A105750 Real part of Product_{k = 0..n} (1 + k*i), i = sqrt(-1).

1, 1, -1, -10, -10, 190, 730, -6620, -55900, 365300, 5864300, -28269800, -839594600, 2691559000, 159300557000, -238131478000, -38894192662000, -15194495654000, 11911522255750000, 29697351895900000, -4477959179352100000, -21683886333440500000, 2029107997508660900000

Offset: 0

Author

Paul Barry, Apr 18 2005

Keywords

• easy
• sign

Comments

Define u(n) as in A220448 and set f(n) = u(n)*f(n-1) for n >= 2, with f(1)=1 (this defines A220449). Then a(0)=1; a(n) = (-1)^(n+1)*f(n) for n >= 1. - N. J. A. Sloane, Dec 22 2012

From Peter Bala, Jun 03 2023: (Start)

Moll (2012) studied the prime divisors of the terms of A105750 and divided the primes into three classes.

Type 1: primes p that do not divide any element of the sequence {a(n)}. The first few type 1 primes appear to be {3, 7, 11, 23, 31, 47, 59}.

Type 2: primes p such that the p-adic valuation v_p(a(n)) has asymptotically linear behavior. The first few type 2 primes appear to be {2, 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97}.

We conjecture that the set of type 2 primes consists of primes p == 1 (mod 4), equivalently, rational primes that split in the field extension Q(sqrt(-1)) of Q, together with the prime p = 2, which ramifies in Q(sqrt(-1)). See A002144.

Type 3: primes p such that the sequence of p-adic valuations {v_p(a(n)) : n >= 0} exhibits an oscillatory behavior (this phrase is not precisely defined).

We conjecture that the sets of type 1 and type 3 primes taken together consist of primes p == 3 (mod 4), equivalently, rational primes that remain inert in the field extension Q(sqrt(-1)) of Q. See A002145. (End)

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..100
• V. H. Moll, An arithmetic conjecture on a sequence of arctangent sums, 2012, see f_n.

Crossrefs

Cf. A003703, A009454, A105751, A220448, A220449, A363409 - A363416.

Programs

Maple

A105750 := proc(n)
    mul(1-k*I,k=0..n) ;
    Re(%) ;
end proc: # R. J. Mathar, Jan 04 2013

Mathematica

x[n_] := x[n] = If[n == 1, 1, (x[n-1]+n)/(1-n*x[n-1])];
u[n_] := n*x[n-1]-1;
f[n_] := f[n] = If[n == 1, 1, u[n]*f[n-1]];
a[n_] := If[n == 0, 1, (-1)^(n+1)*f[n]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 17 2023, after N. J. A. Sloane *)

Python

from sympy.functions.combinatorial.numbers import stirling
def A105750(n): return sum(stirling(n+1,n+1-(k<<1),kind=1)*(-1 if k&1 else 1) for k in range((n+1>>1)+1)) # Chai Wah Wu, Feb 22 2024

Formula

a(n) = Re( Product_{k = 0..n} (1 - k*i) ).

Conjecture: a(n) -3*a(n-1) +(n^2-n+3)*a(n-2) +(-n^2+4*n-5)*a(n-3)=0. - R. J. Mathar, May 23 2014

From Peter Bala, May 28 2023: (Start)

a(n) = Sum_{k = 0..floor((n + 1)/2)} (-1)^k*|Stirling1(n+1, n-2*k+1)|, where Stirling1(n, k) = A048994(n,k).

(n - 1)*a(n) = (2*n - 1)*a(n-1) - n*(n^2 - 2*n + 2)*a(n-2) with a(0) = a(1) = 1 (see Moll, equation 1.16). Mathar's third-order recurrence above follows easily from this.

a(2*n) = (-1)^n*A009454(2*n+1) for n >= 0.

a(2*n-1) = (-1)^n*A003703(2*n) for n >= 1. (End)

Extensions

Corrected by N. J. A. Sloane, Nov 05 2005