In the papers given in the links below, we define factoriangular number (Ftn) as the sum of corresponding factorial and triangular number, that is, Ftn = n! + Tn where Tn = n(n+1)/2. Notice that Ft1 = 2T1 = 2 and Ft3 = 2T3 = 12. Aside from 2 and 12, is there any other factoriangular number that is twice a triangular number?
Article Sums of Two Triangulars and of Two Squares Associated with S...
Article On the Sum of Corresponding Factorials and Triangular Number...
Article On the Sum of Corresponding Factorials and Triangular Number...
Seems rather obvious. Since n! + Tsub(n) must equal 2Tsub(n) for this, then n! must equal Tsub(n). But obviously n! exceeds Tsub(n) for n greater than 3, i.e., (n-1)! exceeds (n+1)/2 for n greater than 3. Or, the difference between (n+1)! and n!, viz., nn!, increases faster than the difference between (n+1)(n+2)/2 and n(n+1)/2 = n+1. What am I missing?
It is easy to see that for n>3 there are no solutions to Ft_n = 2T_n. But what if we replace n! with Gamma(n+1) and allow n to be negative. Then the solution exists: n \approx -4.026...
What I missed is letting n be negative, I guess. I've always thought that n! was only defined for non-negative integers. This mindset again shows that implicit assumptions can be very dangerous!
I think Romer is asking if there exist solutions to the equation Ftm=2Tn apart from (m,n)=(1,1), (3,3).
That's not the way I (re)read his statement. And another implicit assumption: I'm assuming that n must be the same value for both the factoriangular number and the triangular number of his posed problem. :-)
Dr. Ioulia got my question right; m and n are natural numbers there and they may not be equal.
Thank you for clarifying. Now I think I know what I'm missing, so m and n are non-negative integers, correct?
That's correct Dr. White. In the above question, m and n should be non-negative integers. And thank you for your previous answers. When n = m, the only solutions are (n,m) = (1,1), (3,3), but we are also interested on finding solutions when m and n are not equal.
Thank you, Romer: I played with it a little more yesterday, and set up a quadratic equation in m, mexp(2) + m - [n! + n(n+1)/2] = 0, and observed that 4[n! +n(n+1)/2] must be one less than a perfect square, e.g., kexp(2), where k is odd, for m to be a natural number. That's as far as I have gotten so far in answering your question, which now seems rather profound! :-( Brian
Dear George,
In Case 1, (k-2p-1)|(2p)! does not imply k-2p-1 ≤ 2p.
In fact, we have k-2p-1=1*2*...*(p-1)*(p+1)*...*2p>2p for p>1.
Similarly, in Case 2 (k-2p-1)|(2p+1)! does not imply k-2p-1 ≤ 2p+1.
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