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A319544 a(n) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13*14*15*16 + ... - (up to n).

%I A319544 #13 Oct 05 2018 08:01:54
%S A319544 1,2,6,24,19,-6,-186,-1656,-1647,-1566,-666,10224,10211,10042,7494,
%T A319544 -33456,-33439,-33150,-27642,82824,82803,82362,72198,-172200,-172175,
%U A319544 -171550,-154650,319200,319171,318330,292230,-543840,-543807,-542718,-504570,869880
%N A319544 a(n) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13*14*15*16 + ... - (up to n).
%C A319544 In general, for alternating sequences that multiply the first k natural numbers, and subtract/add the products of the next k natural numbers (preserving the order of operations up to n), we have a(n) = (-1)^floor(n/k) * Sum_{i=1..k-1} (1-sign((n-i) mod k)) * (Product_{j=1..i} (n-j+1)) + Sum_{i=1..n} (-1)^(floor(i/k)+1) * (1-sign(i mod k)) * (Product_{j=1..k} (i-j+1)). Here k=4.
%C A319544 An alternating version of A319205.
%F A319544 a(n) = (-1)^floor(n/4) * Sum_{i=1..3} (1-sign((n-i) mod 4)) * (Product_{j=1..i} (n-j+1)) + Sum_{i=1..n} (-1)^(floor(i/4)+1) * (1-sign(i mod 4)) * (Product_{j=1..4} (i-j+1)).
%e A319544 a(1) = 1;
%e A319544 a(2) = 1*2 = 2;
%e A319544 a(3) = 1*2*3 = 6;
%e A319544 a(4) = 1*2*3*4 = 24;
%e A319544 a(5) = 1*2*3*4 - 5 = 19;
%e A319544 a(6) = 1*2*3*4 - 5*6 = -6;
%e A319544 a(7) = 1*2*3*4 - 5*6*7 = -186;
%e A319544 a(8) = 1*2*3*4 - 5*6*7*8 = -1656;
%e A319544 a(9) = 1*2*3*4 - 5*6*7*8 + 9 = -1647;
%e A319544 a(10) = 1*2*3*4 - 5*6*7*8 + 9*10 = -1566;
%e A319544 a(11) = 1*2*3*4 - 5*6*7*8 + 9*10*11 = -666;
%e A319544 a(12) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 = 10224;
%e A319544 a(13) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13 = 10211;
%e A319544 a(14) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13*14 = 10042;
%e A319544 a(15) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13*14*15 = 7494; etc.
%t A319544 a[n_]:=(-1)^Floor[n/4]*Sum[(1-Sign[Mod[n-i,4]])*Product[n-j+1,{j,1,i}],{i,1,3}]+Sum[(-1)^(Floor[i/4]+1)*(1-Sign[Mod[i,4]])*Product[i-j+1,{j,1,3}],{i,1,n}]; Array[a, 30] (* _Stefano Spezia_, Sep 23 2018 *)
%Y A319544 For similar sequences, see: A001057 (k=1), A319373 (k=2), A319543 (k=3), this sequence (k=4), A319545 (k=5), A319546 (k=6), A319547 (k=7), A319549 (k=8), A319550 (k=9), A319551 (k=10).
%Y A319544 Cf. A319205.
%K A319544 sign,easy
%O A319544 1,2
%A A319544 _Wesley Ivan Hurt_, Sep 22 2018