Figurate Numbers

It seems there's a little confusion over exactly what is counted as a figurate number, but it's simple to show there is at most one positive integer that can appear in all figurate number sequences. Simply consider the polygonal numbers , which are a subset of the figurate numbers. The first $n$ -gonal number is always $1$ ; the second is always $n$ .

If any integer $k>1$ were to appear in all figurate number sequences, it would have to appear in all polygonal number sequences; but it doesn't appear in the $k+1$ -gonal number sequence. So the only positive integer that could possibly appear in all figurate sequences is $1$ . Then it comes down to definitions as to whether $0$ should be included as well.

Since the algebraic proof is so long (took me $3$ - $3.5$ hrs to complete it fully):

Figurate Number Wiki

Go to the bottom of the page once you click on the link, press show where it says Figurate Numbers, open all the links in a new tab, note down all of the formulas, set them equal to each other and split it into $9$ stages:

S $1$ : Do the fractions with denominator of $2$ - simplify it to one equation (name it S $1$ finalistic equation)

S $2$ : Do the fractions with denominator of $3$ - simplify it to one equation - then set the S $2$ simplified equation equal to the S $1$ finalistic equation and simplify it (name it S $2$ finalistic equation)

S $3$ : Do the fractions with denominator of $6$ - simplify it to one equation - then set the S $3$ simplified equation equal to the S $2$ finalistic equation and simplify it (name it S $3$ finalistic equation)

S $4$ : Set the S $3$ finalistic equation equal to the pentatope number formula and simplify (name it S $4$ finalistic equation)

S $5$ : Set all of the formulas (excluding the hypercube number formulas and both the pentagon and centered tetrahedral number formulas) equal to each and simplify that to one equation - then set the S $5$ simplified equation equal to the S $4$ finalistic equation and simplify it (name it S $5$ finalistic equation)

S $6$ : Set both the pentagon and centered tetrahedral number formulas equal to each other and simplify it - then set the S $6$ simplified equation equal to the S $5$ finalistic equation and simplify it (name it S $6$ finalistic equation)

S $7$ : Set the hypercube number formulas equal to each other ( $n^2, n^3, n^4, n^5, n^6, n^7, n^8$ and simplify it (this took so long - will explain it in the Figurate Numbers note in case any of you are confused) - then set the S $7$ simplified equation equal to the S $6$ finalistic equation and simplify it (name it S $7$ finalistic equation)

S $8$ : Set the S $7$ finalistic equation equal to $y$ and go to Desmos Polynomial Graph Calculator and note down the turning point (in case you don't know what the equation is: $n^8 - 6n^7 + 15n^6 - 20n^5 + 16n^4 + 143n^3 + 1880n^2 + 1120n + 753$ ). Then set the $x$ co-ordinate equal to $n$ ( $n = -0.308$ ). Then since it needs to be an integer, $n \approx 0$ .

S $9$ : Substitute $n = 0$ into all of the figurate number formulas and you should get: $0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$ .

Therefore, $0$ is the only number that is in all the figurate number sequences. Therefore, using algebraic proof, there is a number that is in all the figurate numbers, therefore, the answer is Yes.

The $r$ -topic number sequence is given by

$n+r-1\choose r$ .

The only member which appears in all these sequences is $n-1\choose 0$ , which is $1$ .