A226008 -id:A226008 - cp's OEIS frontend

A225975 Square root of A226008(n).

0, 2, 2, 6, 1, 10, 6, 14, 4, 18, 10, 22, 3, 26, 14, 30, 8, 34, 18, 38, 5, 42, 22, 46, 12, 50, 26, 54, 7, 58, 30, 62, 16, 66, 34, 70, 9, 74, 38, 78, 20, 82, 42, 86, 11, 90, 46, 94, 24, 98, 50, 102, 13, 106, 54, 110, 28, 114, 58

Offset: 0

Paul Curtz, May 22 2013

Repeated terms of A016825 are in the positions 1,2,3,6,5,10,... (A043547).
From Wolfdieter Lang, Dec 04 2013: (Start)
This sequence a(n), n>=1, appears in the formula 2*sin(2*Pi/n) = R(p(n), x) modulo C(a(n), x), with x = rho(a(n)) = 2*cos(Pi/a(n)), the R-polynomials given in A127672 and the minimal C-polynomials of rho given in A187360. This follows from the identity 2*sin(2*Pi/n) = 2*cos(Pi*p(n)/a(n)) with gcd(p(n), a(n)) = 1. For p(n) see a comment on A106609,
Because R is an integer polynomial it shows that 2*sin(2*Pi/n) is an integer in the algebraic number field Q(rho(a(n))) of degree delta(a(n)) (the degree of C(a(n), x)), with delta(k) = A055034(k). This degree is given in A093819. For the coefficients of 2*sin(2*Pi/n) in the power basis of Q(rho(a(n))) see A231189 . (End)

			For the first formula: a(0)=-1+1=0, a(1)=-3+5=2, a(2)=-1+3=2, a(3)=-1+7=6, a(4)=0+1=1.

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

Cf. A016825, A017089, A017137, A022998, A106609, A145979, A226008.

a[0]=0; a[n_] := Sqrt[Denominator[1/4 - 4/n^2]]; Table[a[n], {n, 0, 58}] (* Jean-François Alcover, May 30 2013 *)
LinearRecurrence[{0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,-1},{0,2,2,6,1,10,6,14,4,18,10,22,3,26,14,30},60] (* Harvey P. Dale, Nov 21 2019 *)

a(n) = A106609(n-4) + A106609(n+4) with A106609(-4)=-1, A106609(-3)=-3, A106609(-2)=-1, A106609(-1)=-1.
a(n) = 2*a(n-8) -a(n-16).
a(2n+1) = A016825(n), a(2n) = A145979(n-2) for n>1, a(0)=0, a(2)=2.
a(4n)   = A022998(n).
a(4n+1) = A017089(n).
a(4n+2) = A016825(n).
a(4n+3) = A017137(n).
G.f.: x*(2 +2*x +6*x^2 +x^3 +10*x^4 +6*x^5 +14*x^6 +4*x^7 +14*x^8 +6*x^9 +10*x^10 +x^11 +6*x^12 +2*x^13 +2*x^14)/((1-x)^2*(1+x)^2*(1+x^2)^2*(1+x^4)^2). [Bruno Berselli, May 23 2013]
From Wolfdieter Lang, Dec 04 2013: (Start)
a(n) = 2*n if n is odd; if n is even then a(n) is n if n/2 == 1, 3, 5, 7 (mod 8), it is n/2 if n/2 == 0, 4 (mod 8) and it is n/4 if n/2 == 2, 6 (mod 8). This leads to the given G.f..
With c(n) = A178182(n), n>=1, a(n) = c(n)/2 if c(n) is even and c(n) if c(n) is odd. This leads to the preceding formula. (End)

Edited by Bruno Berselli, May 24 2013

A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

Original entry on oeis.org

-1, -15, -3, -7, 0, 9, 5, 33, 3, 65, 21, 105, 2, 153, 45, 209, 15, 273, 77, 345, 6, 425, 117, 513, 35, 609, 165, 713, 12, 825, 221, 945, 63, 1073, 285, 1209, 20, 1353, 357, 1505, 99, 1665, 437, 1833, 30, 2009, 525, 2193, 143

Offset: 0

Paul Curtz, May 21 2013

Denominators are in A226008.
Fractions in lowest terms for n>0: -15/4, -3/4, -7/36, 0/1, 9/100, 5/36, 33/196, 3/16, 65/324, 21/100, 105/484, 2/9, 153/676, 45/196, 209/900, 15/64,...
If t(n) is the sequence with period 8: 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, ... (see A226044), then A226008(n) = 4*a(n) + t(n).

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,-3,0,0,0,0,0,0,0,1).

Cf. A000466, A028566, A061037, A061041, A078371, A106609, A142705, A145923, A226008.

[-1] cat [Numerator(1/4-4/n^2): n in [1..50]]; // Bruno Berselli, May 22 2013

Join[{-1}, Table[Numerator[1/4 - 4/n^2], {n, 50}]] (* Bruno Berselli, May 24 2013 *)

concat([-1], vector(100, n, numerator(1/4 - 4/n^2))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-8) -3*a(n-16) +a(n-24).
a(2n) = A061037(n), a(2n+1) = A145923(n-2) for A145923(-2)=-15, A145923(-1)=-7.
a(4n)   = A142705(n) for A142705(0)=-1, a(8n) = A000466(n);
a(4n+1) = A028566(4n-3) for A028566(-3)=-15;
a(4n+2) = A078371(n-1) for A078371(-1)=-3;
a(4n+3) = A028566(4n-1) for A028566(-1)=-7.
a(n+4)  = A106609(n) * A106609(n+8).
G.f.: -(1 +15*x +3*x^2 +7*x^3 -9*x^5 -5*x^6 -33*x^7 -6*x^8 -110*x^9 -30*x^10 -126*x^11 -2*x^12 -126*x^13 -30*x^14 -110*x^15 -3*x^16 -33*x^17 -5*x^18 -9*x^19 +7*x^21 +3*x^22 +15*x^23)/(1-x^8)^3. - Bruno Berselli, May 22 2013
a(n) = (n^2-16)*(6*cos(Pi*n/4)-54*cos(Pi*n/2)+6*cos(3*Pi*n/4)-219*(-1)^n+293)/512. - Bruno Berselli, May 22 2013
a(n+10) = a(n+2)*(n+14)/(n-2) for n=0,1 and n>2. - Bruno Berselli, May 22 2013

Edited by Bruno Berselli, May 22 2013

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

Original entry on oeis.org

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

A226044 Period of length 8: 1, 64, 16, 64, 4, 64, 16, 64.

Original entry on oeis.org

1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64

Offset: 0

Paul Curtz, May 24 2013

A002378(n)/A016754(n) gives 0/1, 2/9, 6/25, 12/49, 20/81, 30/121, 42/169, 56/225,..., where A016754(n) = 4*A002378(n) + 1;
A142705(n)/A154615(n+1) gives 0/1, 3/16,  2/9,  15/64,  6/25,  35/144, 12/49,  63/256,..., where A142705(n) = 4*A154615(n+1) + A010685(n);
A061037(n)/A061038(n) gives 0/1, 5/36,  3/16, 21/100, 2/9,   45/196, 15/64,  77/324,..., where A061038(n) = 4*A061037(n) + A177499(n);
A225948(n)/A226008(n) gives 0/1, 9/100, 5/36, 33/196, 3/16,  65/324, 21/100, 105/484,..., where A226008(n) = 4*A225948(n) + a(n).
See also the triangle in Example lines.

			Triangle in which the terms of each line are repeated:
A000012: 1,   ...
A010685: 1,   4,  ...
A177499: 1,  16,  4,  16,  ...
A226044: 1,  64, 16,  64,  4,  64, 16,  64, ...
         1, 256, 64, 256, 16, 256, 64, 256, 4, 256, 64, 256, 16, 256, 64, 256, ...

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

Cf. A010685, A177499, A205383, A225948.

Table[{1, 64, 16, 64, 4, 64, 16, 64}, {7}] // Flatten (* Jean-François Alcover, May 24 2013 *)

a(n) = A205383(n+7)^2.

G.f.: (1+64*x+16*x^2+64*x^3+4*x^4+64*x^5+16*x^6+64*x^7)/((1-x)*(1+x)*(1+x^2)*(1+x^4)). [Bruno Berselli, May 25 2013]

A226203 a(5n) = a(5n+3) = a(5n+4) = 2n+1, a(5n+1) = 2n-3, a(5n+2) = 2n-1.

Original entry on oeis.org

1, -3, -1, 1, 1, 3, -1, 1, 3, 3, 5, 1, 3, 5, 5, 7, 3, 5, 7, 7, 9, 5, 7, 9, 9, 11, 7, 9, 11, 11, 13, 9, 11, 13, 13, 15, 11, 13, 15, 15, 17, 13, 15, 17, 17, 19, 15, 17, 19, 19, 21, 17, 19, 21, 21, 23, 19, 21, 23, 23, 25, 21, 23, 25, 25

Offset: 0

Paul Curtz, May 31 2013

Given the numerators of A225948/A226008 ordered according to A226096: 0, -15, -3, 2, 3, 6, -7, 5, 12, 15, 20, 9, 21, 30, 35,... = t(n), then (a(n) + t(n)/a(n))^2 =  A226096(n).
First six differences (of period 5):
...-4,   2,   2,   0,   2,  -4,   2,   2,   0,   2, ...
....6,   0,  -2,   2,  -6,   6,   0,  -2,   2,  -6, ...
...-6,  -2,   4,  -8,  12,  -6,  -2,   4,  -8,  12, ...
....4,   6, -12,  20, -18,   4,   6, -12,  20, -18, ...
....2, -18,  32, -38,  22,   2, -18,  32, -38,  22, ...
..-20,  50, -70,  60, -20, -20,  50, -70,  60, -20, ...

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

Cf. A007395, A005408, A130497, A226096.

import Data.List (transpose)
a226203 n = a226203_list !! n
a226203_list = concat $ transpose
               [[1, 3 ..], [-3, -1 ..], [-1, 1 ..], [1, 3 ..], [1, 3 ..]]
-- Reinhard Zumkeller, Jun 02 2013

a[n_] := 2 Quotient[n, 5] + Switch[Mod[n, 5], 0, 1, 1, -3, 2, -1, 3, 1, 4, 1]; Table[a[n], {n, 0, 64}] (* Jean-François Alcover, Jun 22 2017 *)

a(n+5) = a(n) + 2.
G.f.: (1-4*x+2*x^2+2*x^3+x^5)/((1-x)^2*(1+x+x^2+x^3+x^4)). [Bruno Berselli, Jun 01 2013]
a(n) = a(n-1)+a(n-5)-a(n-6) with a(0)=a(3)=a(4)=1, a(1)=-3, a(2)=-1, a(5)=3. [Bruno Berselli, Jun 01 2013]

Edited by Bruno Berselli, Jun 01 2013

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

cp's OEIS Frontend

A225975 Square root of A226008(n).

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A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

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

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A226044 Period of length 8: 1, 64, 16, 64, 4, 64, 16, 64.

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A226203 a(5n) = a(5n+3) = a(5n+4) = 2n+1, a(5n+1) = 2n-3, a(5n+2) = 2n-1.

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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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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).