The non-$k$-gonal pyramidal number formula for $k\geq 9$

Proof, valid for every $k\geq 3$

Let

$$P_k(m)=\frac{m(m+1)\bigl(m(k-2)-(k-5)\bigr)}{6}$$

be the $m$-th $k$-gonal-pyramidal number, and let $a_k(n)$ be the $n$-th positive integer not among these numbers. Set

$$h=\left\lfloor\sqrt[3]{\frac{6n}{k-2}}\right\rfloor.$$

I will prove the following three-branch formula from Equation (5):

• $a_k(n)=n+h+1$ if

$$6n\geq (k-2)h^3+3(k-1)h^2+(2k-1)h+6;$$

• $a_k(n)=n+h-1$ if

$$6n\leq h(h-1)\bigl((k-2)h+k+1\bigr);$$

• $a_k(n)=n+h$ otherwise.

Write $c=k-2\geq1$, and put $Q_m=P_k(m)-m$, with the auxiliary values $P_k(0)=Q_0=0$. Then

$$6P_k(m)=m(m+1)\bigl(c(m-1)+3\bigr).$$

Its first difference is

$$6\bigl(P_k(m+1)-P_k(m)\bigr) =3(m+1)(cm+2)>0,$$

so the pyramidal numbers are strictly increasing.

Suppose $a_k(n)=n+r$. Exactly $r$ pyramidal numbers precede this value, so

$$P_k(r)<n+r<P_k(r+1).$$

Because all quantities are integers, this is equivalent to

$$Q_r<n\leq Q_{r+1}. \qquad (1)$$

Conversely, (1) says that exactly $r$ pyramidal and $n$ complementary values occur through $n+r$. Thus (1) uniquely characterizes the required offset $r$.

The definition of $h$ gives

$$ch^3\leq6n<c(h+1)^3. \qquad (2)$$

For $h\geq1$, direct expansion gives

$$ch^3-6Q_{h-1} =(c-1)h(3h-2)+7h-6>0.$$

For $h\geq0$, similarly,

$$6Q_{h+2}-c(h+1)^3 =(h+1)(3ch+5c+3h+6)>0.$$

Together with (2), these inequalities show that

$$Q_{h-1}<n<Q_{h+2}.$$

Consequently, the offset $r$ can only be $h-1$, $h$, or $h+1$.

It remains to identify which one occurs. The lower expression in the stated formula satisfies

$$h(h-1)(ch+c+3)=6Q_h.$$

Therefore its condition is exactly $n\leq Q_h$, which, by (1), selects $r=h-1$.

The upper expression satisfies

$$ch^3+3(c+1)h^2+(2c+3)h+6 =6Q_{h+1}+6.$$

Its condition is therefore $Q_{h+1}<n$, which selects $r=h+1$.

If neither condition holds, then

$$Q_h<n\leq Q_{h+1},$$

so (1) selects $r=h$.

Finally, when $h=0$, the upper threshold is $6$, so its condition holds automatically because $n\geq1$; it correctly gives $r=1$.

Substituting $c=k-2$ and $a_k(n)=n+r$ yields exactly the three branches above. Hence Conjecture 1 holds, and its range strengthens from $k\geq9$ to every $k\geq3$.

Source: Chai Wah Wu, “Algorithms for complementary sequences,” Equation (5) and Conjecture 1: https://arxiv.org/abs/2409.05844