An arithmetico-geometric sequence is the elementwise product of an arithmetic sequence and a geometric sequence. For example, 1 -4 12 -32 is the product of the arithmetic sequence 1 2 3 4 and the geometric sequence 1 -2 4 -8. The nth term of an integer arithmetico-geometric sequence can be expressed as

$$a_n = r^n \cdot (a_0 + nd)$$

for some real number \$d\$, nonzero real \$r\$, and integer \$a_0\$. Note that \$r\$ and \$d\$ are not necessarily integers.

For example, the sequence 2 11 36 100 256 624 1472 3392 has \$a_0 = 2\$, \$r = 2\$, and \$d = 3.5\$.

Input

An ordered list of \$n \ge 2\$ integers as input in any reasonable format. Since some definitions of geometric sequence allow \$r=0\$ and define \$0^0 = 1\$, whether an input is an arithmetico-geometric sequence will not depend on whether \$r\$ is allowed to be 0. For example, 123 0 0 0 0 will not occur as input.

Output

Whether it is an arithmetico-geometric sequence. Output a truthy/falsy value, or two different consistent values.

Test cases

True:

1 -4 12 -32
0 0 0
-192 0 432 -1296 2916 -5832 10935 -19683
2 11 36 100 256 624 1472 3392
-4374 729 972 567 270 117 48 19
24601 1337 42
0 -2718
-1 -1 0 4 16
2 4 8 16 32 64
2 3 4 5 6 7
0 2 8 24

False:

4 8 15 16 23 42
3 1 4 1
24601 42 1337
0 0 0 1
0 0 1 0 0
1 -1 0 4 16
  • 1
    FYI you can use inline math mode with \$ to write things like \$ a_{0} \$. – FryAmTheEggman Nov 20 at 2:28
  • Are two-term inputs indeed possible? There aren't any in the test cases. – xnor Nov 20 at 3:40
  • @xnor Trivially you can set \$r=1\$ or \$d=0\$ so the sequences aren't unique in that case, but the output should always be truthy – Giuseppe Nov 20 at 3:55
  • 1
    Suggest testcase 0 2 8 24, 0 0 1, 0 0 0 1 – tsh Nov 20 at 12:12
  • 1
    1 -1 0 4 16 would be a useful False case, since it shares four consecutive elements with each of the True cases 1 -1 0 4 -16 and -1 -1 0 4 16. – Anders Kaseorg Nov 20 at 22:33

Perl 6, 184 128 135 bytes

{3>$_||->\x,\y,\z{?grep ->\r{min (x,{r&&r*$_+(y/r -x)*($×=r)}...*)Z==$_},x??map (y+*×sqrt(y²-x*z).narrow)/x,1,-1!!y&&z/y/2}(|.[^3])}

Try it online!

Computes \$r\$ and \$d\$ from the first three elements and checks whether the resulting sequence matches the input. Unfortunately, Rakudo throws an exception when dividing by zero, even when using floating-point numbers, costing ~9 bytes.

Enumerates the sequence using \$a_n=r \cdot a_{n-1}+r^n \cdot d\$.

Some improvements are inspired by Arnauld's JavaScript answer.

Explanation

3>$_||  # Return true if there are less than three elements

->\x,\y,\z{ ... }(|.[^3])}  # Bind x,y,z to first three elements

# Candidates for r
x  # If x != 0
??map (y+*×sqrt(y²-x*z).narrow)/x,1,-1  # then solutions of quadratic equation
!!y&&z/y/2  # else solution of linear equation or 0 if y==0

?grep ->\r{ ... },  # Is there an r for which the following is true?

    ( ,                         ...*)  # Create infinite sequence
     x  # Start with x
       {                       }  # Compute next term
        r&&  # 0 if r==0
                (y/r -x)  # d
           r*$_  # r*a(n-1)
                          ($×=r)  # r^n
                +        *  # r*a(n-1)+d*r^n
                                     Z==$_  # Compare with each element of input
min  # All elements are equal?

JavaScript (ES7), 135 127 bytes

a=>!([x,y,z]=a,1/z)|!a.some(n=>n)|[y/x+(d=(y*y-x*z)**.5/x),y/x-d,z/y/2].some(r=>a.every((v,n)=>(v-(x+n*y/r-n*x)*r**n)**2<1e-9))

Try it online!

How?

We use two preliminary tests to get rid of some special cases. For the main case, we try three different possible values of \$r\$ (and the corresponding values of \$d\$, which can be easily deduced) and test whether all terms of the input sequence are matching the guessed ones. Because of potential rounding errors, we actually test whether all squared differences are \$<10^{-9}\$.

Special case #1: less than 3 terms

If there are less than 3 terms, it's always possible to find a matching sequence. So we force a truthy value.

Special case #2: only zeros

If all terms are equal to \$0\$, we can use \$a_0=0\$, \$d=0\$ and any \$r\neq 0\$. So we force a truthy value.

Main case with \$a_0=0\$

If \$a_0=0\$, the sequence can be simplified to:

$$a_n=r^n\times n\times d$$

Which gives:

$$a_1=r\times d\\ a_2=2r^2\times d$$

We know that \$d\$ is not equal to \$0\$ (otherwise, we'd be in the special case #2). So we have \$a_1\neq 0\$ and:

$$r=\frac{a_2}{2a_1}$$

Main case with \$a_0\neq 0\$

We have the following relation between \$a_{n+1}\$ and \$a_n\$:

$$a_{n+1}=r.a_n+r^{n+1}d$$

For \$a_{n+2}\$, we have:

$$\begin{align}a_{n+2}&=r.a_{n+1}+r^{n+2}d\\ &=r(r.a_n+r^{n+1}d)+r^{n+2}d\\ &=r^2a_n+2r.r^{n+1}d\\ &=r^2a_n+2r(a_{n+1}-r.a_n)\\ &=-r^2a_n+2r.a_{n+1} \end{align}$$

We notably have:

$$a_2=-r^2a_0+2r.a_1$$

Leading to the following quadratic:

$$r^2a_0-2r.a_1+a_2=0$$

Whose roots are:

$$r_0=\frac{a_1+\sqrt{{a_1}^2-a_0a_2}}{a_0}\\ r_1=\frac{a_1-\sqrt{{a_1}^2-a_0a_2}}{a_0}$$

Wolfram Language (Mathematica), 55 bytes

{}!=Solve[i=Range@Tr[1^#];(a+i*d)r^i==#,{r,a,d},Reals]&

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Solve return all solution forms. The result is compared with {} to check if there is any solution.

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