On the nearest integer to $\zeta^{(k)}(1-1/B),B \ge 2$

Question

Let $k \ge 1,B \ge 2$ integers and $\zeta^{(k)}(s)$ denote the $k$-th derivative of Riemann zeta function. For real $x$, let $[x]$ denote the nearest integer to $x$.

Conjecture 1: For all $n \ge 1,[\zeta^{(n)}(1/2)]= -2^{n+1} n!$

Conjecture 2: For all $k \ge 1,B \ge 2$ we have $[\zeta^{(k)}(1-1/B)]= -B^{k+1}\cdot k!$ and $\zeta^{(k)}(1-1/B)$ is very close to integer.

sagemath code per request:

def zetania2(LK=10,prec=100,bb=[3,5]):
    """
    zeta^(k)(1-1/B), near to integers
    """
    import mpmath
    mpmath.mp.pretty=1
    mpmath.mp.dps=prec
    pre2=10
    for B in bb:
        l0=[]
        lerr=[]
        for k in range(1,LK+1):
            T=mpmath.zeta(1-1/mpmath.mpf(B),derivative=k)
            N=mpmath.nint(T)
            N2= -B**(k+1) * factorial(k)
            l0 += [N-N2]
        print(B,l0)

Answer 1

Both conjectures are false. Suppose your conjectured identity holds for all $k\geq 1$ and some $B\geq 2$. It is known that $$ f(s)=\zeta(s)-\frac{1}{s-1} $$ is an entire function. Your identity for fixed $B$ implies, for instance, that $|f^{(k)}(1-1/B)|\leq 1$. By Taylor expansion, from this we get for all $s\in \mathbb C$ that $$ |f(s)|=\left|\sum_{k=0}^{+\infty}\frac{f^{(k)}(1-1/B)}{k!}(s-1+1/B)^k\right|\leq |f(1/B)|+e^{|s-1+1/B|}. $$ The last bound cannot hold, because $f(-n)$ grows more than exponentially for odd integer $n\to+\infty$.

UPDATE: as for the numerical evidence, $k=41$ seems to be the least value of $k$ with $$ |\zeta^{(k)}(1-1/B)+k!B^{k+1}|>1 $$ for $B=2,3,4$. This was computed using the function $\it{derivnum}$ in Pari/GP.