A061037 -id:A061037 - cp's OEIS frontend

A142590 First trisection of A061037 (Balmer line series of the hydrogen atom).

0, 21, 15, 117, 12, 285, 99, 525, 42, 837, 255, 1221, 90, 1677, 483, 2205, 156, 2805, 783, 3477, 240, 4221, 1155, 5037, 342, 5925, 1599, 6885, 462, 7917, 2115, 9021, 600, 10197, 2703, 11445, 756, 12765, 3363, 14157, 930, 15621, 4095, 17157, 1122, 18765, 4899, 20445

Offset: 0

Paul Curtz, Sep 22 2008

All terms are multiples of 3.

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

Cf. A010872, A061037.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(3*x*(-15*x^8 -18*x^5 -74*x^4 -39*x^2 -5*x-7 -4*x^3 +x^10 -2*x^7 -x^9 -58*x^6)/((x-1)^3*(1+x)^3*(x^2+1)^3))); // G. C. Greubel, Sep 19 2018

Table[Numerator[1/4 - 1/#^2] &[2 + 3 n], {n, 0, 47}] (* Michael De Vlieger, Apr 02 2017 *)
CoefficientList[Series[3*x*(-15*x^8 -18*x^5 -74*x^4 -39*x^2 -5*x-7 -4*x^3 +x^10 -2*x^7 -x^9 -58*x^6)/ ((x-1)^3*(1+x)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)

my(x='x+O('x^50)); concat([0], Vec(3*x*(-15*x^8 -18*x^5 -74*x^4 -39*x^2 -5*x-7 -4*x^3 +x^10 -2*x^7 -x^9 -58*x^6)/((x-1)^3*(1+x)^3*(x^2+1)^3))) \\ G. C. Greubel, Sep 19 2018

a(n) = A061037(2+3n).
a(n) mod 9 = 3*A010872(n).
G.f.: 3*x*(-15*x^8 -18*x^5 -74*x^4 -39*x^2 -5*x-7 -4*x^3 +x^10 -2*x^7 -x^9 -58*x^6)/ ((x-1)^3*(1+x)^3*(x^2+1)^3). - R. J. Mathar, Sep 22 2008
a(n) = 3*n*(3*n+4)*(37-27*cos(n*Pi)-6*cos(n*Pi/2))/64. - Luce ETIENNE, Mar 31 2017
Sum_{n>=1} 1/a(n) = 5/4 - 5*Pi/(48*sqrt(3)) - 11*log(3)/16. - Amiram Eldar, Sep 11 2022

Edited and extended by R. J. Mathar, Sep 22 2008

A165943 a(n) = A061037(7*n+2).

Original entry on oeis.org

0, 77, 63, 525, 56, 1365, 483, 2597, 210, 4221, 1295, 6237, 462, 8645, 2499, 11445, 812, 14637, 4095, 18221, 1260, 22197, 6083, 26565, 1806, 31325, 8463, 36477, 2450, 42021, 11235, 47957, 3192, 54285, 14399, 61005, 4032, 68117, 17955, 75621, 4970

Offset: 0

Paul Curtz, Oct 01 2009

The (2k+1)-sections A061037((2*k+1)*n+2) are multiples of 2k+1:
0,...21,...15,..117,...12,..285,...99,..525,...42,..837,..255, k=1, A142590
0,...45,...35,..285,...30,..725,..255,.1365,..110,.2205,..675, k=2, A165248
0,...77,...63,..525,...56,.1365,..483,.2597,..210,.4221,.1295, k=3, here
0,..117,...99,..837,...90,.2205,..783,.4221,..342,.6885,.2115, k=4,
0,..165,..143,.1221,..132,.3245,.1155,.6237,..506,10197,.3135, k=5
0,..221,..195,.1677,..182,.4485,.1599,.8645,..702,14157,.4355, k=6
After division by 2k+1 these define a table T'(k,c) :
0,....7,....5,...39,....4,...95,...33,..175,...14,..279,...85, k=1, A142883
0,....9,....7,...57,....6,..145,...51,..273,...22,..441,..135, k=2
0,...11,....9,...75,....8,..195,...69,..371,...30,..603,..185, k=3
0,...13,...11,...93,...10,..245,...87,..469,...38,..765,..235, k=4
0,...15,...13,..111,...12,..295,..105,..567,...46,..927,..285, k=5
0,...17,...15,..129,...14,..345,..123,..665,...54,.1089,..335, k=6
Differences downwards each second column in this second table are 2 = 7-5 = 9-7..; 18 = 57-39 = 75-57..; 50 = 145-95 = 195-145... = A077591(n+1) = 2*A016754.
The difference T'(k+1,c)-T'(k,c) is 0, 2, 2, 18, 2, 50, 18, 98, 8 ... = 2*A181318(c) =A061037(c-2)+A061037(c+2). - Paul Curtz, Mar 12 2012
Let b(n)= a(n) mod 11. The sequence b(n) has the property b(n+44) = b(n) with the first 43 values being {0, 0 , 8, 1, 1, 10, 1, 1, 8, 8, 0, 0, 10, 5, 5, 9, 7, 1, 5, 6, 10, 7, 0, 2, 1, 1, 8, 1, 5, 8, 2, 0, 8, 10, 6, 5, 10, 7, 9, 5, 0, 10, 0}. - G. C. Greubel, Apr 18 2016

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

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018

seq(numer(1/4 - 1/(7*n+2)^2), n=0..50); # Robert Israel, Apr 20 2016

Table[Numerator[1/4 - 1/(7 n + 2)^2], {n, 0, 40}] (* Michael De Vlieger, Apr 19 2016 *)
CoefficientList[Series[7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)

a(n) = numerator(1/4 - 1/(7*n+2)^2); \\  Altug Alkan, Apr 18 2016

x='x+O('x^50); concat([0], Vec(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>12. - Conjectured by R. J. Mathar, Mar 02 2010, proved by Robert Israel, Apr 20 2016
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -7*n*(7*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

Partially edited and extended by R. J. Mathar, Mar 02 2010

Removed division by 7 in definition and formula - R. J. Mathar, Mar 23 2010

A165248 Quintisection A061037(5*n+2).

Original entry on oeis.org

0, 45, 35, 285, 30, 725, 255, 1365, 110, 2205, 675, 3245, 240, 4485, 1295, 5925, 420, 7565, 2115, 9405, 650, 11445, 3135, 13685, 930, 16125, 4355, 18765, 1260, 21605, 5775, 24645, 1640, 27885, 7395, 31325, 2070, 34965, 9215, 38805, 2550, 42845, 11235, 47085

Offset: 0

Paul Curtz, Sep 10 2009

A trisection of A061037 is in A142590. These (2k+1)-sections A061037(2+n*(2k+1)) are multiples of 2k+1.

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

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018

CoefficientList[Series[5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 +4*x^7 + 33*x^8 + 3*x^9 +x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,45,35,285,30,725,255,1365,110,2205,675,3245},50] (* Harvey P. Dale, Sep 18 2021 *)

a(n) = numerator(1/4 - 1/(5*n+2)^2); \\ Altug Alkan, Apr 19 2016

x='x+O('x^50); concat([0], Vec(5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018

Conjecture: a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11. - R. J. Mathar, Mar 02 2010
The conjecture is equivalent to a(4n) = 5n*(5n+1), a(4n+1) = 5*(20n+9)*(4n+1), a(4n+2) = 5*(10n+7)*(2n+1) and a(4n+3) = 5*(20n+19)*(4n+3). - R. J. Mathar, Feb 13 2011
The conjectures can be proved by taking the closed form of A061037, and writing up the quadrisections case by case. - Bruno Berselli, Feb 20 2011
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 5*x*(9 + 7*x + 57*x^2 + 6*x^3 + 118*x^4 + 30*x^5 + 102*x^6 + 4*x^7 + 33*x^8 + 3*x^9 + x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -5*n (5*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

Extended by R. J. Mathar, Mar 02 2010

A165342 a(n) = A061037(n+2)/A000265(n+4).

Original entry on oeis.org

0, 1, 1, 3, 2, 5, 3, 7, 2, 9, 5, 11, 12, 13, 7, 15, 4, 17, 9, 19, 10, 21, 11, 23, 6, 25, 13, 27, 56, 29, 15, 31, 8, 33, 17, 35, 18, 37, 19, 39, 10, 41, 21, 43, 44, 45, 23, 47, 12, 49, 25, 51, 26, 53, 27, 55, 14, 57, 29, 59, 240, 61, 31, 63, 16, 65, 33, 67, 34, 69, 35, 71, 18, 73, 37

Offset: 0

Paul Curtz, Sep 15 2009

nonn

Coincides with A026741 at many places. Apparently the ratio between the two sequences is always one of 1/2, 1, 2, 4, 8, 16, ...

The distance between antecedents of 1 is 1, between antecedents of 3 is 3, and more generally, for 1+2*k this distance is 1+2*k. Their respective ranks are 1,2,3,5,6,7,9,10,11,... (A042968). - Paul Curtz, Apr 08 2011

G. C. Greubel, Table of n, a(n) for n = 0..10000

A000265 := proc(n) nshft := n ; while nshft mod 2 = 0 do nshft := nshft/2 ; od: nshft ; end:
A061037 := proc(n) numer ( 1/4-1/n^2) ; end: A165342 := proc(n) A061037(n+2)/A000265(n+4) ; end: seq( A165342(n),n=0..120) ; # R. J. Mathar, Sep 16 2009

A061037[n_] := Numerator[(n - 2) (n + 2)/(4 n^2)]; A000265[n_] := If[n == 0, 0, n/2^IntegerExponent[n, 2]]; Table[A061037[n + 2]/A000265[n + 4], {n, 0, 100}] (* G. C. Greubel, Sep 19 2018 *)

Edited, extended by R. J. Mathar, Sep 16 2009

A175628 a(2n+1) = A005563(n). a(2n) = A061037(n+1).

Original entry on oeis.org

0, 0, 3, 5, 8, 3, 15, 21, 24, 2, 35, 45, 48, 15, 63, 77, 80, 6, 99, 117, 120, 35, 143, 165, 168, 12, 195, 221, 224, 63, 255, 285, 288, 20, 323, 357, 360, 99, 399, 437, 440, 30, 483, 525, 528, 143, 575, 621, 624, 42, 675, 725, 728, 195, 783, 837, 840, 56, 899, 957, 960, 255, 1023, 1085, 1088, 72, 1155, 1221, 1224, 323, 1295, 1365

Offset: 1

Paul Curtz, Dec 04 2010

Mingles the numerators of the Lyman and Balmer series of the hydrogen problem.

G. C. Greubel, Table of n, a(n) for n = 1..5000
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. A005563, A061037.

R:= RealField(20);
a:= func< n | (n mod 2) eq 1 select (n-1)*(n+3)/4 else Round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*Cos((n+2)*Pi(R)/4))/2^8) >;
[a(n): n in [1..90]]; // G. C. Greubel, Sep 19 2018; Dec 04 2019

seq( `if`( (n mod 2) = 1, (n-1)*(n+3)/4, (n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*cos((n+2)*Pi/4))/2^8 ), n=1..90); # G. C. Greubel, Dec 04 2019

a[n_]:= If[OddQ[n], (n-1)*(n+3)/4, (n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*Cos[(n + 2)*Pi/4])/2^8]; Table[a[n], {n, 90}] (* G. C. Greubel, Sep 19 2018; Dec 04 2019 *)

a(n) = if(n%2==1, (n-1)*(n+3)/4, round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*cos((n+2)*Pi/4))/2^8) ); \\ G. C. Greubel, Sep 19 2018; Dec 04 2019

def a(n):
    if (mod(n,2)==1): return (n-1)*(n+3)/4
    else: return round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*cos((n+2)*pi/4))/2^8)
[a(n) for n in (1..90)] # G. C. Greubel, Dec 04 2019

a(n) = 3*a(n-8) - 3*a(n-16) + a(n-24). - R. J. Mathar, Dec 08 2010
G.f.: x^3*(3*x^21 + x^20 + x^19 + 3*x^17 - 3*x^16 - 8*x^14 - 14*x^13 - 18*x^12 - 6*x^11 - 24*x^10 - 30*x^9 - 26*x^8 - 2*x^7 - 24*x^6 - 21*x^5 - 15*x^4 - 3*x^3 - 8*x^2 - 5*x -3) / ((x-1)^3*(x+1)^3*(x^2+1)^3*(x^4+1)^3). - Colin Barker, Jan 26 2014
a(n) = (n-1)*(n+3)/4 when n is odd, otherwise (n^2+4*n-12)*(37 + 27*(-1)^(n/2) + 6*cos((n+2)*Pi/4))/2^8. - G. C. Greubel, Dec 04 2019
Sum_{n>=3} 1/a(n) = 31/12. - Amiram Eldar, Aug 14 2022

A142600 Third trisection of A061037.

Original entry on oeis.org

3, 45, 6, 165, 63, 357, 30, 621, 195, 957, 72, 1365, 399, 1845, 132, 2397, 675, 3021, 210, 3717, 1023, 4485, 306, 5325, 1443, 6237, 420, 7221, 1935, 8277, 552, 9405, 2499, 10605, 702, 11877, 3135, 13221, 870, 14637, 3843, 16125, 1056, 17685, 4623, 19317

Offset: 1

Paul Curtz, Sep 23 2008

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. A061037, A174325.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(3*x*(x^11 -7*x^9 -5*x^8 -42*x^7 -4*x^6 -74*x^5 -18*x^4 -55*x^3 -2*x^2 -15*x -1)/((x-1)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 19 2018

Table[Numerator[(n-2)*(n+2)/(4*n^2)],{n,4,100,3}] (* Vaclav Kotesovec, Oct 15 2014 *)
Rest[CoefficientList[Series[3*x*(x^11 -7*x^9 -5*x^8 -42*x^7 -4*x^6 -74*x^5 -18*x^4 -55*x^3 -2*x^2 -15*x -1)/((x-1)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x]] (* G. C. Greubel, Sep 19 2018 *)

Vec(3*x*(x^11-7*x^9-5*x^8-42*x^7-4*x^6-74*x^5-18*x^4-55*x^3 -2*x^2-15*x-1)/((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Oct 15 2014

G.f.: 3*x*(x^11 -7*x^9 -5*x^8 -42*x^7 -4*x^6 -74*x^5 -18*x^4 -55*x^3 -2*x^2 -15*x -1) / ((x -1)^3*(x +1)^3*(x^2 +1)^3). - Colin Barker, Oct 15 2014

Sum_{n>=1} 1/a(n) = 11*log(3)/16 - 5*Pi/(48*sqrt(3)) + 1/12. - Amiram Eldar, Sep 11 2022

Edited by N. J. A. Sloane, Jan 04 2009

More terms from Colin Barker, Oct 15 2014

A174325 Trisection A061037(3*n-2) of the Balmer spectrum numerators extended to negative indices.

Original entry on oeis.org

0, -3, 3, 45, 6, 165, 63, 357, 30, 621, 195, 957, 72, 1365, 399, 1845, 132, 2397, 675, 3021, 210, 3717, 1023, 4485, 306, 5325, 1443, 6237, 420, 7221, 1935, 8277, 552, 9405, 2499, 10605, 702, 11877, 3135, 13221, 870, 14637, 3843, 16125, 1056, 17685, 4623, 19317

Offset: 0

Paul Curtz, Nov 27 2010

All terms are multiples of 3.

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

Cf. A061037, A142590, A142600.

I:=[0,-3,3,45,6,165,63,357,30,621,195,957]; [n le 12 select I[n] else 3*Self(n-4)-3*Self(n-8)+Self(n-12): n in [1..50]]; // Vincenzo Librandi, Oct 15 2014

Table[Numerator[(n-2)*(n+2)/(4*n^2)],{n,-2,300,3}] (* Vaclav Kotesovec, Oct 15 2014 *)

concat(0, Vec(-3*x*(7*x^10 +5*x^9 +39*x^8 +4*x^7 +74*x^6 +18*x^5 +58*x^4 +2*x^3 +15*x^2 +x -1) / ((x -1)^3*(x +1)^3*(x^2 +1)^3) + O(x^100))) \\ Colin Barker, Oct 15 2014

a(n) = A142600(n-1), n>1.
G.f.: -3*x*(7*x^10 +5*x^9 +39*x^8 +4*x^7 +74*x^6 +18*x^5 +58*x^4 +2*x^3 +15*x^2 +x -1) / ((x -1)^3*(x +1)^3*(x^2 +1)^3). - Colin Barker, Oct 15 2014
Sum_{n>=1} 1/a(n) = 11*log(3)/16 - 5*Pi/(48*sqrt(3)) - 1/4. - Amiram Eldar, Sep 11 2022

A142599 Second trisection of A061037.

Original entry on oeis.org

5, 2, 77, 35, 221, 20, 437, 143, 725, 56, 1085, 323, 1517, 110, 2021, 575, 2597, 182, 3245, 899, 3965, 272, 4757, 1295, 5621, 380, 6557, 1763, 7565, 506, 8645, 2303, 9797, 650, 11021, 2915, 12317, 812, 13685, 3599, 15125, 992, 16637, 4355, 18221, 1190, 19877, 5183, 21605, 1406, 23405

Offset: 1

Paul Curtz, Sep 23 2008

From Balmer spectrum of hydrogen. First trisection is A142590.

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

Cf. A061037, A142590.

A061037[n_] := Numerator[(n-2)*(n+2)/(4*n^2)]; Table[A061037[n], {n, 1, 300}][[3 ;;  ;; 3]] (* G. C. Greubel, Sep 19 2018 *)

Sum_{n>=1} 1/a(n) = 5*Pi/(24*sqrt(3)) + 1/2. - Amiram Eldar, Sep 11 2022

Edited by N. J. A. Sloane, Sep 27 2008

Terms a(18) onward added by G. C. Greubel, Sep 19 2018

A174850 Quintisection A061037(5*n-2).

Original entry on oeis.org

0, 5, 15, 165, 20, 525, 195, 1085, 90, 1845, 575, 2805, 210, 3965, 1155, 5325, 380, 6885, 1935, 8645, 600, 10605, 2915, 12765, 870, 15125, 4095, 17685, 1190, 20445, 5475, 23405, 1560, 26565, 7055, 29925, 1980, 33485, 8835, 37245, 2450

Offset: 0

Paul Curtz, Dec 01 2010

All entries are multiples of 5. Like A061037(n+2), the (2k+1)-sections A061037((2*k+1)*n-2) are multiples of 2k+1; see A165248, A165943.

The sequence contains 4 interlaced second-order polynomials: a(4n) = 5n*(5n-1), a(4n+1) = 5*(4n+1)*(20n+1), a(4n+2)= 5*(2n+1)*(10n+3), a(4n+3)= 5*(4n+3)*(20n+11). - R. J. Mathar, Feb 10 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

[ Numerator(1/4-1/(5*n-2)^2): n in [0..40] ]; // Bruno Berselli, Feb 10 2011

f[n_] := n/GCD[n, 4]; Table[ f[n] f[n + 4], {n, -4, 200, 5}] (* Robert G. Wilson v, Feb 03 2011 *)

vector(50, n, n--; numerator( 1/4 - 1/(5*n-2)^2 )) \\ G. C. Greubel, Sep 19 2018

a(n) = numerator( 1/4 - 1/(5*n-2)^2 ).
From R. J. Mathar, Feb 10 2011: (Start)
a(n) = +3*a(n-4) -3*a(n-8) +a(n-12).
G.f. ( -5*x*(1+3*x+33*x^2+4*x^3+102*x^4+30*x^5+118*x^6+6*x^7+57*x^8 +7*x^9+9*x^10) )/( (x-1)^3*(1+x)^3*(x^2+1)^3 ). (End)
a(n) = 5*n*(5*n-4)*(37-27*(-1)^n-3*(-i)^n-3*i^n)/64, where i=sqrt(-1).  - Bruno Berselli, Feb 10 2011

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

cp's OEIS Frontend

A142590 First trisection of A061037 (Balmer line series of the hydrogen atom).

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A165943 a(n) = A061037(7*n+2).

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A165248 Quintisection A061037(5*n+2).

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A165342 a(n) = A061037(n+2)/A000265(n+4).

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A175628 a(2*n+1) = A005563(n). a(2*n) = A061037(n+1).

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A142600 Third trisection of A061037.

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A174325 Trisection A061037(3*n-2) of the Balmer spectrum numerators extended to negative indices.

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A175628 a(2n+1) = A005563(n). a(2n) = A061037(n+1).