                                                           |__ __|
      a(2) = 2                a(3) = 4                  a(4) = 8
   __    __ __ __     __    __    __         __    __    __    __
  |__|__|__ __ __|   |__|__|__|  |__|__     |__|__|__|  |__|__|__|
   __|__|__    __     __|__|__    __|__|     __|__|__    __|__|__
  |  |  |__|__|__|   |__|  |__|__|__|       |__|  |__|__|__|  |__|
  |  |   __|__|__     __    __|__|__         __    __|__|__    __
  |__|  |__|  |__|   |__|__|__|  |__|__     |__|__|__|  |__|__|__|
                        |__|        |__|     __|__|__    __|__|__
                                            |__|  |__|  |__|  |__|
      a(5) = 12           a(6) = 22               a(7) = 28

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000290, A056594, A309038, A326118, A327479.

I:=[0, 0, 2, 4, 8, 12, 22, 28, 40, 48, 64]; [n le 11 select I[n] else 2*Self(n-1)-Self(n-2)+Self(n-4)-2*Self(n-5)+Self(n-6): n in [1..53]];

gf := (1/24)*exp(-x)*(33+9*exp(2*x)*(2*x^2-2*x+7)-2*exp(x)*(x^4+12*x^2+48)-12*exp(x)*sin(x)); ser := series(gf, x, 53):
seq(factorial(n)*coeff(ser, x, n), n = 0 .. 52)

Join[{0,0,2,4,8},Table[(1/8)*(21-12n+6n^2+11*(-1)^n-4*Sin[n*Pi/2]),{n,5,52}]]

concat([0, 0], Vec(2*x^2*(1+x^2+2*x^4-2*x^5+2*x^6-2*x^7+x^8)/((1-x)^3*(1+x)*(1+x^2))+O(x^53)))

O.g.f.: 2*x^2*(1 + x^2 + 2*x^4 - 2*x^5 + 2*x^6 - 2*x^7 + x^8)/((1 - x)^3*(1 + x)*(1 + x^2)).
E.g.f.: (1/24)*exp(-x)*(33 + 9*exp(2*x)*(7 - 2*x + 2*x^2) - 2*exp(x)*(48 + 12*x^2 + x^4) - 12*exp(x)*sin(x)).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n > 10.
a(n) = (1/8)*(21 - 12*n + 6*n^2 + 11*(-1)^n + 4*A056594(n+1)) for n > 4, a(0) = 0, a(1) = 0, a(2) = 2, a(3) = 4, a(4) = 8. [corrected by Jason Yuen, Dec 17 2024]
Limit_{n->oo} a(n)/A000290(n) = 3/4.

A327479 a(n) is the minimum number of squares of unit area that must be removed from an n X n square to obtain a connected figure without holes and of the longest perimeter.

Original entry on oeis.org

0, 0, 0, 4, 6, 12, 16, 28, 32, 44, 52, 68, 76, 92, 104, 124, 136, 156, 172, 196, 212, 236, 256, 284, 304, 332, 356, 388, 412, 444, 472, 508, 536, 572, 604, 644, 676, 716, 752, 796, 832, 876, 916, 964, 1004, 1052, 1096, 1148, 1192, 1244, 1292, 1348, 1396, 1452, 1504

Offset: 0

Stefano Spezia, Sep 16 2019

a(n) is equal to h_1(n) as defined in A309038.

All the terms are even numbers (A005843).

			Illustrations for n = 3..8:
      __    __               __    __.__             __    __.__.__
     |__|__|__|             |__|__|__.__|           |__|__|__.__.__|
      __|__|__               __|__|__.__             __|__|__    __
     |__|  |__|             |  |  |     |           |  |  |__|__|__|
                            |__|  |__.__|           |  |   __|__|__
                                                    |__|  |__|  |__|
      a(3) = 4                a(4) = 6                  a(5) = 12
   __    __    __.__     __    __    __    __     __    __    __    __.__
  |__|__|__|  |__   |   |__|__|__|  |__|__|__|   |__|__|__|  |__|__|__   |
   __|__|__    __|  |    __|__|__    __|__|__     __|__|__    __|  |  |__|
  |__|  |__|__|__.__|   |__|  |__|__|__|  |__|   |__|  |__|__|__.__|   __
   __    __|__|__.__     __    __|__|__    __     __    __|__|__    __|  |
  |  |__|  |  |     |   |__|__|__|  |__|__|__|   |__|__|  |  |__|__|__.__|
  |__.__.__|  |__.__|    __|__|__    __|__|__     __|__.__|   __|__|__.__
                        |__|  |__|  |__|  |__|   |  |__    __|  |  |     |
                                                 |__.__|  |__.__|  |__.__|
     a(6) = 16                a(7) = 28                 a(8) = 32

Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000290, A005843, A309038, A326118, A327480.

I:=[0, 0, 0, 4, 6, 12, 16, 28, 32, 44, 52]; [n le 11 select I[n] else 2*Self(n-1)-Self(n-2)+Self(n-4)-2*Self(n-5)+Self(n-6): n in [1..55]];

gf := 8+4*x+2*x^2+(1/12)*x^4+1/4*(-7*exp(-x)+exp(x)*(2*x^2+6*x-25)-4*sin(x)):
ser := series(gf, x, 55): seq(factorial(n)*coeff(ser, x, n), n = 0..54);

Join[{0,0,0,4,6},Table[(1/4)*(-25+2n*(2+n)-7*Cos[n*Pi]-4*Sin[n*Pi/2]),{n,5,54}]]

concat([0, 0, 0], Vec(2*x^3*(-2+x-2*x^2+x^3-2*x^4+3*x^5-2*x^6+x^7)/((-1+x)^3*(1+x+x^2+x^3))+O(x^55)))

O.g.f.: 2*x^3*(-2 + x - 2*x^2 + x^3 - 2*x^4 + 3*x^5 - 2*x^6 + x^7)/((-1 + x)^3*(1 + x + x^2 + x^3)).
E.g.f.: 8 + 4*x + 2*x^2 + x^4/12 + (1/4)*(-7*exp(-x) + exp(x)*(-25 + 6*x + 2*x^2) - 4*sin(x)).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n > 10.
a(n) = (1/4)*(- 25 + 2*n*(2 + n) - 7*cos(n*Pi) - 4*sin(n*Pi/2)) for n > 4, a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 4, a(4) = 6.
Lim_{n->inf} a(n)/A000290(n) = 1/2.

A328685 Row sums of A309038.

Original entry on oeis.org

0, 4, 28, 120, 320, 716, 1380, 2464, 3984, 6196, 9124, 13128, 18048, 24476, 32244, 42096, 53440, 67460, 83604, 103192, 124944, 150892, 179908, 214080, 251184, 294356, 341700, 396264, 454624, 521276, 593364, 675088, 761568, 858916, 963124, 1079736, 1202160, 1338380

Offset: 0

Stefano Spezia, Oct 25 2019

nonn

All the terms are even.

Stefano Spezia, Table of n, a(n) for n = 0..100

Cf. A000583, A309038, A326118, A327479, A327480.

(* The function T is defined in A309038. *)
Flatten[Table[Sum[T[n, k], {k, 0, n^2}], {n, 0, 37}]]

Conjectures from Colin Barker, Oct 25 2019: (Start)
G.f.: 4*x*(1 + 5*x + 17*x^2 + 27*x^3 + 46*x^4 + 52*x^5 + 54*x^6 + 28*x^7 + 29*x^8 - 7*x^9+ 5*x^10 - 17*x^11 + 4*x^12 - 6*x^13 + 12*x^14 - 14*x^15 + 8*x^16 - 4*x^17) / ((1 - x)^5*(1 + x)^3*(1 + x^2)^3).
a(n) = 2*a(n-1) - a(n-2) + 3*a(n-4) - 6*a(n-5) + 3*a(n-6) - 3*a(n-8) + 6*a(n-9) - 3*a(n-10) + a(n-12) - 2*a(n-13) + a(n-14) for n > 18.
(End)
a(n) ~ 5*n^4/8. - Conjectured by Stefano Spezia, Sep 08 2021

cp's OEIS Frontend

A338329 First differences of A326118.

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A309038 Irregular triangle T read by rows: given a square made of n^2 squares of unit area, T(n, k) is the longest perimeter that can be obtained by removing k of n^2 squares such that the modified figure remains connected and without holes (n >= 0 and 0 <= k <= n^2).

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A327480 a(n) is the maximum number of squares of unit area that can be removed from an n X n square while still obtaining a connected figure without holes and of the longest perimeter.

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A327479 a(n) is the minimum number of squares of unit area that must be removed from an n X n square to obtain a connected figure without holes and of the longest perimeter.

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A328685 Row sums of A309038.

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