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A129194 a(n) = (n/2)^2*(3 - (-1)^n). 14

%I #93 Feb 03 2024 10:16:00

%S 0,1,2,9,8,25,18,49,32,81,50,121,72,169,98,225,128,289,162,361,200,

%T 441,242,529,288,625,338,729,392,841,450,961,512,1089,578,1225,648,

%U 1369,722,1521,800,1681,882,1849,968,2025,1058,2209,1152,2401,1250,2601,1352

%N a(n) = (n/2)^2*(3 - (-1)^n).

%C The numerator of the integral is 2,1,2,1,2,1,...; the moments of the integral are 2/(n+1)^2. See 2nd formula.

%C The sequence alternates between twice a square and an odd square, A001105(n) and A016754(n).

%C Partial sums of the positive elements give the absolute values of A122576. - _Omar E. Pol_, Aug 22 2011

%C Partial sums of the positive elements give A212760. - _Omar E. Pol_, Dec 28 2013

%C Conjecture: denominator of 4/n - 2/n^2. - _Wesley Ivan Hurt_, Jul 11 2016

%C Multiplicative because both A000290 and A040001 are. - _Andrew Howroyd_, Jul 25 2018

%D G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 1, Sect. 7, Problem 73.

%H Vincenzo Librandi, <a href="/A129194/b129194.txt">Table of n, a(n) for n = 0..960</a>

%H Olivier Bordelles, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Bordelles/bord21.html">A Multidimensional Cesaro Type Identity and Applications</a>, J.

%H Int. Seq. 18 (2015) # 15.3.7.

%H John M. Campbell, <a href="http://arxiv.org/abs/1105.3399">An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences</a>, arXiv preprint arXiv:1105.3399 [math.GM], 2011.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1).

%F G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4)/(1-x^2)^3.

%F a(n+1) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*n*t)(-((Pi-t)/i)^2)), i=sqrt(-1).

%F a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 5. - _Paul Curtz_, Mar 07 2011

%F a(n) is the numerator of the coefficient of x^4 in the Maclaurin expansion of exp(-n*x^2). - _Francesco Daddi_, Aug 04 2011

%F O.g.f. as a Lambert series: x*Sum_{n >= 1} J_2(n)*x^n/(1 + x^n), where J_2(n) denotes the Jordan totient function A007434(n). See Pólya and Szegő. - _Peter Bala_, Dec 28 2013

%F From _Ilya Gutkovskiy_, Jul 11 2016: (Start)

%F E.g.f.: x*((2*x + 1)*sinh(x) + (x + 2)*cosh(x))/2.

%F Sum_{n>=1} 1/a(n) = 5*Pi^2/24. [corrected by _Amiram Eldar_, Sep 11 2022] (End)

%F a(n) = A000290(n) / A040001(n). - _Andrew Howroyd_, Jul 25 2018

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - _Amiram Eldar_, Sep 11 2022

%F From _Peter Bala_, Jan 16 2024: (Start)

%F a(n) = Sum_{1 <= i, j <= n} (-1)^(1 + gcd(i,j,n)) = Sum_{d | n} (-1)^(d+1) * J_2(n/d), that is, the Dirichlet convolution of the pair of multiplicative functions f(n) = (-1)^(n+1) and the Jordan totient function J_2(n) = A007434(n). Hence this sequence is multiplicative. Cf. A193356 and A309337.

%F Dirichlet g.f.: (1 - 2/2^s)*zeta(s-2). (End)

%F a(n) = Sum_{1 <= i, j <= n} (-1)^(n + gcd(i, n)*gcd(j, n)) = Sum_{d|n, e|n} (-1)^(n+e*d) * phi(n/d)*phi(n/e). - _Peter Bala_, Jan 22 2024

%p A129194:=n->n^2*(3-(-1)^n)/4: seq(A129194(n), n=0..80); # _Wesley Ivan Hurt_, Jul 11 2016

%t Table[n^2*(3-(-1)^n)/4, {n,0,60}] (* _Wesley Ivan Hurt_, Jul 11 2016 *)

%t LinearRecurrence[{0,3,0,-3,0,1},{0,1,2,9,8,25},60] (* _Harvey P. Dale_, Dec 27 2023 *)

%o (Magma) [n^2*(3-(-1)^n)/4: n in [0..60]]; // _Vincenzo Librandi_, Apr 26 2011

%o (PARI) a(n) = lcm(2, n^2)/2; \\ _Andrew Howroyd_, Jul 25 2018

%o (SageMath) [n^2*(1+(n%2))/2 for n in range(61)] # _G. C. Greubel_, Apr 04 2023

%Y Cf. A000290, A001105, A007434, A010713, A016742, A016754, A040001.

%Y Cf. A061038, A061040, A061050, A105398, A129204, A152020, A222171, A309337.

%K easy,frac,nonn,mult

%O 0,3

%A _Paul Barry_, Apr 02 2007

%E More terms from _Michel Marcus_, Dec 28 2013

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Last modified March 28 08:22 EDT 2024. Contains 371236 sequences. (Running on oeis4.)