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Real part of a complex operation analogous to the factorials.
1

%I #39 Jan 26 2025 09:05:33

%S 1,0,-12,-96,-480,0,40320,645120,5806080,0,-1277337600,-30656102400,

%T -398529331200,0,167382319104000,5356234211328000,91055981592576000,0,

%U -62282291409321984000,-2491291656372879360000,-52317124783830466560000,0

%N Real part of a complex operation analogous to the factorials.

%C Conjecture (1): Current nonzero term divides next nonzero term; as an operation, getting a new sequence: 12, 8, 5, 84, 16, 9, 220, 24, 13, ...; using unsigned terms). Example: a(5) = -480 and a(7)= 40320, the next nonzero term in the sequence. Then 40320/480 = 84. Conjecture (2): If the term in (12, 8, 5, ...) is generated from a(n)/a(n-1) or a(n)/a(n-2), then n divides each (12, 8, 5, ...). Example: a(11) = -1277337600 and the previous nonzero term = a(9) = 5806080. Then a(11)/a(9) = 220 and 11 divides 220: 220/11 = 20.

%C From _Hieronymus Fischer_, Nov 20 2011: (Start)

%C Due to the statements (1) and (2) below a(n) divides a(n+k) for all k>=0, only provided both the residues of n (mod 8) and (n+k) (mod 8) are 0, 1, 3, 4, 5 or 7. As a consequence it follows that conjecture (1) plainly holds true (set k=1).

%C From statements (1) and (2) it further follows that n! divides a(n) (for n with a(n)<>0) and therefore conjecture(2) also holds true.

%C Statement(1): |a(n)| = 2^((n-1)/2)*n! for odd n (or n==1, 3, 5 or 7 (mod 8) respectively).

%C Statement(2): |a(n)| = 2^(n/2)*n! for n==0 or 4 (mod 8). (End)

%F Let a(1) = 1, real part of (1 + i) = k(1); then k(n) = (n + ni) * k(n-1), n>1. a(n) = real part of each k(n).

%F From _Hieronymus Fischer_, Nov 20 2011: (Start)

%F a(n) = 2^(n/2)*n!*((-1)^floor((n+1)/4)*(1-(-1)^n)/sqrt(8)+(-1)^floor((n+2)/4)*(1-(-1)^floor((n+2)/2))*(1+(-1)^n)/4).

%F a(n) = 2^(n/2)*n!*cos(n*Pi/4).

%F a(n) = 2^(n/2)*n!*Re(exp(i*n*Pi/4)).

%F a(n) = 0 iff n==2 (mod 8) or n==6 (mod 8);

%F a(n) > 0 iff n==0 (mod 8) or n==1 (mod 8) or n==7 (mod 8);

%F a(n) < 0 iff n==3 (mod 8) or n==4 (mod 8) or n==5 (mod 8).

%F |a(n)| = 2^floor(n/2)*n! for n==0,1,3,4,5 or 7 (mod 8).

%F Recursion: a(1)=1, a(n+1) = (n+1)*(a(n)-b(n)), where b(n) is the appropriately defined imaginary part with b(1)=1, b(n+1) = (n+1)*(a(n)+b(n)). (End)

%F E.g.f.: x*(1 - 2*x)/(1 - 2*x + 2*x^2). - _Stefano Spezia_, Jan 26 2025

%e a(3) = -12 since a(1) = (1 + i); (2 + 2i)* (1 + i) = (0 +4i); (3 + 3i)*(0 + 4i) = (-12 + 12i).

%t a[1] = 1 + I; a[n_] := a[n] = a[n - 1]*(n + n*I); Table[Re[a[n]], {n, 22}] (* Or *)

%t f[n_] := Fold[(#2 + #2*I)*#1 &, 1 + I, Range@n + 1]; Table[ Re[f[n]], {n, 0, 22}] (* _Robert G. Wilson v_ *)

%o (Magma) [Integers()!Real((1+Sqrt(-1))^n)*Factorial(n): n in [1..22]]; // _Bruno Berselli_, Nov 21 2011

%o (Maxima) makelist(expand(((1-%i)^n+(1+%i)^n)*n!/2), n,1,22); /* _Bruno Berselli_, Nov 21 2011 */

%o (PARI) my(x='x + O('x^23)); Vec(serlaplace(x*(1 - 2*x)/(1 - 2*x + 2*x^2))) \\ _Stefano Spezia_, Jan 26 2025

%K sign

%O 1,3

%A _Gary W. Adamson_, Aug 12 2006

%E More terms from _Robert G. Wilson v_, Aug 17 2006