# On a unique(?) binomial property of $3003$

Given the triangular number,

$$T_k = \frac{k(k+1)}{2}$$

and remembering that,

$$\binom{n}{m}=\binom{n}{n-m}$$

Excluding $a_0=1$, we then have the six-fold (at least) equalities,

\begin{aligned} a_1&=\binom{a_1}{1}=\binom{16}{2}=\binom{10}{3}=T_{15}=120\\[1.5mm] a_2&=\binom{a_2}{1}=\binom{21}{2}=\binom{10}{4}=T_{20}=210\\[1.5mm] a_3&=\binom{a_3}{1}=\binom{56}{2}=\binom{22}{3}=T_{55}=1540\\[1.5mm] \color{blue}{a_4}&=\binom{a_4}{1}=\binom{78}{2}=\binom{15}{5}=\binom{14}{6}=T_{77}=\color{blue}{3003}\\[1.5mm] a_5&=\binom{a_5}{1}=\binom{120}{2}=\binom{36}{3}=T_{119}=7140\\[1.5mm] a_6&=\binom{a_6}{1}=\binom{153}{2}=\binom{19}{5}=T_{152}=11628\\[1.5mm] a_7&=\binom{a_7}{1}=\binom{221}{2}=\binom{17}{8}=T_{220}=24310\\[1.5mm] \color{blue}{a_8}&=\binom{a_8}{1}=\binom{104}{39}=\binom{103}{40}\neq T_k\, \approx\, 6.12\times10^{28}\\[1.5mm] \color{blue}{a_9}&\overset{\color{red}?}{=}\binom{a_9}{1}=\binom{714}{272}=\binom{713}{273}\neq T_k\, \approx\, 3.53\times10^{204}\\ \end{aligned}

(Assuming Weger and Noe's results are conclusive, then there is no other $T_k$ with $k<3.49\times 10^{14}$ in this list.)

Questions:

1. The first eight $a_i$ is A003015. Since $a_9$ is so big, is it really the ninth, or is there a smaller term?
2. The terms $a_4, a_8, a_9$ belong to an infinite family involving Fibonacci numbers, $$a_i = B(m)=\binom{F_{2m} F_{2m+1}}{F_{2m-1}F_{2m}-1}$$ Other than $a_4=3003$, is there another triangular number in this family? (I checked that $B(5) \approx 4.59\times10^{1411}$ is not triangular, but $B(6)$ seemed already too big for my computer.)
• Where is Weger and Noe's result? – Elaqqad Jul 26 '15 at 12:45
• @Elaqqad: Kindly see Noe's comment (2004) in the OEIS link above. – Tito Piezas III Jul 26 '15 at 12:51
• Wolfie was about to determine that $B(6)$ is not a triangular number as well. That leaves an infinite more to check. :) – Tito Piezas III Jul 26 '15 at 12:53
• Do you know Singmaster's conjecture? – Elaqqad Jul 26 '15 at 12:53
• This seems tied to the fact that $1001, 2002, 3003$ are consecutive values in the $14$ row of Pascal's triangle, at positions $4,5,6.$ This means that $5$ is close to the point of inflection in the appropriate inflated Gaussian normal curve. I expect there are three term arithmetic progressions later, but not necessarily consecutive, and not necessarily of the very restrictive form $A,2A,3A.$ – Will Jagy Jul 26 '15 at 18:57