A319545 a(n) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12*13*14*15 - ... + (up to n).

1, 2, 6, 24, 120, 114, 78, -216, -2904, -30120, -30109, -29988, -28404, -6096, 330240, 330224, 329968, 325344, 237216, -1530240, -1530219, -1529778, -1519614, -1275216, 4845360, 4845334, 4844658, 4825704, 4275336, -12255360, -12255329, -12254368, -12222624

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=5.

An alternating version of A319206.

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 = 120;
a(6) = 1*2*3*4*5 - 6 = 114;
a(7) = 1*2*3*4*5 - 6*7 = 78;
a(8) = 1*2*3*4*5 - 6*7*8 = -216;
a(9) = 1*2*3*4*5 - 6*7*8*9 = -2904;
a(10) = 1*2*3*4*5 - 6*7*8*9*10 = -30120;
a(11) = 1*2*3*4*5 - 6*7*8*9*10 + 11 = -30109;
a(12) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12 = -29988;
a(13) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12*13 = -28404;
a(14) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12*13*14 = -6096;
a(15) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12*13*14*15 = 330240;
a(16) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12*13*14*15 - 16 = 330224;
a(17) = 1*2*3*4*5 - 6*7*8*9*10 + 11*12*13*14*15 - 16*17 = 329968; etc.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

For similar sequences, see: A001057 (k=1), A319373 (k=2), A319543 (k=3), A319544 (k=4), this sequence (k=5), A319546 (k=6), A319547 (k=7), A319549 (k=8), A319550 (k=9), A319551 (k=10).

Cf. A319206.

Programs

Mathematica

a[n_]:=(-1)^Floor[n/5]*Sum[(1-Sign[Mod[n-i,5]])*Product[n-j+1,{j,1,i}],{i,1,4}]+Sum[(-1)^(Floor[i/5]+1)*(1-Sign[Mod[i,5]])*Product[i-j+1,{j,1,4}],{i,1,n}]; Array[a, 30] (* Stefano Spezia, Sep 23 2018 *)
Table[Total[Times@@@Partition[Riffle[Times@@@Partition[Range[n],UpTo[5]],{1,-1},{2,-1,2}],2]],{n,40}] (* Harvey P. Dale, Mar 30 2023 *)

Formula

a(n) = (-1)^floor(n/5) * Sum_{i=1..4} (1-sign((n-i) mod 5)) * (Product_{j=1..i} (n-j+1)) + Sum_{i=1..n} (-1)^(floor(i/5)+1) * (1-sign(i mod 5)) * (Product_{j=1..5} (i-j+1)).