A185138 - OEIS

%I #57 Aug 12 2022 09:22:26

%S 0,0,1,1,3,3,15,6,14,10,45,15,33,21,91,28,60,36,153,45,95,55,231,66,

%T 138,78,325,91,189,105,435,120,248,136,561,153,315,171,703,190,390,

%U 210,861,231,473,253,1035,276,564,300,1225,325

%N a(4*n) = n*(4*n-1); a(2*n+1) = n*(n+1)/2; a(4*n+2) = (2*n+1)*(4*n+1).

%C a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).

%C a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.

%C (n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).

%C The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.

%C A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:

%C 0  1  1   3   3   15...

%C 0  3  6  45  21   60...

%C 0 15 15  60  55  325...

%C 0 14 28 231 105  315...

%C 0 45 45 189 171 1035...

%C The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:

%C 0 1 1  3  3  15  6  14   k=0

%C 0 1 2 15  7  20 15  77   k=1

%C 0 3 3 12 11  65 24  63   k=2

%C 0 2 4 33 15  45 33 175   k=3

%C 0 5 5 21 19 115 42 112   k=4

%C 0 3 6 51 23  70 51 273   k=5

%C The entries T'(k,c) are divisible by A060819(c).

%C Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.

%C Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is  A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.

%C The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

%H G. C. Greubel, <a href="/A185138/b185138.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).

%F a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).

%F a(n) = 3*A208950(n)/A109613(n).

%F a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).

%F G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - _R. J. Mathar_, Mar 22 2012

%F a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - _Luce ETIENNE_, May 13 2016

%F Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - _Amiram Eldar_, Aug 12 2022

%p A185138 := proc(n)

%p         if n mod 4 = 0 then

%p                 return n/4*(n-1) ;

%p         elif n mod 2 = 1 then

%p                 return (n-1)*(n+1)/8 ;

%p         else

%p                 return (n-1)*n/2 ;

%p         end if;

%p end proc: # _R. J. Mathar_, Apr 05 2012

%t Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))

%t a = Table[b[n], {n, 1, 52}] (* _Roger L. Bagula_, Mar 14 2012 *)

%t LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* _Harvey P. Dale_, Nov 23 2015 *)

%o (PARI) x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ _G. C. Greubel_, Jun 23 2017

%Y Cf. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

%K nonn,easy,less

%O 0,5

%A _Paul Curtz_, Mar 12 2012