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

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

Author

Paul Curtz, Dec 04 2010

Keywords

• nonn
• easy
• less

Comments

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

Links

• 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).

Crossrefs

Cf. A005563, A061037.

Programs

Magma

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

Maple

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

Mathematica

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 *)

PARI

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

Sage

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

Formula

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