Proof. Write \(N = k + 1\), with \(N \geq 4\) being even. For an integer \[\label{srange} s \in \left[ 1, \frac{N-2}{2} \right],\tag{6}\] we define \[\label{defineTs} T_{N, s} = T_{s} = 2 \frac{(N-1)!}{(N-2s)!} B_{N-2s}.\tag{7}\] Since every even integer \(h \in [2, k-1]\) can be written uniquely in the form \(h = N - 2s\) for \(N\) and \(s\) as specified, it remains to prove that \(T_{s} \in \mathbb{Z}\).

Now, define \[\label{defineFN} F_{N}(x) = x^{1-N} \big( B_{N}(x) - B_{N} \big),\tag{8}\] recalling the definition of Bernoulli polynomials in 4 . We claim that \[\label{nestedD} C_{N - 1}(2s) = \frac{2}{N} \big( D^{2s} F_{N} \big)(1).\tag{9}\] To prove the relation in 9 , we begin by manipulating the formula in 4 to obtain that \[\label{manipulatedBNx} x^{1-N} B_{N}(x) = \sum_{r=0}^{N} \binom{N}{r} B_{r} x^{1-r},\tag{10}\] so that 10 gives us that \[\label{sumtoreduce} D^{2s}\big( x^{1-N} B_{N}(x) \big) \Big|_{x=1} = \sum_{r=0}^{N} \binom{N}{r} B_{r}(1-r)^{2s}.\tag{11}\] From the vanishing of the \(r = 1\) case of the summand on the right of 11 together with the vanishing of Bernoulli numbers with odd indices greater than \(1\), we find that \[\label{combinewithFN} D^{2s}\big( x^{1-N} B_{N}(x) \big) \Big|_{x=1} = \frac{N}{2} C_{N-1}(2s) + B_{N}(1-N)^{2s},\tag{12}\] so that an application of the definition of \(F_{N}(x)\) in 8 via 12 gives us an equivalent version of 9 .

Now, we rewrite \(F_{N}(x)\), as defined in 8 , by setting \(g(x) = x^{1-N}\) and \(H(x) = B_{N}(x) - B_{N}\), with \(F_{N}(x) = g(x) H(x)\). Observe that \(B_{N}(1) = B_{N}\) and \(H(1) = 0\). The product rule for derivatives then gives us that \[\label{diffprod} F_{N}^{(\ell)}(1) = \sum_{r=1}^{\ell} \binom{\ell}{r} g^{(\ell - r)}(1) B_{N}^{(r)}(1),\tag{13}\] noting that we obtain the vanishing of \(\binom{\ell}{r} g^{(\ell - r)}(1) H^{(r)}(1)\) for the \(r = 0\) case. Observe that \(g^{(\ell - r)}(1) \in \mathbb{Z}\), recalling the definition of \(g(x)\) as an integer power of \(x\). From the definition of Bernoulli polynomials in 4 , one may verify that \[\label{setx1} B_{N}^{(r)}(x) = \frac{N!}{(N-r)!} B_{N - r}(x)\tag{14}\] for positive integers \(r \leq N\), and we specialize 14 to \[\label{specializeto1} B_{N}^{(r)}(1) = \frac{N!}{(N-r)!} B_{N - r}(1).\tag{15}\] According to the product rule expansion in 13 , we have that sum in this product rule is restricted to indices \(r \in [1, \ell]\). In view of the identity in 9 , which involves the application of \(D^{2s}\) to \(F_{N}(1)\), it is implicit that \(\ell \leq 2 s\), and we recall the bounds on \(s\) in 6 , giving us that \(2 \leq N - r\). Now, if \(r\) is odd, then \(N - r \geq 2\) is odd. So, if \(r\) is odd, then \(B_{N-r}(1) = B_{N-r} = 0\). So, from 15 , we have that if \(r\) is odd, then \(B_{N}^{(r)}(1)\) vanishes. So, for an even value \(r = 2i\), we find that \[\label{BntoT} \frac{2}{N} B_{N}^{(2i)}(1) = T_{i},\tag{16}\] recalling the definition in 7 . An application of the expression in 9 of \(C_{N-1}(2s)\) in terms of \(\big( D^{2s}F_{N} \big)(1)\) together with the classical Stirling number identity in 5 give us that \[\label{separateStirling} \big( D^{2s} F_{N} \big)(1) = F_{N}^{(2s)}(1) + \sum_{\ell = 1}^{2s-1} \genfrac\{\}{0pt}{}{2s}{\ell} F_{N}^{(\ell)}(1).\tag{17}\] Through an application of 13 to 17 , we obtain \[\begin{gather} \frac{2}{N} \big( D^{2s} F_{N} \big)(1) = \sum_{r=1}^{2s} \binom{2s}{r} g^{(2s - r)}(1) \frac{2}{N} B_{N}^{(r)}(1) + \\ \sum_{\ell = 1}^{2s-1} \genfrac\{\}{0pt}{}{2s}{\ell} \sum_{r=1}^{\ell} \binom{\ell}{r} g^{(\ell - r)}(1) \frac{2}{N} B_{N}^{(r)}(1), \end{gather}\] and we rewrite this as \[\begin{gather} \frac{2}{N} \big( D^{2s} F_{N} \big)(1) = \frac{2}{N} B_{N}^{(2s)}(1) + \sum_{r=1}^{2s-1} \binom{2s}{r} g^{(2s - r)}(1) \frac{2}{N} B_{N}^{(r)}(1) + \\ \sum_{\ell = 1}^{2s-1} \genfrac\{\}{0pt}{}{2s}{\ell} \sum_{r=1}^{\ell} \binom{\ell}{r} g^{(\ell - r)}(1) \frac{2}{N} B_{N}^{(r)}(1). \end{gather}\] For indices \(r\) involved in the above sums, we have that \(1 \leq r \leq 2s \leq N-2\), with \(N - r \geq 2\). So, if \(r\) is odd, then \(N - r\) is odd and \(\geq 3\), with \(B_{N-r}(1) = B_{N - r} = 0\), and hence \[\begin{gather} \frac{2}{N} \big( D^{2s} F_{N} \big)(1) = \frac{2}{N} B_{N}^{(2s)}(1) + \sum_{r=1}^{s-1} \binom{2s}{2r} g^{(2s - 2r)}(1) \frac{2}{N} B_{N}^{(2r)}(1) + \\ \sum_{\ell = 1}^{2s-1} \genfrac\{\}{0pt}{}{2s}{\ell} \sum_{ r = 1 }^{ \lfloor {\ell}/{2} \rfloor} \binom{\ell}{2r} g^{(\ell - 2r)}(1) \frac{2}{N} B_{N}^{(2r)}(1). \end{gather}\] From the formula for \(C_{N - 1}(2s)\) in 9 together with the definition in 16 , we find that there exist integers \(a_{s, r}\) such that \[\label{inductionT} T_{s} = C_{N-1}(2s) - \sum_{r=1}^{s-1} a_{s, r} T_{r}.\tag{18}\] For the \(s = 1\) case of 18 , we find that \(C_{k}(2) = T_{1}\), and \(C_{k}(2)\) is an integer by assumption. So, from the assumption that \(C_{k}(v)\) is an integer for every \(v \geq 1\) and for \(k > 1\) odd, and since \(N \geq 4\) is even, an inductive argument gives us, from 18 , that \(T_{s}\) is an integer for \(s\) in the interval indicated in 6 , as desired. ◻

Proof. Let \(k \in \mathcal{D}_{1}\). To evaluate \[\mathcal{D}_{1} = \{ k \geq 1 : \forall n \geq 1 \;\varphi_{1}(n) \mid \varphi_k(n) \},\] we begin by noting that \(\varphi_1(3) = 3\) and that \(\varphi_k(3) = 1 + 2^k\). So, in order for \(\varphi_1(3) \mid \varphi_k(3)\) to hold, the integer \(k\) is required to be odd. So, the integer \(k\) is henceforth assumed to be odd.

Now, if \(k = 1\), then \(k\) is in the desired containing set. So, we henceforth let \(k > 1\), (again while working under the assumption that \(k \in \mathcal{D}_{1}\)).

We proceed to make use of a standard formulation of the Faulhaber formula whereby \[\sum_{a=1}^{N} a^{k} = \frac{1}{k+1} \sum_{r=0}^{k} (-1)^{r} \binom{k+1}{r} B_{r} N^{k+1-r}.\] It was proved by Singh [2] that \[\label{inclusionex} \varphi_{k}(n) = \sum_{d \mid n} \mu(d) d^{k} \sum_{b=1}^{n/d} b^k,\tag{19}\] and the relation in 19 can be obtained via the Möbius inversion formula. Write \[\alpha_j(n) = \sum_{d \mid n} \mu(d) d^{j-1} = \prod_{p \mid n} (1 - p^{j - 1})\] for \(n > 1\). Singh also obtained that \[\label{Singhalpha} \varphi_{k}(n) = \frac{1}{k+1} \sum_{j = 0}^{k} (-1)^{j} \binom{k+1}{j} B_{j} \alpha_{j}(n) n^{k+1-j}\tag{20}\] as a consequence of 19 .

Now, let \(n > 1\). Since \(B_{j}\) vanishes for odd \(j > 1\), and since \(\alpha_1(n)\) also vanishes, we obtain from 20 that \[\label{bisection} \varphi_{k}(n) = \frac{n^k \varphi(n)}{k+1} + \frac{1}{k+1} \sum_{j=1}^{(k-1)/2} \binom{k+1}{2j} B_{2j} n^{k+1-2j} \prod_{p \mid n} (1-p^{2j-1}),\tag{21}\] recalling the assumption that \(k > 1\) is odd.

Now, suppose that \(n\) is squarefree and that \(m\) is an odd and positive integer. An application of Euler’s product formula for \(\varphi\) then gives us that \[\label{prodgeom} \prod_{p \mid n} (1 - p^m) = (-1)^{\omega(n)} \varphi(n) \prod_{p \mid n} (1 + p + \cdots + p^{m-1}).\tag{22}\] Dividing both sides of 21 by \(\varphi_{1}(n)\), the relations in 2 and 22 then give us that \[\begin{gather} \frac{\varphi_k(n)}{\varphi_1(n)} = \\ \frac{2n^{k-1}}{k+1} + \frac{2 (-1)^{\omega(n)}}{k+1} \sum_{j=1}^{(k-1)/2} \binom{k+1}{2j} B_{2j} n^{k-2j} \prod_{p \mid n} (1 + p + \cdots + p^{2j-2}), \end{gather}\] again with the assumptions that \(n\) is squarefree. Since we are working under the assumption that \(k \in \mathcal{D}_{1}\), we have that the quotient \(\frac{\varphi_k(n)}{\varphi_1(n)}\) is an integer.

Again with the ongoing assumption that \(k \in \mathcal{D}_{1}\), we claim that \(C_{k}(u) \in \mathbb{Z}\) for each \(u \geq 1\). In this direction, we let \(L\) denote the least common multiple of the denominators, in lowest terms, among all expressions of the form \(\frac{2}{k+1} \binom{k+1}{2j} B_{2j}\) for \(0 \leq j \leq \frac{k-1}{2}\), noting that the \(j=0\) case reduces to \(\frac{2}{k+1}\). Dirichlet’s theorem on primes in arithmetic progressions then gives us that there are infinitely many primes congruent to \(1\) modulo \(L\). For fixed \(u \geq 1\), choose \(u\) distinct primes \(p_1\), \(p_2\), \(\ldots\), \(p_u\) that are all congruent to \(1\) modulo \(L\), and set \(n = p_1 p_2 \cdots p_u\). Then \(n\) is squarefree, and \(u = \omega(n)\), and \[\label{n1modL} n \equiv 1 \bmod L.\tag{23}\] Moreover, the expansion of the product \(\prod_{p \mid n} (1 + p + \cdots + p^{m-1})\) reveals that \[\label{mtoomega} \prod_{p \mid n} (1 + p + \cdots + p^{m-1}) \equiv m^{\omega(n)} \pmod L.\tag{24}\] From 23 and 24 together, we find that \[\label{summandmod} (-1)^{\omega(n)} n^{k-2j} \prod_{p \mid n} (1 + p + \cdots + p^{2j-2}) \equiv (1-2j)^{\omega(n)} \pmod L,\tag{25}\] from the construction of \(L\), letting \(j \in \big[1, \frac{k-1}{2} \big]\). Now, we proceed to rewrite \(\frac{ \varphi_k(n) }{ \varphi_1(n) } - C_k(\omega(n))\) as \[\begin{gather} \frac{2n^{k-1}}{k+1} + \frac{2 }{k+1} \sum_{j=1}^{(k-1)/2} \binom{k+1}{2j} B_{2j} (-1)^{\omega(n)} n^{k-2j} \prod_{p \mid n} (1 + p + \cdots + p^{2j-2}) - \\ \frac{2}{k+1} \sum_{j=0}^{(k-1)/2} \binom{k+1}{2j} B_{2j} (1 - 2 j)^{\omega(n)}, \end{gather}\] and this may, in turn, and via 25 , be rewritten as \[\begin{gather} \frac{2n^{k-1}}{k+1} + \frac{2 }{k+1} \sum_{j=1}^{(k-1)/2} \binom{k+1}{2j} B_{2j} (z_j L + (1- 2 j)^{\omega(n)}) - \\ \frac{2}{k+1} \sum_{j=0}^{(k-1)/2} \binom{k+1}{2j} B_{2j} (1 - 2 j)^{\omega(n)}, \end{gather}\] for some integers \(z_j\) for \(j \in \big\{ 1, 2, \ldots, \frac{k-1}{2} \big\}\), so that the above combination of Bernoulli sums reduces to \[\label{intBernoulli} \frac{2 }{k+1} (n^{k-1}-1) + \sum_{j=1}^{(k-1)/2} \left( \frac{2 }{k+1} \binom{k+1}{2j} B_{2j} L \right) \, z_j.\tag{26}\] From our construction, we have that \(\operatorname{den}\big( \frac{2}{k+1} \big)\) divides \(L\). Recalling 23 , we find that \(n^{k-1}-1\) is a multiple of \(L\), so that \(\frac{2}{k+1} (n^{k-1}-1)\) is an integer. In a similar spirit, our construction of \(L\) is such that each expression of the form \(\frac{2 }{k+1} \binom{k+1}{2j} B_{2j} L\) for \(j \in \big\{ 1, 2, \ldots, \frac{k-1}{2} \big\}\) is an integer. So, since 26 is an integer, we have that \(\frac{ \varphi_k(n) }{ \varphi_1(n) } - C_k(\omega(n))\) is an integer, and, in turn, since \(\frac{ \varphi_k(n) }{ \varphi_1(n) }\) is an integer, we can conclude that \(C_{k}(\omega(n))\) is an integer.

Observe that \(C_k(u)\) is defined, as in 3 , independently of the value \(n\) given above. Moreover, for an arbitrary integer \(v \geq 1\), we can construct an admissible squarefree integer \(n\) (in the above manner) such that \(\omega(n) = v\). So, since \(C_{k}(\omega(n)) \in \mathbb{Z}\), we can conclude that \(C_{k}(v) \in \mathbb{Z}\) for an arbitrary integer \(v \geq 1\).

So, we have shown that if \(k \in \mathcal{D}_{1}\) and \(k > 1\), then \(k\) is odd and \(C_{k}(v) \in \mathbb{Z}\) for all \(v \geq 1\). By Lemma 1, we thus have that: For each even integer \(h \in [2, k-1]\), we have that \[\label{2fracBinZ} 2 \frac{k!}{h!} B_{h} \in \mathbb{Z}.\tag{27}\]

As above, let \(h\) be an even integer such that \(h \in [2, k-1]\). Now, let \(p\) be an odd prime such that \((p-1) \mid h\). By the von Staudt–Clausen theorem, this same prime \(p\) is a factor in the denominator of \(B_{h}\). Moreover, since \(p \neq 2\), the factor \(2\) in \(2 \frac{k!}{h!} B_{h}\) does not cancel with the factor \(p\) in the denominator of \(B_{h}\). From 27 , we can conclude that \(p\) divides \(\frac{k!}{h!} = (h+1)(h+2) \cdots k\). So, we have proved the following property: For an even integer \(h \in [2, k-1]\), and for an odd prime \(p\) such that \((p-1) \mid h\), there exists a multiple of \(p\) in \[\label{displayinterval} \{ h+1, h+2, \ldots, k \}.\tag{28}\]

Now, let \(p \leq k\) denote an odd prime. We write \(\rho_p\) in place of the least nonnegative residue of \(k\) modulo \(p-1\). Since \(k\) is odd and \(p-1\) is even, it is not the case that \(k \equiv 0 \bmod {(p-1)}\), and hence the bounds \[\label{usewithq} 1 \leq \rho_p \leq p - 2.\tag{29}\] Set \(h = k - \rho_p\). Since \(k\) is odd and since \(\rho_p\) is the residue of an odd number modulo an even number, we see that \(h\) is even. Moreover, since \(\rho_p \equiv k \bmod {(p-1)}\), we find that \((p-1) \mid h\). By assumption that \(p \leq k\), we see that \(h = k - \rho_p \geq k - (p-2) \geq 2\). So, our constructed value \(h\) satisfies the above conditions concerning the interval on display in 28 , i.e., so that there is at least one value among \[\label{wobrackets} k - \rho_p + 1, k - \rho_p+2, \ldots, k\tag{30}\] that is a multiple of \(p\).

Now, let \(r_p\) denote the least nonnegative residue of \(k\) modulo \(p\) (noting the contrast to the definition of \(\rho_p\)). The greatest multiple of \(p\) not exceeding \(k\) is \(k - r_p\). So, since there is a multiple of \(p\) among the values in 30 , we can conclude that there exists an index \[\label{iotain} \iota \in \{ 0, 1, \ldots, \rho_p - 1 \}\tag{31}\] such that \[\label{kminusrp} k - r_{p} = k - \iota.\tag{32}\] So, from 31 and 32 together, we find that \(r_p < \rho_p\), i.e., so that: For each odd prime \(p \leq k\), the relation \[\label{comparepmod} k \pmod {p} < k \pmod {(p-1)}\tag{33}\] holds.

By way of contradiction, suppose that there exists an odd prime \(q\) dividing \(k + 1\). Since \(k+1\) is even, we see that \(q \leq k\). Since \(k \equiv -1 \bmod q\), we find that the least nonnegative residue of \(k\) modulo \(q\) is \(q-1\). Being consistent with notation given above, we have that \(\rho_q\) is the least nonnegative residue of \(k\) modulo \(q - 1\), and, from 29 , we have that \(\rho_{q} \leq q-2\). The relation in 33 would then give us that \(q - 1 < q - 2\), providing a desired contradiction.

So, since \(k + 1\) does not have any odd prime divisor, we write \(k + 1 = 2^{a}\) for some integer \(a\), noting that \(a \geq 2\) since \(k\) is odd and since \(k > 1\).

Recall that 33 holds for an odd prime \(p \leq k\). Since \(k > 1\) is odd, we are permitted to let \(p = 3\). So, from 33 (recalling that \(k\) is odd), since \(k \bmod {3} < k \bmod {2}\), we can conclude that \(3 \mid k\). So, since \(2^{a} \equiv 1 \bmod 3\), we can, in turn, conclude that \(a\) is even.

By way of contradiction, suppose that \(a = 2^{t} u\) for an odd integer \(u > 1\) and for \(t \geq 1\). We proceed to make use of the property of cyclotomic polynomials such that \(X^{v} + 1 = (X+1)(X^{v - 1} - X^{v - 2} + X^{v-3} - \cdots - X + 1)\) for an odd, positive integer \(v\). This factorization gives us that \[\label{fromcyclotomic} \big( 2^{2^{t}} + 1 \big) \mid \big( 2^{2^{t} u} + 1 \big),\tag{34}\] and we find that 34 is equivalent to \[\label{rewritecyclo} \big( 2^{2^{t}} + 1 \big) \mid \left( k + 2 \right).\tag{35}\] Now, let \(q\) denote an odd prime factor of \(2^{2^{t}} + 1\). Recalling that \(u > 1\), we find that \(q \leq 2^{2^{t}} + 1 < 2^{2^{t} u} - 1 = k\). So, since \(q\) is an odd prime satisfying \(q \leq k\), we have, from 33 , that \[\label{modmodcyclo} k \pmod {q} < k \pmod {(q-1)}.\tag{36}\] Also, the relation in 35 gives us that \(q \mid (k+2)\), i.e., so that \[\label{givesqminus2} k \equiv -2 \pmod {q}.\tag{37}\] From 37 , we see that the least nonnegative residue of \(k\) modulo \(q\) is \(q-2\), but the right-hand side of 36 is in \(\{ 0, 1, \ldots, q - 2 \}\), so that the strict inequality in 36 is contradicted.

So, we have shown that \(a\) is necessarily a power of \(2\). Recalling that \(a \geq 2\), we write \(a = 2^{\lambda}\) for an integer \(\lambda \geq 1\), with \[\label{kpowertower} k = 2^{2^{\lambda}} - 1.\tag{38}\] Since the cases for \(\lambda = 1\) and \(\lambda = 2\) produce the values \(k = 3\) and \(k = 15\), respectively, it remains to consider the cases for \(\lambda \geq 3\).

We henceforth assume that \(\lambda \geq 3\).

Now, consider the case whereby \(\lambda\) is odd. Recalling that \(p \leq k\) denotes an odd prime, and recalling 38 , we let \(p = 7\). The parity of \(2^{\lambda}\) then gives us that \(2^{2^{\lambda}} - 1 \equiv 3 \bmod 6\), i.e., so that \(k \equiv 3 \bmod 6\). The integer sequence given by expressions of the form \(2^z \bmod 7\) for positive integers \(z\) is \(3\)-periodic, so that the parity of \(\lambda\) gives us that \(2^{2^{\lambda}} \equiv 4 \bmod 7\), i.e. so that \(k \equiv 3 \bmod 7\), contradicting the strict inequality in 33 .

So, we have that \(\lambda \geq 3\) is even. Now, suppose that \(\lambda \equiv 0 \bmod 4\). Again recalling 38 and that \(p \leq k\) is an odd prime, we proceed to take \(p = 11\). The residue class of \(2^{\lambda}\) modulo \(4\) then gives us that \(2^{2^{\lambda}} \equiv 6 \bmod 10\), i.e., so that \(k \equiv 5 \bmod 10\). The residue class of \(2^{\lambda}\) modulo \(10\) then gives us that \(2^{2^{\lambda}} \equiv 9 \bmod 11\), i.e., so that \(k \equiv 8 \bmod 11\), and we again find that 33 is contradicted.

From the foregoing arguments, we see that \(\lambda \geq 3\) satisfies \(\lambda \equiv 2 \bmod 4\), with \(\lambda \geq 6\). In this case, we set \(p = 29\). Omitting details, the periodicity of powers of \(2\) modulo \(28\) gives us that \(k \equiv 15 \bmod 28\), and the periodicity of powers of \(2\) modulo \(29\) gives us that \(k \bmod 29 \in \{ 15, 23, 24 \}\), and we again find that the strict inequality in 33 is contradicted.

So, we have shown that it is not the case that \(\lambda \geq 3\). So, if \(k > 1\), then \(k \in \{ 3, 15 \}\), and we can then conclude that the desired containment \(\mathcal{D}_{1} \subseteq \{ 1, 3, 15 \}\) holds. ◻