#triangular numbers

[lordandgodoftheobvious]

What one typically imagines when someone says something "came to [them] in a dream" is that it was, for all intents and purposes, handed down to them like some divine revelation. I was not so lucky; I battled in a half-lucid state to form an equation for triangular numbers for reasons that I cannot now recall.

So that we're all on the same page, the Xth triangular number--hereafter ΔX for simplicity; this almost certainly isn't the convention, but if the math nerds out there wanted me to care you shouldn't have made e a number and theta a variable--is the sum of the numbers from 1 to X. E.g. the ninth triangular number (Δ9) is 45, since 1+2+3+4+5+6+7+8+9=45. (The smartasses out there will notice that this is already an equation, but it'd quickly get tedious if I wanted to find Δ100 or something.)

I had the intuition that the way to attack the problem was to play with unit representations like when we were children first learning what numbers were.

^ Artist's interpretation of the bullshit I was getting up to.

Now, the first thing I noticed was that if you spin the Δ(X-1) piece around and drop it on the ΔX piece like this is Tetris, you get a square of side length X--

--and I really thought I was on to something here. But the problem with this idea is that you now have to calculate Δ(X-1). Well that's okay--according to the equation, that's just (X-1)²-Δ(X-2).

And if I'd been lucid, that's the point where I'd be like Hey, wait a second.... But I was not, so I got all the way to ΔX=X²-(X-1)²+(X-2)²...±1 before I realized that I'd created a more tedious equation.

My next thought was that the bigger ΔX is, the more the shape approximates a triangle, so I struggled with trying to find a way to average the zigzag into a proper hypotenuse--you know, in my head, while half-asleep.

Shockingly, this was not a productive use of my time.

But eventually this led to an actually good idea: if we just separate the triangle with height and base X from the rest of the shape, what do we get?

We get one triangle with a base and height of X (and thus an area of X²/2) and X triangles with a base and height of 1 (and thus areas of 1/2). Or to simplify it down:

ΔX=(X²+X)/2

This surely exists out there somewhere already--it'd be wild if it doesn't--but I deduced it myself from first principles.

[religion-is-a-mental-illness]

184 birds, 140 house guests and 40 cows are not gifts you give to someone you like, much less your “true love.”

[trupurt]

Was investigating triangular numbers in excel in English class because I was bored and found that in the first 30 triangular numbers, 3 was the only one that was prime. I thought I was so cool for figuring this out, and tried to find if three was the only prime triangular number. So I put the first 100 triangular numbers in Excel using a formula I knew: Tn= [n(n+1)]/2. And then I spent a little looking at it and got very stuck because golly they were all composite. Searched it up to check if anyone figured out why.

It's in the freaking formula. n MULTIPLIED BY n+1 (divided by 2). So if n is even, then it's composite, and if n is odd, then it's composite (bc n*(n+1) will always be an even number by an odd number, so it's always divisible by 2, leading to a composite result).

Thanks Mathematics Stack Exchange.

[riseball33]

Triangular Numbers There are four sets of triangular numbers in this pattern.

[riseball33]

Pascal’s Triangle

Sum on the combinations makes for interesting patterns in each column.

[collective-evolution]

A triangular number or triangle number counts the objects that can form an equilateral triangle. The formula is ½(n²+n). The early Greeks used pebbles arranged in patterns to learn arithmetic and geometry.

[wordswilling]

The math side of Wikipedia:

666 is the sum of the first 36 natural numbers (i.e. 1 + 2 + 3 + ... + 34 + 35 +36 = 666), and thus it is a triangular number. Notice that 36 = 15 + 21; 15 and 21 are also triangular numbers; and 15² + 21² = 225 + 441 = 666. (In addition: the sum of the 15th triangular number (120) and the 21st triangular number (231) is the 26th triangular number (351).)

666 is part of a Pythagorean quadruple (320, 410, 416, 666), since 320² + 410² + 416² = 102,400 + 168,100 + 173,056 = 443,556 = 666².

Since 36 is both square and triangular, 666 is the sixth number of the form n²(n² + 1)/2 (sequence A037270 in OEIS) and the eighth number of the form n (n + 1)(n² + n + 2)/8 (doubly triangular numbers) (sequence A002817 in OEIS).

The number of prime numbers up to 666 is 121, which is the square of 11, which is the number of prime numbers up to 36.

666 is an abundant number.

The harmonic mean of the decimal digits of 666 is (trivially – all repdigit natural numbers have this property) an integer: 3/(1/6 + 1/6 + 1/6) = 6, making 666 the 54th number with this property.

In base 10, 666 is a rep digit (and therefore a palindromic number) and a Smith number. A prime reciprocal magic square based on 1/149 in base 10 has a magic total of 666.

The Roman numeral for 666, DCLXVI, has exactly one occurrence of all symbols whose value is less than 1000 in decreasing order (D=500, C=100, L=50, X=10, V=5, I=1).

(60 and 360 were used as bases in Babylonian mathematics, as in the degrees of a circle; 360=10x36 and 666 is the sum of the numbers from 1 to 36.)

666 is 1010011010 in the binary numeral system which is the same as its one's complement in reverse digit order.

666 is a member of the indices of prime Padovan numbers: 3, 4, 5, 7, 8, 14, 19, 30, 37, 84, 128, 469, 666, 1262, 1573, 2003, 2210, ... (sequence A112882 in OEIS).

The sum of the squares of the first seven prime numbers: 2² + 3² + 5² + 7² + 11² + 13² + 17² = 666.

[clubsandwedge-blog]

Coded Triangle Numbers (Project Euler problem 42)

#include<stdio.h>

#include<math.h> int tri(int); int main() {     FILE *wordf;     wordf=fopen("words1.txt","r");     int count=0,i,j;     char words[2800][20],ch;     for(i=0;i<1786;i++)     {         j=0;         do         {             ch=(char)getc(wordf);             words[i][j]=ch;             j++;         }while(ch!='\n');     }     int sum=0,notris=0;     for(i=0;i<1786;i++)     {         sum=0;         for(j=0;words[i][j]!='\n';j++)         {             sum+=(int)words[i][j]-64;                 }         if(tri(sum))             notris++;     }     printf("%d\n",notris);     fclose(wordf);     return 0; } int tri(int n) {     double cp=(double)n;     double res=(sqrt(8.0*cp + 1.0) - 1.0)/2.0;     if(res==ceil(res))         return 1;     return 0; }