|
%I #7 Sep 12 2019 04:44:45
%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.
%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).
%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 a(n) = floor((n+1)/4)*floor(n/2); a(2n+1) = A093005(n); a(2n) = A093353(n-1) = floor(n/2)*n. - _M. F. Hasler_, May 03 2009
%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}.
%o (PARI) A159915(n)=polcoeff((1-x+x^2)/(1-3*x+5*x^2-7*x^3+7*x^4-5*x^5+3*x^6-x^7)+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
%K easy,nonn
%O 0,5
%A _M. F. Hasler_, May 01 2009, May 03 2009
|
