OFFSET
1,2
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.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
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)).
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Wesley Ivan Hurt, Sep 22 2018
STATUS
approved