A226096 -id:A226096 - cp's OEIS frontend

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

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

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) = 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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cp's OEIS Frontend

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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Programs

Haskell

Mathematica

Formula

Extensions

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).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

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).