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a(n) = floor((n+1)/4)*floor(n/2).
2

%I #12 Sep 18 2024 02:28:04

%S 0,0,0,1,2,2,3,6,8,8,10,15,18,18,21,28,32,32,36,45,50,50,55,66,72,72,

%T 78,91,98,98,105,120,128,128,136,153,162,162,171,190,200,200,210,231,

%U 242,242,253,276,288,288,300,325,338,338,351,378,392,392,406,435,450,450

%N a(n) = floor((n+1)/4)*floor(n/2).

%C Half the number of (n-2)-element subsets of {1,...,n} with odd sum of the elements.

%C This is half the antepenultimate column of A159916, cf. formula.

%C The number of subsets of {1,...,n} with n-2 elements, adding up to an odd integer, is always even (cf. examples), so we divide it by 2.

%C We prefer to include a(0)=a(1)=a(2)=0, even if it might seem more natural to start only at n=2 or n=3.

%C From the rational g.f. it can be seen that the sequence is a linear recurrence with constant coefficients (3,-5,7,-7,5,-3,1) of order 7.

%C A quasipolynomial of order 4 and degree 2. - _Charles R Greathouse IV_, Sep 18 2024

%H G. C. Greubel, <a href="/A159915/b159915.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (3,-5,7,-7,5,-3,1).

%F G.f.: x^3*(1 - x + x^2)/(1 - 3*x + 5*x^2 - 7*x^3 + 7*x^4 - 5*x^5 + 3*x^6 - x^7) = x^3*(1-x+x^2)/((1-x)^3*(1+x^2)^2).

%F a(n) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7) for n > 7.

%F For n > 2, a(n) = A159916(n*(n-1)/2 + n - 2)/2 = T(n,n-2)/2 as defined there.

%F From _M. F. Hasler_, May 03 2009: (Start)

%F a(n) = floor((n+1)/4)*floor(n/2).

%F a(2n+1) = A093005(n).

%F a(2n) = A093353(n-1) = floor(n/2)*n. (End)

%F a(n) ~ n^2/8. - _Charles R Greathouse IV_, Sep 18 2024

%e a(0)=a(1)=0 since there are no subsets with -2 or -1 elements.

%e a(2)=0 since the sum of the elements of a 0-element subset is zero.

%e a(3)=1 since for n=3 we have two singleton subsets of {1,2,3}, {1} and {3}, with odd sum of elements.

%e a(4)=2 since for n=4 we have four 2-element subsets of {1,2,3,4} with odd sum: {1,2}, {2,3}, {1,4}, {3,4}.

%t Table[Floor[(n+1)/4]*Floor[n/2], {n,0,70}] (* _G. C. Greubel_, Sep 18 2024 *)

%o (PARI) A159915(n)= polcoeff( (1-x+x^2)/((1-x)^3*(1+x^2)^2) + O(x^(n-2)), n-3);

%o a(n,t=[0,0,0,1,2,2,3],c=[1,-3,5,-7,7,-5,3]~)=while(n-->5,t=concat(vecextract(t,"^1"),t*c));t[n+2] /* Note: a(n+1,[0,0,0,0,1,2,2]) gives the same result as a(n) */

%o (PARI) A159915(n)=(n+1)\4*(n\2) \\ _M. F. Hasler_, May 03 2009

%o (Magma)

%o A159915:= func< n | Floor((n+1)/4)*Floor(n/2) >;

%o [A159915(n): n in [0..70]]; // _G. C. Greubel_, Sep 18 2024

%o (SageMath)

%o def A159915(n): return ((n+1)//4)*(n//2)

%o [A159915(n) for n in range(71)] # _G. C. Greubel_, Sep 18 2024

%Y Cf. A093005, A093353, A159916.

%K easy,nonn

%O 0,5

%A _M. F. Hasler_, May 01 2009, May 03 2009