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 A257942 a(n) = (n+1)*(n+2)/A014695(n+1), where A014695 is repeat (1, 2, 2, 1). 1
 1, 3, 12, 20, 15, 21, 56, 72, 45, 55, 132, 156, 91, 105, 240, 272, 153, 171, 380, 420, 231, 253, 552, 600, 325, 351, 756, 812, 435, 465, 992, 1056, 561, 595, 1260, 1332, 703, 741, 1560, 1640, 861, 903, 1892, 1980, 1035, 1081, 2256, 2352, 1225, 1275, 2652 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Consider, for n >= 0, a sequence s(n). A useful transform is wi(n) = s(0), s(2), s(3), ..., i.e., s(n) without s(1). For s(n) = 1/(n+1), wi(n)= 1, 1/3, 1/4, 1/5, ..., whose inverse binomial transform is f(n) = 1, -2/3, 7/12, -11/20, 8/15, -11/21, 29/56, -37/72, 23/45, -28/55, 67/132, -79/156, 46/91, -53/105, 121/240, -137/272, ... The denominator of f(n) is a(n), for n >= 0. If the numerator of f(n) is b(n), then it can be seen that b(n+1) = -(-1)^n* A226089(n). Alternating a(n) - b(n) with a(n) + b(n) yields 0, 1, 5, 9, ... = A160050(n+1). a(4n+1) is linked to the Rydberg-Ritz spectra of hydrogen. h(n) = 0, 0, 1, 1, 4, 4, 3, 3, 8, 8, 5, 5, ... = duplicated A022998(n). A022998(n) is linked to the Balmer series (see A246943(n)). With an initial 0 and offset=0, a(-n) = a(n). Then (a(n+10) - a(n-10))/10 = 1, 6, 10, 7, 9, 22, 26, ... = g(n). a(n) mod 9 is of period 20. LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1). FORMULA a(4n)    = (2*n+1)*(4*n+1). a(4n+1)  = (2*n+1)*(4*n+3). a(4n+2)  = (4*n+3)*(4*n+4). a(4n+3)  = (4*n+4)*(4*n+5). a(n) = A064038(n+2) * (period 4: repeat 1, 1, 4, 4). From Colin Barker, Jul 14 2015: (Start) a(n) = (-1/8+i/8)*(((-3-i*3)+i*(-i)^n+i^n)*(2+3*n+n^2)) where i=sqrt(-1). G.f.: -(x^6+9*x^4-8*x^3+9*x^2+1) / ((x-1)^3*(x^2+1)^3). (End) a(n) = h(n+2) * A109613(n+1). a(n) = (n+1)*(n+2) * period 4:repeat (1, 1, 2, 2) /2. From Wesley Ivan Hurt, Jul 18 2015: (Start) a(n) = (n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2). a(n) = 3*a(n-1)-6*a(n-2)+10*a(n-3)-12*a(n-4)+12*a(n-5)-10*a(n-6)+6*a(n-7)-3*a(n-8)+a(n-9), n>9. (End) MAPLE A257942:=n->(n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2): seq(A257942(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2015 MATHEMATICA CoefficientList[Series[-(x^6 + 9 x^4 - 8 x^3 + 9 x^2 + 1)/((x - 1)^3 (x^2 + 1)^3), {x, 0, 50}], x] (* Michael De Vlieger, Jul 14 2015 *) (* Using inverse binomial transform *) s[0]=1; s[n_] := 1/(n+2); f[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*s[k], {k, 0, n}]; Table[f[n]//Denominator, {n, 0, 50}] (* Jean-François Alcover, Jul 14 2015 *) LinearRecurrence[{3, -6, 10, -12, 12, -10, 6, -3, 1}, {1, 3, 12, 20, 15, 21, 56, 72, 45}, 55] (* Vincenzo Librandi, Jul 15 2015 *) PROG (PARI) Vec(-(x^6+9*x^4-8*x^3+9*x^2+1)/((x-1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jul 14 2015 (PARI) a(n)=(n+1)*(n+2)/if(n%4<2, 2, 1) \\ Charles R Greathouse IV, Jul 14 2015 CROSSREFS Cf. A002378(n+1), A014695(n+1)=A130658(n+2), A014634, A033567(n+1), A104188(n+1), 4*A007742(n+1), A160050 (in A226089), A022998, A109613, A000217(n+1), A246943. Sequence in context: A117554 A049714 A199129 * A063244 A281813 A063102 Adjacent sequences:  A257939 A257940 A257941 * A257943 A257944 A257945 KEYWORD nonn,easy AUTHOR Paul Curtz, Jul 14 2015 STATUS approved

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