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

1, 2, 6, 24, 19, -6, -186, -1656, -1647, -1566, -666, 10224, 10211, 10042, 7494, -33456, -33439, -33150, -27642, 82824, 82803, 82362, 72198, -172200, -172175, -171550, -154650, 319200, 319171, 318330, 292230, -543840, -543807, -542718, -504570, 869880

Offset: 1

Author

Wesley Ivan Hurt, Sep 22 2018

Keywords

• sign
• easy

Comments

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.

An alternating version of A319205.

Examples

			a(1) = 1;
a(2) = 1*2 = 2;
a(3) = 1*2*3 = 6;
a(4) = 1*2*3*4 = 24;
a(5) = 1*2*3*4 - 5 = 19;
a(6) = 1*2*3*4 - 5*6 = -6;
a(7) = 1*2*3*4 - 5*6*7 = -186;
a(8) = 1*2*3*4 - 5*6*7*8 = -1656;
a(9) = 1*2*3*4 - 5*6*7*8 + 9 = -1647;
a(10) = 1*2*3*4 - 5*6*7*8 + 9*10 = -1566;
a(11) = 1*2*3*4 - 5*6*7*8 + 9*10*11 = -666;
a(12) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 = 10224;
a(13) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13 = 10211;
a(14) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13*14 = 10042;
a(15) = 1*2*3*4 - 5*6*7*8 + 9*10*11*12 - 13*14*15 = 7494; etc.
		

Crossrefs

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).

Cf. A319205.

Programs

Mathematica

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 *)

Formula

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)).