login
a(2*n+1) = A005563(n). a(2*n) = A061037(n+1).
3

%I #26 Aug 14 2022 03:02:26

%S 0,0,3,5,8,3,15,21,24,2,35,45,48,15,63,77,80,6,99,117,120,35,143,165,

%T 168,12,195,221,224,63,255,285,288,20,323,357,360,99,399,437,440,30,

%U 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

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

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

%H G. C. Greubel, <a href="/A175628/b175628.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Rec#order_24">Index entries for linear recurrences with constant coefficients</a>, 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).

%F a(n) = 3*a(n-8) - 3*a(n-16) + a(n-24). - _R. J. Mathar_, Dec 08 2010

%F 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

%F 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

%F Sum_{n>=3} 1/a(n) = 31/12. - _Amiram Eldar_, Aug 14 2022

%p 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

%t 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 *)

%o (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

%o (Magma) R:= RealField(20);

%o 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) >;

%o [a(n): n in [1..90]]; // _G. C. Greubel_, Sep 19 2018; Dec 04 2019

%o (Sage)

%o def a(n):

%o if (mod(n,2)==1): return (n-1)*(n+3)/4

%o else: return round((n^2+4*n-12)*(37 +27*(-1)^(n/2) +6*cos((n+2)*pi/4))/2^8)

%o [a(n) for n in (1..90)] # _G. C. Greubel_, Dec 04 2019

%Y Cf. A005563, A061037.

%K nonn,easy,less

%O 1,3

%A _Paul Curtz_, Dec 04 2010