A225948 -id:A225948 - cp's OEIS frontend

A142705 Numerator of 1/4 - 1/(2n)^2.

0, 3, 2, 15, 6, 35, 12, 63, 20, 99, 30, 143, 42, 195, 56, 255, 72, 323, 90, 399, 110, 483, 132, 575, 156, 675, 182, 783, 210, 899, 240, 1023, 272, 1155, 306, 1295, 342, 1443, 380, 1599, 420, 1763, 462, 1935, 506, 2115, 552, 2303, 600, 2499, 650, 2703, 702

Offset: 1

Paul Curtz, Sep 24 2008

Read modulo 10 (the last digits), a sequence with period length 10 results: 0, 3, 2, 5, 6, 5, 2, 3, 0, 9. Read modulo 9, a sequence with period length 18 results.
Denominators are in A154615.
a(n) is the numerator of (n-1)*(n+1)/4. - Altug Alkan, Apr 19 2018

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

Essentially the same as A070260. Cf. A078371 (second bisection of A061037), A142888 (first differences), A154615 (denominators), A225948.

[-(3/4)*(-1)^n*n-(3/8)*(-1)^n*n^2+(5/8)*n^2+(5/4)*n: n in [0..60]]; // Vincenzo Librandi, Jul 02 2011

Numerator[Table[(1/4)*(1 - 1/n^2), {n,1,50}]] (* G. C. Greubel, Jul 20 2017 *)

for(n=1, 50, print1(numerator((1/4)*(1 - 1/n^2)), ", ")) \\ G. C. Greubel, Jul 20 2017

a(n) = if(n%2,(n^2-1)/4,n^2-1); \\ Altug Alkan, Apr 19 2018

a(n) = A061037(2*n).
a(n) = A070260(n-1), n>1.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(2^(n-1)) = a(1+A000225(n-1)) = 4^(n-1)-1 = A024036(n-1).
G.f.: x^2*(3+2x+6x^2-x^4)/(1-x^2)^3. - R. J. Mathar, Oct 24 2008
E.g.f.: 1 + (1/4)*((4*x^2 + x - 4)*cosh(x) + (x^2 + 4*x -1)*sinh(x)). - G. C. Greubel, Jul 20 2017
Sum_{n>=2} 1/a(n) = 3/2. - Amiram Eldar, Aug 11 2022

Edited by R. J. Mathar, Oct 24 2008

A226008 a(0) = 0; for n>0, a(n) = denominator(1/4 - 4/n^2).

Original entry on oeis.org

0, 4, 4, 36, 1, 100, 36, 196, 16, 324, 100, 484, 9, 676, 196, 900, 64, 1156, 324, 1444, 25, 1764, 484, 2116, 144, 2500, 676, 2916, 49, 3364, 900, 3844, 256, 4356, 1156, 4900, 81, 5476, 1444, 6084, 400, 6724, 1764, 7396, 121, 8100

Offset: 0

Paul Curtz, May 22 2013

Numerators are in A225948.

Repeated terms of A016826 are in the positions 1, 2, 3, 6, 5, 10, ... (A043547).

			a(0) = (-1+1)^2 = 0, a(1) = (-3+5)^2 = 4, a(2) = (-1+3)^2 = 4.

Cf. A016754, A016802, A016826, A017090, A017138, A154615, A225948 (numerators).

Cf. A225975 (associated square roots).

[0] cat [Denominator(1/4-4/n^2): n in [1..50]]; // Bruno Berselli, May 23 2013

Join[{0},Table[Denominator[1/4 - 4/n^2], {n, 49}]] (* Alonso del Arte, May 22 2013 *)

a(n)    = 3*a(n-8) -3*a(n-16) +a(n-24).
a(8n)   = A016802(n), a(8n+4) = A016754(n).
a(4n)   = A154615(n).
a(4n+1) = A017090(n).
a(4n+2) = a(2n+1) = A016826(n); a(2n) = A061038(n).
a(4n+3) = A017138(n).
From Bruno Berselli, May 23 2013: (Start)
G.f.: x*(4 +4*x +36*x^2 +x^3 +100*x^4 +36*x^5 +196*x^6 +16*x^7 +312*x^8 +88*x^9 +376*x^10 +6*x^11 +376*x^12 +88*x^13 +312*x^14 +16*x^15 +196*x^16 +36*x^17 +100*x^18 +x^19 +36*x^20 +4*x^21 +4*x^22)/(1-x^8)^3.
a(n) = n^2*(6*cos(3*Pi*n/4)+6*cos(Pi*n/4)-54*cos(Pi*n/2)-219*(-1)^n+293)/128.
a(n+9) = a(n+1)*((n+9)/(n+1))^2. (End)
Sum_{n>=1} 1/a(n) = 19*Pi^2/96. - Amiram Eldar, Aug 14 2022

Edited by Bruno Berselli, May 23 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

A142705 Numerator of 1/4 - 1/(2n)^2.

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A226008 a(0) = 0; for n>0, a(n) = denominator(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).