A061038 -id:A061038 - cp's OEIS frontend

A246943 a(4n) = 4n , a(2n+1) = 8n+4 , a(4n+2) = 2*n+1.

0, 4, 1, 12, 4, 20, 3, 28, 8, 36, 5, 44, 12, 52, 7, 60, 16, 68, 9, 76, 20, 84, 11, 92, 24, 100, 13, 108, 28, 116, 15, 124, 32, 132, 17, 140, 36, 148, 19, 156, 40, 164, 21, 172, 44, 180, 23, 188, 48, 196, 25, 204, 52, 212, 27, 220, 56, 228

Offset: 0

Paul Curtz, Sep 08 2014

Consider the denominators of the Balmer series A061038(n) = 0, 4, 1, 36, 16, 100,... (a permutation of the squares of the nonnegative numbers i.e. A000290(n)) divided by A028310(n)=1,1,2,... . The numerators are a(n). The denominators are A138191(n).
Note that A061038(3n)=9*A061038(n), n>=1.
a(3n) is divisible by the period 3 sequence: repeat 9, 3, 3.

			Numerators of a(0)=0/1=0, a(1)=4/1=4, a(2)=1/2, a(3)=36/3=12,... .

J. J. Balmer, Notiz über die Spectrallinien des Wasserstoffs, Annalen der Physik, vol. 261, 5 (1885) 80-87. First published June 25 1884 (Basel).
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A145979, A138191, A061038, A028310, A022998.

A246943:=n->n*(19-(-1)^n*13+2*cos(n*Pi/2))/8: seq(A246943(n), n=0..100); # Wesley Ivan Hurt, Apr 18 2017

LinearRecurrence[{0,0,0,2,0,0,0,-1},{0,4,1,12,4,20,3,28},60] (* Harvey P. Dale, Jun 22 2022 *)

concat(0, Vec(x*(4*x^6+x^5+12*x^4+4*x^3+12*x^2+x+4)/((x-1)^2*(x+1)^2*(x^2+1)^2) + O(x^100))) \\ Colin Barker, Sep 08 2014

Numerators of A061038(n)/A028310(n).
a(2n) = A022998(n).
G.f.: x*(4*x^6+x^5+12*x^4+4*x^3+12*x^2+x+4) / ((x-1)^2*(x+1)^2*(x^2+1)^2). - Colin Barker, Sep 08 2014
a(n) = n*(19-13*(-1)^n+(1+(-1)^n)*(-1)^((2*n-1+(-1)^n)/4))/8. - Luce ETIENNE, May 26 2015
a(n) = n*(19-(-1)^n*13+2*cos(n*Pi/2))/8. - Giovanni Resta, May 26 2015

A247829 a(3k) = k(k+1), a(3k+1) = (2k-1)(2k+1), a(3k+2) = (2k-1)(2k+3).

Original entry on oeis.org

0, -1, -3, 2, 3, 5, 6, 15, 21, 12, 35, 45, 20, 63, 77, 30, 99, 117, 42, 143, 165, 56, 195, 221, 72, 255, 285, 90, 323, 357, 110, 399, 437, 132, 483, 525, 156, 575, 621, 182, 675, 725, 210, 783, 837, 240, 899, 957, 272, 1023, 1085, 306, 1155, 1221, 342, 1295

Offset: 0

Paul Curtz, Dec 01 2014

A permutation of A061037(n) = -1, -3, 0, 5, 3, 21, 2, 45, 15, 77, 6, ... and of A214297(n) = -1, 0, -3, 2, 3, 6, 5, 12, ... .
Among consequences: b(n) = 4*a(n) + (sequence of period 3:repeat 1, 4, 16) = 1, 0, 4, 9, 16, 36, 25, 64, 100, ... , is a permutation of the squares of the nonnegative integers A000290(n).
And a(n)/b(n) = 0/1, -1/0, -3/4, 2/9, 3/16, 5/36, ... is a permutation of the Balmer series A061037(n)/A061038(n) = -1/0, -3/4, 0/1, 5/36, 3/16, ... .
a(5n) is divisible by 5.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Cf. A142717 (2 terms out of 3), A000290, A000466, A002378, A061037/A061038, A078371, A214297.

m:=50; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3*x^7-3*x^6-14*x^4-6*x^3-2*x^2+3*x+1)/((x-1)^3*(x^2 +x+1)^3))); // G. C. Greubel, Sep 20 2018

seq(op([k*(k+1),(2*k-1)*(2*k+1),(2*k-1)*(2*k+3)]), k=0..100); # Robert Israel, Dec 01 2014

Table[Sequence @@ {n*(n+1), (2*n-1)*(2*n+1), (2*n-1)*(2*n+3)}, {n, 0, 18}] (* Jean-François Alcover, Dec 16 2014 *)

concat(0, Vec(x*(3*x^7-3*x^6-14*x^4-6*x^3-2*x^2+3*x+1)/((x-1)^3*(x^2+x+1)^3) + O(x^100))) \\ Colin Barker, Dec 02 2014

a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9).
a(3*k) + a(3*k+1) + a(3*k+2) = 9*k^2 + 5*k - 4.
G.f.: x*(3*x^7 - 3*x^6 - 14*x^4 - 6*x^3 - 2*x^2 + 3*x + 1)/((x-1)^3*(x^2 +x+1)^3). - Robert Israel, Dec 01 2014
a(n) = -(n^2 + n + floor(n/3)*(27*floor(n/3)^3 - 18*(n+1)*floor(n/3)^2 + (3*n^2 + 21*n - 14)*floor(n/3) - (5*n^2 - n + 5)))/2. - Luce ETIENNE, Mar 13 2017
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/2.
Sum_{n>=1} (-1)^n/a(n) = Pi/4 + 1 - 2*log(2). (End)

A152018 Denominator of 1/n^2-1/(3n)^2 or of 8/(9n^2).

Original entry on oeis.org

9, 9, 81, 18, 225, 81, 441, 72, 729, 225, 1089, 162, 1521, 441, 2025, 288, 2601, 729, 3249, 450, 3969, 1089, 4761, 648, 5625, 1521, 6561, 882, 7569, 2025, 8649, 1152, 9801, 2601, 11025, 1458, 12321, 3249, 13689, 1800, 15129, 3969, 16641, 2178, 18225

Offset: 1

Paul Curtz, Nov 20 2008

nonn

The associated terms of the n-th main series of the Hydrogen energy spectrum are A000290(3), A061038(6), A061040(9), A061042(12), A061044(15), A061046(18), A061048(21), A061050(24), etc.

All numbers are multiples of 9.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1).

Cf. A143025 with a similar principle of construction.

Cf. A291050.

Denominator/@(8/(9Range[50]^2))  (* Harvey P. Dale, Mar 15 2011 *)

Sum_{n>=1} 1/a(n) = Pi^2/27 (A291050). - Amiram Eldar, Sep 14 2022

Stratified definition, corrected indices, extended, R. J. Mathar, Dec 10 2008

A181763 a(n) = A061037(n)^2.

Original entry on oeis.org

0, 25, 9, 441, 4, 2025, 225, 5929, 36, 13689, 1225, 27225, 144, 48841, 3969, 81225, 400, 127449, 9801, 190969, 900, 275625, 20449, 385641, 1764, 525625, 38025, 700569, 3136, 915849, 65025, 1177225, 5184, 1490841, 104329, 1863225, 8100

Offset: 2

Paul Curtz, Nov 14 2010

A061038(n)/a(n+2) for n >= 2 gives the reduced fractions 1/9, 4/49, 4, 4/81, 1/25, 4/121, 16/9, 4/169, ...

G. C. Greubel, Table of n, a(n) for n = 2..1000

Cf. A061037, A061038.

A061037:=func< n | Numerator(1/4-1/n^2) >; A181763:=func< n | A061037(n)^2 >; [ A181763(n): n in [2..50] ]; // Klaus Brockhaus, Jan 09 2011

A061037[n_] := Numerator[(n - 2)*(n + 2)/(4 n^2)]; Table[A061037[n]^2, {n, 2, 100}] (* G. C. Greubel, Sep 19 2018 *)

a(n) = numerator((n-2)*(n+2)/(4*n^2))^2;
for(n=2, 50, print1(a(n), ", ")) \\ G. C. Greubel, Sep 19 2018

Sum_{n>=3} 1/a(n) = 79*Pi^2/192 - 65/18. - Amiram Eldar, Aug 14 2022

A187541 a(4n+2) = 2n+1, otherwise a(n) = 4n.

Original entry on oeis.org

0, 4, 1, 12, 16, 20, 3, 28, 32, 36, 5, 44, 48, 52, 7, 60, 64, 68, 9, 76, 80, 84, 11, 92, 96, 100, 13, 108, 112, 116, 15, 124, 128, 132, 17, 140, 144, 148, 19, 156, 160, 164, 21, 172, 176, 180, 23, 188, 192, 196, 25, 204, 208, 212, 27, 220, 224, 228, 29, 236, 240, 244, 31, 252

Offset: 0

Paul Curtz, Mar 11 2011

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

Cf. A008586, A010855, A060819, A061038.

A187541:=n->16*n/(11+7*(I^(2*n)-I^(-n)-I^n)): seq(A187541(n), n=0..100); # Wesley Ivan Hurt, Jul 05 2016

Table[16n/(11+7*(I^(2*n)-I^(-n)-I^n)), {n, 0, 80}] (* Wesley Ivan Hurt, Jul 05 2016 *)

a(n) = 2*a(n-4) - a(n-8) for n>7.
G.f.: x*(4+x+12*x^2+16*x^3+12*x^4+x^5+4*x^6)/(1-x^4)^2; a(n) = (n/8)*(32 -7*(1+(-1)^n)*(1-i^n)) where i=sqrt(-1). - Bruno Berselli, Mar 15 2011
From Paul Curtz, Mar 22 2011: (Start)
A060819(n)*a(n) = 0,4,1,36,16,100, = 0,4, followed by A061038(n+2).
a(n) = a(n-4) + period 4: repeat [16, 16, 2, 16]. Note that a(n) = 4*n/(period 4: repeat [1, 1, 8, 1]), Hence 16's = A010855. (End)
a(n) = 16*n/(11+7*(I^(2*n)-I^(-n)-I^n)). - Wesley Ivan Hurt, Jul 05 2016

Edited by N. J. A. Sloane, Mar 15 2011

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

A226279 a(4n) = a(4n+2) = 2n , a(4n+1) = a(4n+3) = 2n-1.

Original entry on oeis.org

0, -1, 0, -1, 2, 1, 2, 1, 4, 3, 4, 3, 6, 5, 6, 5, 8, 7, 8, 7, 10, 9, 10, 9, 12, 11, 12, 11, 14, 13, 14, 13, 16, 15, 16, 15, 18, 17, 18, 17, 20, 19, 20, 19, 22, 21, 22, 21, 24, 23, 24, 23, 26, 25, 26, 25, 28, 27, 28, 27, 30, 29, 30, 29

Offset: 0

Paul Curtz, Jun 02 2013

a(n)=c(n) in A214297(n).
In A214297 d(n)=-1,1,1,3,1,3,3,... = mix (-A186422(2n) , A186422(2n+1)).
A214297 is the (reduced) numerator of 1/4 - 1/A061038(n).
(i.e. (1/4 -(1/0, 1/4, 1, 1/36, 1/16,...)) = -1/0, 0/1, -3/4, 2/9, 3/16,... )
1/0 is a convention.
n^2=(a(n+1)+d(n+1))^2 are the denominators.

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

Cf. A134967, A162330, A103889, A000290.

Table[{0, -1} + 2*Floor[n/2], {n, 0, 31}] // Flatten (* Jean-François Alcover, Jun 03 2013 *)

a(n)=n\4*2-n%2 \\ Charles R Greathouse IV, Sep 15 2013

a(0) = a(2)=0, a(1)=a(3)=-1, a(4)=2.
a(n) = a(n-4) + 2, n > 3.
a(n) = a(n-1) + a(n-4) - a(n-5), n > 4.
A214297(n) = a(n+1) * d(n+1).
G.f.: x*(3*x^3-x^2+x-1) / ((x-1)^2*(x+1)*(x^2+1)). - Colin Barker, Sep 22 2013

A227168 a(n) = gcd(2n, n(n+1)/2)^2.

Original entry on oeis.org

1, 1, 36, 4, 25, 9, 196, 16, 81, 25, 484, 36, 169, 49, 900, 64, 289, 81, 1444, 100, 441, 121, 2116, 144, 625, 169, 2916, 196, 841, 225, 3844, 256, 1089, 289, 4900, 324, 1369, 361, 6084, 400

Offset: 1

Paul Curtz, Jul 03 2013

a(n) is defined as A062828(n)^2 for n >= 1. If we extend the sequence to n=0 and negative n by use of the recurrence that relates a(n) to a(n+12), a(n+8) and a(n+4), we obtain a(0)=0, a(-1)=4 and a(-n) = A176743(n-2)^2 for n >= 2.
Define c(n) = a(n+2) - a(n-2) for c >= 0. Because a(n) is a shuffle of three interleaved 2nd-order polynomials, c(n) is a shuffle of three interleaved 1st-order polynomials: c(n) = 4* A062828(n)*(periodically repeated 1, 8, 1, 1).
The sequence a(n) is case p=0 of the family A062828(n)*A062828(n+p):
0, 1, 1, 36,  4, 25,  9, 196, ... = a(n).
0, 1, 6, 12, 10, 15, 42,  56, ... = A130658(n)*A000217(n) = A177002(n-1)*A064038(n+1).
0, 6, 2, 30,  6, 70, 12, 126, ... = 2*A198148(n)
0, 2, 5, 18, 28, 20, 27,  70, ... = A177002(n+2)*A160050(n+1) = A014695(n+2)*A000096(n).

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A062828, A016814, A017138, A005565, A006752, A061037, A061038.

[GCD(2*n, n*(n+1)/2)^2: n in [1..50]]; // G. C. Greubel, Sep 20 2018

A227168 := proc(n)
    A062828(n)^2 ;
end proc: # R. J. Mathar, Jul 25 2013

a[n_] := GCD[2*n, n*(n + 1)/2]^2; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Jul 03 2013 *)

a(n)=if(n%2, n*if(n%4>2, 2, 1), n/2)^2 \\ Charles R Greathouse IV, Jul 07 2013

a(n) = A062828(n)^2.
a(4n) = (4*n+1)^2; a(2n+1) = (n+1)^2; a(4n+2) = 4*(4*n+3)^2.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) * (period 4: repeat 4, 1, 1, 4) = A061038(n).
A005565(n-3) = a(n+1) * A061037(n). - Corrected by R. J. Mathar, Jul 25 2013
a(n) = A130658(n-1)^2 * A181318(n). - Corrected by R. J. Mathar, Aug 01 2013
G.f.: -x*(1 + x + 36*x^2 + 4*x^3 + 22*x^4 + 6*x^5 + 88*x^6 + 4*x^7 + 9*x^8 + x^9 + 4*x^10) / ( (x-1)^3*(1+x)^3*(x^2+1)^3 ). - R. J. Mathar, Jul 20 2013
Sum_{n>=1} 1/a(n) = 47*Pi^2/192 + 3*G/8, where G is Catalan's constant (A006752). - Amiram Eldar, Aug 21 2022

A236203 Interleave A005563(n), A028347(n).

Original entry on oeis.org

0, 0, 3, 5, 8, 12, 15, 21, 24, 32, 35, 45, 48, 60, 63, 77, 80, 96, 99, 117, 120, 140, 143, 165, 168, 192, 195, 221, 224, 252, 255, 285, 288, 320, 323, 357, 360, 396, 399, 437, 440, 480, 483, 525, 528, 572, 575, 621, 624, 672, 675, 725, 728, 780, 783, 837, 840, 896

Offset: 2

Paul Curtz, Jan 20 2014

A175628 gives the numerators of interleaved Lyman and Balmer series, i.e., A005563(n)/A000290(n+1) and A061037(n+2)/A061038(n+2).
Difference table of a(n):
   -1,  -3,  0,  0,  3,   5,   8,  12,  15,  21,  24, ...
   -2,   3,  0,  3,  2,   3,   4,   3,   6,   3,   8, ...
    5,  -3,  3, -1,  1,   1,  -1,   3,  -3,   5,  -5, ...
   -8,   6, -4,  2,  0,  -2,   4,  -6,   8, -10,  12, ...
   14, -10,  6, -2, -2,   6, -10,  14, -18,  22, -26, ...
  -24,  16, -8,  0,  8, -16,  24, -32,  40, -48,  56, ... .
a(n+2) gives the numerators of 0/1, 0/16, 3/4, 5/36, 8/9, 12/64, 15/16, 21/100, 24/25, 32/144, ... . The denominators are A097362(n+1)^2. (Compare A097362 to A029578.)
Note the particular distribution of a(-n). Example:
a(n-9) = 12,15, 5,8, 0,3, -3,0, -4,-1, -3,0, 0,3, 5,8, 12,15, ... .
a(2n) + a(2n+1) = a(-2n-1) + a(-2n-2) = -4,0,8,20,36,56,80,... = 4*A000096(n-1).
a(2n) + a(2n-1) = a(-2n) + a(-2n-1) = -5,-3,3,13,... = A001105(n) - A010716(n).

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

Cf. A005843, A008590, A016825, A109613, A147658.

List([2..60], n-> (2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8 ); # G. C. Greubel, Dec 04 2019

[(2*n^2 + 2*n - 19 - (2*n - 11)*(-1)^n)/8: n in [2..60]]; // Vincenzo Librandi, Jul 27 2014

seq( (2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8, n=2..60); # G. C. Greubel, Dec 04 2019

CoefficientList[Series[x^2(3x^2-2x-3)/((x-1)^3(x+1)^2), {x, 0, 60}], x] (* Vincenzo Librandi, Jul 27 2014 *)
LinearRecurrence[{1,2,-2,-1,1},{0,0,3,5,8},60] (* Harvey P. Dale, Aug 30 2018 *)

concat([0,0], Vec(x^4*(3*x^2-2*x-3)/((x-1)^3*(x+1)^2) + O(x^60))) \\ Colin Barker, Jan 26 2014

[(2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8 for n in (2..60)] # G. C. Greubel, Dec 04 2019

a(n+2) = (period 8: repeat 1, 16, 1, 1, 1, 4, 1, 1)*A175628(n+1).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n+4) - a(n-4) = 0, 8, 8, ... = A168397.
From Colin Barker, Jan 26 2014: (Start)
a(n) = (n^2 -4)/4 for n even, a(n) = (n^2 +2*n -15)/4 for n odd.
G.f.: x^4*(3 + 2*x - 3*x^2)/ ((1-x)^3*(1+x)^2). (End)
a(n) = (2*n^2 + 2*n - 19 - (2*n - 11)*(-1)^n)/8. - Luce ETIENNE, Jul 26 2014
Sum_{n>=4} (-1)^n/a(n) = 11/48. - Amiram Eldar, Aug 21 2022

More terms from Colin Barker, Jan 26 2014

A251091 a(n) = n^2 / gcd(n+2, 4).

Original entry on oeis.org

0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121, 72, 169, 49, 225, 128, 289, 81, 361, 200, 441, 121, 529, 288, 625, 169, 729, 392, 841, 225, 961, 512, 1089, 289, 1225, 648, 1369, 361, 1521, 800, 1681, 441, 1849, 968, 2025, 529, 2209, 1152, 2401, 625, 2601, 1352

Offset: 0

Paul Curtz, May 08 2015

A061038(n), which appears in 4*a(n) formula, is a permutation of n^2.
Origin. In December 2010, I wrote in my 192-page Exercise Book no. 5, page 41, the array (difference table of the first row):
   1     0,   1/3,     1,   9/5,    8/3,   25/7,    9/2,   49/9, ...
  -1,  1/3,   2/3,   4/5, 13/15,  19/21,  13/14,  17/18,  43/45, ...
Numerators are listed in A176126, denominators are in A064038, and denominator - numerator = 2, 2, 1, 1,... (A014695).
  4/3, 1/3,  2/15,  1/15, 4/105,   1/42,   1/63,   1/90,  4/495, ...
  -1, -1/5, -1/15, -1/35, -1/70, -1/126, -1/210, -1/330, -1/495, ...
where the denominators of the second row are listed in A000332.
Also for those of the inverse binomial transform
  1, -1, 4/3, -1, 4/5, -2/3, 4/7, -1/2, 4/9, -2/5, 4/11, -1/3, ... ?
a(n) is the (n+1)-th term of the numerators of the first row.

			a(0) = 0/2, a(1) = 1/1, a(2) = 4/4, a(3) = 9/1.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000290, A000332, A016754, A026741, A060819, A061038, A109008, A139098, A168077, A176895, A181900.

[(1-(1/16)*(1+(-1)^n)*(5-(-1)^(n div 2)) )*n^2: n in [0..60]]; // Vincenzo Librandi, Jun 12 2015

seq(seq((4*i+j-1)^2/[2,1,4,1][j],j=1..4),i=0..30); # Robert Israel, May 14 2015

f[n_] := Switch[ Mod[n, 4], 0, n^2/2, 1, n^2, 2, n^2/4, 3, n^2]; Array[f, 50, 0] (* or *) Table[(4 i + j - 1)^2/{2, 1, 4, 1}[[j]], {i, 0, 12}, {j, 4}] // Flatten (* after Robert Israel *) (* or *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121}, 53] (* or *) CoefficientList[ Series[-((x (1 + x (1 + x (9 + x (8 + x (22 + x (6 + x (22 + x (8 + x (9 + x + x^2))))))))))/(-1 + x^4)^3), {x, 0, 52}], x] (* Robert G. Wilson v, May 19 2015 *)

concat(0, Vec(-x*(x^10 + x^9 + 9*x^8 + 8*x^7 + 22*x^6 + 6*x^5 + 22*x^4 + 8*x^3 + 9*x^2 + x + 1) / ((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100))) \\ Colin Barker, May 14 2015

a(n) = n^2/(period 4: repeat 2, 1, 4, 1).
a(4n) = 8*n^2, a(2n+1) = a(4n+2) = (2*n+1)^2.
a(n+4) = a(n) + 8*A060819(n).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11.
4*a(n) = (period 4: repeat 2, 1, 4, 1) * A061038(n).
G.f.: -x*(x^10+x^9+9*x^8+8*x^7+22*x^6+6*x^5+22*x^4+8*x^3+9*x^2+x+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, May 14 2015
a(2n) = A181900(n), a(2n+1) = A016754(n). [Bruno Berselli, May 14 2015]
a(n) = ( 1 - (1/16)*(1+(-1)^n)*(5-(-1)^(n/2)) )*n^2. - Bruno Berselli, May 14 2015
Sum_{n>=1} 1/a(n) = 13*Pi^2/48. - Amiram Eldar, Aug 12 2022

Missing term (1521) inserted in the sequence by Colin Barker, May 14 2015
Definition uses a formula by Jean-François Alcover, Jul 01 2015
Keyword:mult added by Andrew Howroyd, Aug 06 2018

cp's OEIS Frontend

A246943 a(4n) = 4*n , a(2n+1) = 8*n+4 , a(4n+2) = 2*n+1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A247829 a(3*k) = k*(k+1), a(3*k+1) = (2*k-1)*(2*k+1), a(3*k+2) = (2*k-1)*(2*k+3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A152018 Denominator of 1/n^2-1/(3n)^2 or of 8/(9n^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A181763 a(n) = A061037(n)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A187541 a(4n+2) = 2n+1, otherwise a(n) = 4n.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A226279 a(4n) = a(4n+2) = 2*n , a(4n+1) = a(4n+3) = 2*n-1.

Views

Author

Keywords

Comments

Links

A246943 a(4n) = 4n , a(2n+1) = 8n+4 , a(4n+2) = 2*n+1.

A247829 a(3k) = k(k+1), a(3k+1) = (2k-1)(2k+1), a(3k+2) = (2k-1)(2k+3).

A226279 a(4n) = a(4n+2) = 2n , a(4n+1) = a(4n+3) = 2n-1.

A227168 a(n) = gcd(2n, n(n+1)/2)^2.