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A185138 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). 2
0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

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

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.

(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).

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

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

0  1  1   3   3   15...

0  3  6  45  21   60...

0 15 15  60  55  325...

0 14 28 231 105  315...

0 45 45 189 171 1035...

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

0 1 1  3  3  15  6  14   k=0

0 1 2 15  7  20 15  77   k=1

0 3 3 12 11  65 24  63   k=2

0 2 4 33 15  45 33 175   k=3

0 5 5 21 19 115 42 112   k=4

0 3 6 51 23  70 51 273   k=5

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

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.

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).

LINKS

G. C. Greubel, 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).

FORMULA

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

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

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

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

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

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

MAPLE

A185138 := proc(n)

        if n mod 4 = 0 then

                return n/4*(n-1) ;

        elif n mod 2 = 1 then

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

        else

                return (n-1)*n/2 ;

        end if;

end proc: # R. J. Mathar, Apr 05 2012

MATHEMATICA

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)))

a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)

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 *)

PROG

(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

CROSSREFS

Cf. A033991, A000217, A014634.

Sequence in context: A260078 A163590 A114320 * A285947 A285843 A285822

Adjacent sequences:  A185135 A185136 A185137 * A185139 A185140 A185141

KEYWORD

nonn,easy,less

AUTHOR

Paul Curtz, Mar 12 2012

STATUS

approved

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Last modified July 21 04:40 EDT 2019. Contains 325189 sequences. (Running on oeis4.)