A377825 - OEIS

%I #32 Dec 25 2024 04:26:52

%S 1,2,3,24,30,720,630,40320,22680,3628800,1247400,479001600,97297200,

%T 87178291200,10216206000,20922789888000,1389404016000,

%U 6402373705728000,237588086736000,2432902008176640000,49893498214560000,1124000727777607680000,12623055048283680000

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

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

%H Paolo Xausa, <a href="/A377825/b377825.txt">Table of n, a(n) for n = 0..400</a>

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

%F 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

%F a(n) = (n+1)!/A072345(n-1) for n > 0. - _Stefano Spezia_, Nov 09 2024

%F Sum_{n>=0} 1/a(n) = cosh(1) + sinh(sqrt(2))/sqrt(2) - 1. - _Amiram Eldar_, Dec 25 2024

%e For n = 0, a(0) = 1 since there is just one term.

%e For n = 1, the signed row terms are {1, -1} so a(1) = 2 permutations.

%e For n = 2, the signed row terms are {1, -2, 1} which have only a(2) = 3 distinct permutations.

%e For n = 3, the signed row terms are {1, -3, 3, -1} which have a(3) = 24 permutations.

%p seq((n+1)! / (2^((n*(1+(-1)^n)) / 4)), n=0..22); # _Georg Fischer_, Dec 19 2024

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

%Y Bisections are: A007019, A010050.

%Y Cf. A007318, A072345, A142150.

%K nonn,easy

%O 0,2

%A _Ryan Jean_, Nov 08 2024

%E a(22) corrected by _Georg Fischer_, Dec 19 2024