Triangular numbers

[sole carques]

"Triangular number, 6 is the double of another, viz 3; there are just 6 instances of this phenomenon in the first billion natural numbers."

Can you please tell me what the other five numbers are?

Many thanks.

 

[sole carques]

I've found one other number, the hard way, 3570 & 7140.

Any takers for the next?

 

[Denis]

You missed 105 : 210

4th: 121278 : 242556

 

[soroban]

Hello, sole carques!

A triangular number, 6, is the double of another, 3.
There are just 6 instances of this phenomenon in the first billion natural numbers.

Can you please tell me what the other five numbers are?

Thanks to Denis' input, I discovered the pattern . . .

. . \(\displaystyle \begin{array}{|c|c|} m & T_m \\ \hline T_2 & 3 \\ T_{14} & 105 \\ T_{84} & 3570 \\ T_{492} & 121,\!278 \\ T_{2870} & 4,\!119,\!885 \\ T_{16,730} & 139,\!954,\!815 \end{array} \qquad \begin{array}{|c|c|} n & T_n \\ \hline T_3 & 6 \\ T_{20} & 210 \\ T_{119} & 7140 \\ T_{696} & 242,\!556 \\ T_{4059} & 8,\!239,\!770 \\ T_{23,660} & 279,\!909,\!630 \end{array}\)


\(\displaystyle \text{The }m\text{'s form this sequence: }\:2, 14, 84, 492, 2870, 16,\!730, \hdots\)

\(\displaystyle \text{The recursion is: }\;m_1 = 2,\;m_2=14,\;\;m_k \;=\;6\!\cdot\! m_{k-1} - m_{k-2} + 2\)


\(\displaystyle \text{The }n\text{'s form this sequence: }\:3, 20, 119, 696, 4059, 23,\!660, \hdots\)

\(\displaystyle \text{The recursion is: }\;n_1 = 3,\;n_2=20,\;\;n_k \;=\;6\!\cdot\! n_{k-1} - n_{k-2} + 2\)


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I could ride off into the sunset, saying "My work is done here",
. . leaving the townfolks muttering, "Who was that masked man?"
But I will share my thoughts with you . . .

\(\displaystyle \text{The equation: }\:\frac{a(a+1)}{2} \:=\:2\cdot\frac{b(b+1)}{2}\,\text{ is a Pellian equation.}\)

What little I know of Pellians includes the fact that the solutions are recursive.
. . So I looked for the recursive relationship . . . that's it!


For those of you who believe the rumor that I am from Gamma Hydra IV
(where preschoolers solve partial differential equations in their heads),
sorry . . . I'm just an average carbon-based lifeform like you.

 

[Denis]

\(\displaystyle \text{The equation: }\:\frac{a(a+1)}{2} \:=\:2\cdot\frac{b(b+1)}{2}\,\text{ is a Pellian equation.}\)

Yep. So:
a = {sqrt[8b(b + 1) + 1] - 1} / 2
So, clearly, b is even.
I simply looped b from 2 increment by 2: less than 1 second to find the 1st 6.

 

[sole carques]

Many thanks for your speedy replies, Denis & soroban.

Please note that there's a typo in soroban's table: The Tm for T84 should read 3570 (I haven't got round to checking further).

 

[Denis]

Please note that there's a typo in soroban's table: The Tm for T84 should read 3570

Huh? Look again :shock: