cp's OEIS Frontend

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

%I A377825 #32 Dec 25 2024 04:26:52
%S A377825 1,2,3,24,30,720,630,40320,22680,3628800,1247400,479001600,97297200,
%T A377825 87178291200,10216206000,20922789888000,1389404016000,
%U A377825 6402373705728000,237588086736000,2432902008176640000,49893498214560000,1124000727777607680000,12623055048283680000
%N A377825 Number of distinct permutations of the terms of the n-th row of Pascal's triangle with alternating signs.
%C A377825 Note that for any given n, there are n+1 terms in that row.
%H A377825 Paolo Xausa, <a href="/A377825/b377825.txt">Table of n, a(n) for n = 0..400</a>
%F A377825 a(n) = (n+1)! / (2^((n*(1+(-1)^n)) / 4)).
%F A377825 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 A377825 a(n) = (n+1)!/A072345(n-1) for n > 0. - _Stefano Spezia_, Nov 09 2024
%F A377825 Sum_{n>=0} 1/a(n) = cosh(1) + sinh(sqrt(2))/sqrt(2) - 1. - _Amiram Eldar_, Dec 25 2024
%e A377825 For n = 0, a(0) = 1 since there is just one term.
%e A377825 For n = 1, the signed row terms are {1, -1} so a(1) = 2 permutations.
%e A377825 For n = 2, the signed row terms are {1, -2, 1} which have only a(2) = 3 distinct permutations.
%e A377825 For n = 3, the signed row terms are {1, -3, 3, -1} which have a(3) = 24 permutations.
%p A377825 seq((n+1)! / (2^((n*(1+(-1)^n)) / 4)), n=0..22); # _Georg Fischer_, Dec 19 2024
%t A377825 A377825[n_] := (n+1)!/2^((n*(1 + (-1)^n))/4); Array[A377825, 25, 0] (* _Paolo Xausa_, Dec 20 2024 *)
%Y A377825 Bisections are: A007019, A010050.
%Y A377825 Cf. A007318, A072345, A142150.
%K A377825 nonn,easy
%O A377825 0,2
%A A377825 _Ryan Jean_, Nov 08 2024
%E A377825 a(22) corrected by _Georg Fischer_, Dec 19 2024