A377825 Number of distinct permutations of the terms of the n-th row of Pascal's triangle with alternating signs.

1, 2, 3, 24, 30, 720, 630, 40320, 22680, 3628800, 1247400, 479001600, 97297200, 87178291200, 10216206000, 20922789888000, 1389404016000, 6402373705728000, 237588086736000, 2432902008176640000, 49893498214560000, 1124000727777607680000, 12623055048283680000

Offset

0,2

Comments

Note that for any given n, there are n+1 terms in that row.

Links

Paolo Xausa, Table of n, a(n) for n = 0..400

Formula

a(n) = (n+1)! / (2^((n*(1+(-1)^n)) / 4)).

E.g.f.: 2*(x^6+x^5-4*x^3-3*x^2+4*x+2)/((x-1)^2*(x+1)^2*(x^2-2)^2). - Alois P. Heinz, Nov 09 2024

a(n) = (n+1)!/A072345(n-1) for n > 0. - Stefano Spezia, Nov 09 2024

Sum_{n>=0} 1/a(n) = cosh(1) + sinh(sqrt(2))/sqrt(2) - 1. - Amiram Eldar, Dec 25 2024

Example

For n = 0, a(0) = 1 since there is just one term.
For n = 1, the signed row terms are {1, -1} so a(1) = 2 permutations.
For n = 2, the signed row terms are {1, -2, 1} which have only a(2) = 3 distinct permutations.
For n = 3, the signed row terms are {1, -3, 3, -1} which have a(3) = 24 permutations.

Maple

seq((n+1)! / (2^((n*(1+(-1)^n)) / 4)), n=0..22); # Georg Fischer, Dec 19 2024

Mathematica

A377825[n_] := (n+1)!/2^((n*(1 + (-1)^n))/4); Array[A377825, 25, 0] (* Paolo Xausa, Dec 20 2024 *)

Crossrefs

Bisections are: A007019, A010050.

Cf. A007318, A072345, A142150.

Sequence in context: A032811 A092049 A257789 * A336616 A354278 A061778

Adjacent sequences: A377821 A377823 A377824 * A377826 A377827 A377828

Keyword

nonn,easy

Author

Ryan Jean, Nov 08 2024

Extensions

a(22) corrected by Georg Fischer, Dec 19 2024

Status

approved