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 A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1). 9
 0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS From Paul Curtz, Mar 26 2011: (Start) Successive A026741(n) * A026741(n+p):   p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),   p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,   p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,   p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,   p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,   p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,   p=6:  0, 7,  4, 27, 10, 55, 18,   p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,   p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),   p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End) The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016 Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 John M. Campbell, An Integral Representation of KekulĂ© Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011. Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1). FORMULA From R. J. Mathar, Jan 22 2011: (Start) G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3). a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6). a(n) = n^2*(5 - 3*(-1)^n)/8. (End) a(n) = A026741(n)^2. a(2n) = A000290(n); a(2n+1) = A016754(n). a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011 4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011 a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018 From Peter Bala, Feb 19 2019: (Start) a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2. a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n. O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3. Cf. A181318. (End) MAPLE a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016 MATHEMATICA LinearRecurrence[{0, 3, 0, -3, 0, 1}, {0, 1, 1, 9, 4, 25}, 60] (* Harvey P. Dale, May 14 2011 *) f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *) CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x, 0, 60}], x] (* Vincenzo Librandi, Jul 10 2016 *) PROG (PARI) concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016 (PARI) a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018 (MAGMA) I:=[0, 1, 1, 9, 4, 25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016 (Sage) (x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019 CROSSREFS Cf. A040001, A168068, A181318. Sequence in context: A100077 A317739 A174679 * A173536 A014717 A104728 Adjacent sequences:  A168074 A168075 A168076 * A168078 A168079 A168080 KEYWORD nonn,easy,mult AUTHOR Paul Curtz, Nov 18 2009 STATUS approved

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Last modified May 19 17:48 EDT 2019. Contains 323395 sequences. (Running on oeis4.)