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
Examples
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.
Links
- Paolo Xausa, Table of n, a(n) for n = 0..400
Programs
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Maple
seq((n+1)! / (2^((n*(1+(-1)^n)) / 4)), n=0..22); # Georg Fischer, Dec 19 2024
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Mathematica
A377825[n_] := (n+1)!/2^((n*(1 + (-1)^n))/4); Array[A377825, 25, 0] (* Paolo Xausa, Dec 20 2024 *)
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
Extensions
a(22) corrected by Georg Fischer, Dec 19 2024
Comments