A226023 -id:A226023 - cp's OEIS frontend

A226096 Squares with doubled (4*n+2)^2.

1, 4, 4, 9, 16, 25, 36, 36, 49, 64, 81, 100, 100, 121, 144, 169, 196, 196, 225, 256, 289, 324, 324, 361, 400, 441, 484, 484, 529, 576, 625, 676, 676, 729, 784, 841, 900, 900, 961, 1024, 1089, 1156, 1156, 1225, 1296, 1369

Offset: 0

Paul Curtz, May 26 2013

Also nondecreasing ordered values of A226008 (except 0).
Consider A225948/A226008 ordered according to a(n): 0/1, -15/4, -3/4, 2/9, 3/16, 6/25, -7/36, 5/36, 12/49, 15/64, 20/81, ... = b(n)/a(n), and consider the sequence with period 5: 1, 64, 16, 1, 4, ... = t(n); then a(n) = 4*b(n) + t(n).
The recurrences in Formula lines are also valid for b(n).
Note that the fractions b(n)/a(n) of rank 0, 3,4,5, 8,9,10, ... = A047205:
0,  2/9, 3/16, 6/25, 12/49, 15/64, 20/81, ... are all in A226023(n).

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

Cf. A016826, A226008.

MapIndexed[ If [Mod[First[#2], 4] == 2, Sequence @@ {#1, #1}, #1] &, Range[40]]^2 (* Jean-François Alcover, May 28 2013 *)

a(n+5) - a(n) = 8*A090223(n+4).
a(n) = 1 followed by (A090223(n) + 2)^2.
a(n) = 3*a(n-5) -3*a(n-10) +a(n-15).
G.f.: (x^9 + 3*x^8 + 5*x^6 + 7*x^5 + 7*x^4 + 5*x^3 + 3*x + 1)/((1 - x)*(1 - x^5)^2). [Ralf Stephan, May 30 2013]
a(n) = a(n-1) +2*a(n-5) -2*a(n-6) -a(n-10) +a(n-11). [Bruno Berselli, May 30 2013]
a(n) = (24*(16*floor(n/5)^2 + 8*floor(n/5) + 1) - (11 + 24*floor(n/5))*(n - 5*floor(n/5))^4 + 2*(49 + 104*floor(n/5))*(n - 5*floor(n/5))^3 - 23*(11 + 24*floor(n/5))*(n - 5*floor(n/5))^2 + 2*(119 + 280*floor(n/5))*(n - 5*floor(n/5)))/24. - Luce ETIENNE, May 08 2017

A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).

Original entry on oeis.org

0, -15, -3, 2, 3, 6, -7, 5, 12, 15, 20, 9, 21, 30, 35, 42, 33, 45, 56, 63, 72, 65, 77, 90, 99, 110, 105, 117, 132, 143, 156, 153, 165, 182, 195, 210, 209, 221, 240, 255, 272, 273, 285, 306, 323, 342, 345, 357, 380, 399, 420, 425, 437

Offset: 0

Paul Curtz, Jun 05 2013

The sequence is the fifth row of the following array:
0,   6,  20, 42, 72, 110, 156, 210, 272, ...   A002943
0,   3,   6, 15, 20,  35,  42,  63,  72, ...   bisections A002943, A000466
0,   2,   3,  6, 12,  15,  20,  30,  35, ...   A226023 (trisections A002943, A000466, A002439)
0,  -3,   2,  3,  6,   5,  12,  15,  20, ...   A214297 (quadrisections A078371)
0, -15,  -3,  2,  3,   6,  -7,   5,  12, ...   a(n)
0, -63, -15, -3,  2,   3,   6, -55,  -7, ...
The principle of construction is that (i) the lower left triangular portion has constant values down the diagonals (6, 3, 2, -3, -15, ...), defined from row 4 on by the negated values of A024036. (ii) The extension along the rows is defined by maintaining bisections, trisections, quadrisections etc of the form (2*n+x)*(2*n+y) with some constants x and y. In the fifth line this needs the quintisections shown in the NAME.
Each row in the array has the subsequences of the previous row plus another subsequence of the format (2*n+1)*(2*n+y) shuffled in; the first A002943, the second also A000466, the third also A002439, the fourth also A078371, and the fifth (2*n+3)*(2*n-5).
Only the first three rows are monotonically increasing everywhere.
a(n) is divisible by A226203(n).
Numerators of: 0, -15/4, -3/4, 2/9, 3/16, 6/25, -7/36, 5/36, 12/49, 15/64, 20/81, ... = a(n)/A226096(n). A permutation of A225948(n+1)/A226008(n+1).
Is the sequence increasing monotonically from 221 on?

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2,-2,0,0,0,-1,1).

Cf. A000466, A002439, A002939, A002943, A024036, A078371, A145923.

Cf. A214297, A225948, A226008, A226023, A226096, A226203.

R:=PowerSeriesRing(Integers(), 60); [0] cat Coefficients(R!( -x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9)/((1-x^5)^2*(1-x)) )); // G. C. Greubel, Mar 23 2024

CoefficientList[Series[x*(15 - 12*x - 5*x^2 - x^3 - 3*x^4 - 17*x^5 + 12*x^6 + 3*x^7 - x^8 + x^9)/((x^4 + x^3 + x^2 + x + 1)^2*(x - 1)^3), {x, 0, 80}], x] (* Wesley Ivan Hurt, Oct 03 2017 *)

def A226379_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( -x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9)/((1-x^5)^2*(1-x)) ).list()
A226379_list(50) # G. C. Greubel, Mar 23 2024

4*a(n) = A226096(n) - period 5: repeat [1, 64, 16, 1, 4].
G.f.: x*(15-12*x-5*x^2-x^3-3*x^4-17*x^5+12*x^6+3*x^7-x^8+x^9) / ( (x^4+x^3+x^2+x+1)^2 *(x-1)^3 ). - R. J. Mathar, Jun 13 2013
a(n) = a(n-1)+2*a(n-5)-2*a(n-6)-a(n-10)+a(n-11) for n > 10. - Wesley Ivan Hurt, Oct 03 2017

A226122 Expansion of (1+2x+x^2+x^3+2x^4+x^5)/(1-2*x^3+x^6).

Original entry on oeis.org

1, 2, 1, 3, 6, 3, 5, 10, 5, 7, 14, 7, 9, 18, 9, 11, 22, 11, 13, 26, 13, 15, 30, 15, 17, 34, 17, 19, 38, 19, 21, 42, 21, 23, 46, 23, 25, 50, 25, 27, 54, 27, 29, 58, 29, 31, 62, 31, 33, 66, 33, 35, 70, 35, 37, 74, 37, 39, 78, 39

Offset: 0

Paul Curtz, May 27 2013

A226023 (starting from A226023(-2)=0) and successive differences:
0,    -1,   0,    2,     3,   6,   12,   15,   20,   30,...
-1,    1,   2,    1,     3,   6,    3,    5,   10,    5,...  = a(n-1)
2,     1,  -1,    2,     3,  -3,    2,    5,   -5,    2,...
-1,   -2,   3,    1,    -6,   5,    3,  -10,    7,    5,...
-1,    5,  -2,   -7,    11,  -2,  -13,   17,   -2,  -19,...
6,    -7,  -5,   18,   -13, -11,   30,  -19,  -17,   42,...
-13,   2,  23,  -31,    2,   41,  -49,    2,   59,   67,...
15,   21, -54,   33,   39,  -90,   51,   57, -126,   69,... multiples of 3
6,   -75,  87,    6, -129,  141,    6, -183,  195,    6,... multiples of 3
-81, 162, -81, -135,  270, -135, -189,  378, -189, -243,... multiples of 27
The last line is -27*a(n+3)*A131561(n+1).
The recurrences in the Formula field hold for the array.

			Given A130823 = 1,1,1,3,3,3,5,5,5,7,7,7,... and A131534 = 1,2,1,1,2,1,1,2,1,1,2,1,..., then a(0)=1*1=1, a(1)=1*2=2, a(2)=1*1=1, a(3)=3*1=3, a(4)=3*2=6, etc.
Given A226023(n) from A226023(-1)=-1, then a(0)=0-(-1)=1, a(1)=2-0=2, a(2)=3-2=1, a(3)=6-3=3, a(4)=12-6=6, etc.

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

Cf. A005408, A016754, A130823, A131534, A226023.

repeat=20; Table[{1, 2, 1}, {repeat}]*(2*Range[repeat]-1) // Flatten
(* or *) Table[Floor[(2*n+1)/3]*Floor[(2*n+5)/3], {n, -1, 59}] // Differences (* Jean-François Alcover, May 29 2013 *)

a(n) = A130823(n-1) * A131534(n).
a(n) = A226023(n) - A226023(n-1) with A226023(-1)=-1.
a(n) = 3*a(n-3) -3*a(n-6) +a(n-9) = a(n-1) +2*a(n-3) -2*a(n-4) -a(n-6) +a(n-7). [Ralf Stephan]
From Bruno Berselli, May 29 2013: (Start)
G.f.: (1+x)^3*(1-x+x^2)/((1-x)^2*(1+x+x^2)^2).
a(n) = 2*a(n-3)-a(n-6).
a(3n)*a(3n-1)-a(3n-2) = A016754(n-1), n>0. (End)

cp's OEIS Frontend

A226096 Squares with doubled (4*n+2)^2.

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A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).

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A226122 Expansion of (1+2x+x^2+x^3+2x^4+x^5)/(1-2*x^3+x^6).

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cp's OEIS Frontend

A226096 Squares with doubled (4*n+2)^2.

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Formula

A226379 a(5n) = 2*n*(2*n+1), a(5n+1) = (2*n-3)*(2*n+5), a(5n+2) = (2*n-1)*(2*n+3), a(5n+3) = (2*n+2)*(2*n+1), a(5n+4) = (2*n+1)*(2*n+3).

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A226122 Expansion of (1+2*x+x^2+x^3+2*x^4+x^5)/(1-2*x^3+x^6).

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A226379 a(5n) = 2n(2n+1), a(5n+1) = (2n-3)(2n+5), a(5n+2) = (2n-1)(2n+3), a(5n+3) = (2n+2)(2n+1), a(5n+4) = (2n+1)(2*n+3).

A226122 Expansion of (1+2x+x^2+x^3+2x^4+x^5)/(1-2*x^3+x^6).