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 A191567 Four interlaced 2nd order polynomials: a(4*k) = k*(1+2*k); a(1+2*k) = 2*(1+2*k)*(3+2*k); a(2+4*k) = 4*(1+k)*(1+2*k). 3
 0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286, 21, 390, 112, 510, 36, 646, 180, 798, 55, 966, 264, 1150, 78, 1350, 364, 1566, 105, 1798, 480, 2046, 136, 2310, 612, 2590, 171, 2886, 760, 3198, 210, 3526, 924, 3870, 253, 4230, 1104, 4606, 300, 4998, 1300, 5406, 351 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) = T(0,n) and differences T(n,k) = T(n-1,k+1) - T(n-1,k) define the array 0,   6,  4,  30,    3,  70,   24,  126,   10,  198,   60,  286,   21,  390,  .. 6,  -2, 26, -27,   67, -46,  102, -116,  188, -138,  226, -265,  369, -278, .. -8, 28 -53,  94, -113, 148, -218,  304, -326,  364, -491,  634, -647,  676, ... T(3,n) mod 9 is the sequence 1, 1, 1, 4, 4, 4, 7, 7, 7, 4, 4, 4 (and periodically repeated with period 12). A064680(2+n) divides a(n), where b(n) = a(n)/A064680(2+n) = 0, 1, 2, 3, 1, 5, 6, 7, 2,... for n>=0, obeys b(4*k) = k and has recurrence b(n) = 2*b(n-4) - b(n-8). LINKS Vincenzo Librandi, 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(n) = A061037(n+2) + A181318(n). - Paul Curtz, Jul 19 2011 a(n) = A060819(n) * A145979(n). - Paul Curtz, Sep 06 2011 G.f.: x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10) /( (1-x)^3 *(1+x)^3 *(1+x^2)^3 ). - R. J. Mathar, Jun 17 2011 Let BEB(n) = a(n)/A061038(n+2) = A060819(n)/A145979(n). Then (BEB(n))^2 = A181318(n)/A061038(n+2) = BEB(n) - A061037(n+2)/A061038(n+2). - Paul Curtz, Jul 19 2011, index corrected by R. J. Mathar, Sep 09 2011 From Luce ETIENNE, Apr 18 2017: (Start) a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*((-1)^((2*n + 1 - (-1)^n)/4) + (-1)^((2*n - 1 + (-1)^n)/4)))/32. a(n) = n*(n+2)*(37-27*cos(n*Pi) - 6*cos(n*Pi/2))/32. a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*(i^n + (-i)^n))/32, where i=sqrt(-1). (End) MATHEMATICA Table[Which[OddQ@ n, 2 (1 + 2 #) (3 + 2 #) &[(n - 1)/2], Mod[n, 4] == 0, # (1 + 2 #) &[n/4], True, 4 (1 + #) (1 + 2 #) &[(n - 2)/4]], {n, 0, 60}] (* or *) CoefficientList[Series[x(6 +4x +30x^2 +3x^3 +52x^4 +12x^5 +36x^6 +x^7 +6x^8 -2x^10)/((1-x)^3*(1+x)^3*(1+x^2)^3), {x, 0, 60}], x] (* Michael De Vlieger, Apr 22 2017 *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286}, 80] (* Vincenzo Librandi, Apr 23 2017 *) PROG (MAGMA) I:=[0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286]; [n le 12 select I[n] else 3*Self(n-4)-3*Self(n-8)+Self(n-12): n in [1..60]]; // Vincenzo Librandi, Apr 23 2017 (PARI) m=60; v=concat([0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286], vector(m-12)); for(n=13, m, v[n]=3*v[n-4]-3*v[n-8]+v[n-12]); v \\ G. C. Greubel, Feb 26 2019 (Sage) (x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10)/((1-x)^3 *(1+x)^3*(1+x^2)^3 )).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019 (GAP) a:=[0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286];; for n in [13..60] do a[n]:= 3*a[n-4]-3*a[n-8]+a[n-12]; od; a; # G. C. Greubel, Feb 26 2019 CROSSREFS Cf. A014105, A000466, A000384, A177427. Sequence in context: A211945 A121682 A237425 * A274707 A163934 A163939 Adjacent sequences:  A191564 A191565 A191566 * A191568 A191569 A191570 KEYWORD nonn,easy AUTHOR Paul Curtz, Jun 12 2011 STATUS approved

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Last modified June 25 05:41 EDT 2019. Contains 324346 sequences. (Running on oeis4.)