# Is there a (conjectured) relationship between triangular numbers and friendly numbers?

This question is related to these earlier questions [1] and [2].

Here is the crux of my inquiry:

Is there a (conjectured) relationship between triangular numbers and friendly numbers?

My Try

Numbers Known To Be Friendly $$\underline{6}, 12, 24, \underline{28}, 30, 40, 42, 56, 60, 66, \underline{78}, 80, 84, 96, 102, 108, 114, \underline{120},$$ $$132, 135, 138, 140, 150, 168, 174, 186, 200, 204, \underline{210}, 222, 224, 228, 234,$$ $$240, 246, 252, 258, 264, 270, 273, \underline{276}, 280, 282, 294, \underline{300}, 308, 312, 318,$$ $$330, 348, 354, 360, 364, 366, 372, \ldots, \underline{496}$$

Triangular Numbers $$0, 1, 3, \overline{6}, 10, 15, 21, \overline{28}, 36, 45, 55, 66, \overline{78}, 91, 105, \overline{120}, 136, 153, 171,$$ $$190, \overline{210}, 231, 253, \overline{276}, \overline{300}, 325, 351, 378, 406, 435, 465, \overline{496}, 528, 561,$$ $$595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128,$$ $$1176, 1225, 1275, 1326, 1378, 1431$$

Here are the numbers common to both lists:

Even Perfect Numbers $$6, 28, 496, \ldots$$

Triangular Numbers $T(p)$ Where $p$ Is A Prime $$78 = \dfrac{{12}\cdot{13}}{2}$$

Triangular Numbers $T(pq)$ Where $pq$ Is A Semiprime $$120 = \dfrac{{15}\cdot{16}}{2}$$

Triangular Numbers $T({p^2}q)$ Where $p$ And $q$ Are Primes $$210 = \dfrac{{20}\cdot{21}}{2}$$

Triangular Numbers $T({p^3}q)$ Where $p$ And $q$ Are Primes $$300 = \dfrac{{24}\cdot{25}}{2}$$

As a final check, I compared the list of triangular numbers above with the following:

A095739 Numbers known to be solitary but not coprime to sigma. $$18, \overline{45}, 48, 52, \overline{136}, 148, 160, 162, 176, 192, 196, 208, 232, 244, 261, 272,$$ $$292, 296, 297, 304, 320, 352, 369$$

A095738 Numbers that are coprime to sigma but are not prime powers. $$\overline{21}, 35, 36, 39, 50, \overline{55}, 57, 63, 65, 75, 77, 85, 93, 98, 100, 111, 115, 119,$$ $$129, 133, 143, 144, 155, 161, 171, 175, 183, 185, 187, 189, 201, 203, 205,$$ $$209, 215, 217, 219, 221, 225, 235, 237, 242, 245, 247, \overline{253}, 259, 265, 275,$$ $$279, 291, 299, 301, 305, 309, 319$$

A000961 Powers of primes. Alternatively, 1 and the prime powers (p^k, p prime, k >= 1). $$\overline{1}, 2, \overline{3}, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 37, 41,$$ $$43, 47, 49, 53, 59, 61, 64, 67, 71, 73, 79, 81, 83, 89, 97, 101, 103, 107,$$ $$109, 113, 121, 125, 127, 128, 131, 137, 139, 149, 151, 157, 163, 167, 169,$$ $$173, 179, 181, 191, 193, 197, 199, 211, 223, 227$$

I would have added more details to this question, but I have to eat for now. I hope what I have written is enough!

## 1 Answer

With this I try do feedback about your question, and I am waiting if you or other users want to argue if my calculations are related or are useful to your question. Since I don't know if this contribution, a set of calculations without a concise consequence, is a good answer I am waiting such discussion, comments or votes, thanks.

I've combined the definition of friendly numbers and a consequence of the abc conjecture, that is explained in the first paragraph in the proof of Proposition 2, in page 26 of [1].

After I've read the explanation I did the specialization to work with the polynomial $$f(X)=X(X+1).$$ Thus on assumption that the abc conjecture holds, $\forall\epsilon>0$ there exists a constant $0<C=C(\epsilon,f)$ such that for (a fixed pair of) friendly numbers $\frac{\sigma(n)}{n}=\frac{\sigma(m)}{m}:=\iota$ one has $$\left(\frac{\sigma(n)}{\iota}\right)^{1-\epsilon}=\left( m\frac{\sigma(n)}{\sigma(m)}\right)^{1-\epsilon}\leq C\cdot\operatorname{rad} \left( n(n+1) \right) .$$ Here $n(n+1)$ is two times a triangle number, but,as I said, I don't know if is a good statement or has mathematical meaning.

## References:

[1] Luca and Pomerance, On the radical of a perfect number, New York J. Math. 16 (2010).

• Reference for the consequence of the abc conjecture: [2] Elkies, ABC implies Mordell, IMRN 1991, 99–109 – user243301 Jan 25 '17 at 19:58