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Numbers of the form k*(k+1)/6 for k = 2 or 3 modulo 6.
6

%I #49 Feb 22 2022 03:37:19

%S 1,2,12,15,35,40,70,77,117,126,176,187,247,260,330,345,425,442,532,

%T 551,651,672,782,805,925,950,1080,1107,1247,1276,1426,1457,1617,1650,

%U 1820,1855,2035,2072,2262,2301,2501,2542,2752,2795,3015,3060,3290,3337,3577,3626

%N Numbers of the form k*(k+1)/6 for k = 2 or 3 modulo 6.

%C Numbers with an odd number of partitions with an extra odd partition; coefficient of z^p in Product_{n >= 1}(1-z^n) has coefficient (-1).

%C n such that the number of partitions of n into distinct parts with an odd number of parts exceed by 1 the number of partitions of n into distinct parts with an even number of parts. [Euler's 1754/55 pentagonal number theorem, see, e.g., the Freitag-Busam reference (in German). This reference is from _Wolfdieter Lang_, Jan 18 2016]

%C In formal power series, A010815=(product(1-x^k),k>0), ranks of coefficients -1. (A001318=ranks of nonzero (1 or -1) in A010815=ranks of odds terms in A000009).

%C Quasipolynomial of order 2. - _Charles R Greathouse IV_, Dec 08 2011

%C Union of A033568 and A033570. - _Ray Chandler_, Dec 09 2011

%D Eberhard Freitag and Rolf Busam, Funktionentheorie 1, Springer, Vierte Auflage, 2006, p. 410.

%H Robert Israel, <a href="/A036499/b036499.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F a(n) = (3*n*n-5*n+2)/2 + (2*n-1)*(n mod 2). - _Frank Ellermann_, Mar 16 2002

%F G.f.: (1+x+8*x^2+x^3+x^4)/((1-x)^3*(1+x)^2). - _Ray Chandler_, Dec 09 2011

%F Bisection: a(2*k+1) = A001318(1+4*k) = (2*k+1)*(3*k+1) = A033570(k), a(2*(k+1)) = A001318(2+4*k) = (2*k+1)*(3*k+2) = A033568(k+1), k >= 0. - _Wolfdieter Lang_, Jan 18 2016

%F a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>5. - _Wesley Ivan Hurt_, Jan 18 2016

%F From _Amiram Eldar_, Feb 22 2022: (Start)

%F Sum_{n>=1} 1/a(n) = Pi/sqrt(3).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(3) - 4*log(2). (End)

%p seq(seq((6*k+i)*(6*k+i+1)/6,i=2..3),k=0..50); # _Robert Israel_, Jan 18 2016

%t Table[ 1/8*(3 + (-1)^k - 6*k)*(1 + (-1)^k - 2*k), {k, 64} ]

%t LinearRecurrence[{1,2,-2,-1,1},{1,2,12,15,35},50] (* or *)

%t CoefficientList[Series[(1+x+8x^2+x^3+x^4)/((1-x)^3(1+x)^2),{x,0,100}],x] (* or *)

%t Table[(2n+1)(3n+{1,2}),{n,0,24}]//Flatten (* _Ray Chandler_, Dec 09 2011 *)

%o (PARI) a(n)=n*(3*n-5)/2+1+n%2*(2*n-1) \\ _Charles R Greathouse IV_, Dec 08 2011

%o (Magma) [(3*n*n-5*n+2)/2+(2*n-1)*(n mod 2): n in [1..50]]; // _Vincenzo Librandi_, Jan 19 2016

%Y Cf. A000009, A001318, A010815, A033568, A033570, A036498.

%K nonn,easy

%O 1,2

%A _Wouter Meeussen_

%E Better description from Claude Lenormand (claude.lenormand(AT)free.fr), Feb 12 2001

%E Edited by _Ray Chandler_, Dec 09 2011