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I'm interested in solving the following difference equation: $x[k-1]+(k^2+k+a)/x[k]=b$, $k=1,2,\ldots$, where $a,b$ are fixed positive numbers; let's say $x[1]=c>0$. Mathematica's RSolve function doesn't yield the general solution.

RSolve[{x[k - 1] + (k^2 + k + a)/x[k] == b}, x, k]

I tried RecurrenceTable. It produced something that looks like a cont fraction. Does anyone recognize it?

RecurrenceTable[{x[k - 1] + (k^2 + k + a)/x[k] == b, 
  x[1] == c}, x, {k, 10}]
{c, (6 + a)/(b - c), (12 + a)/(b - (6 + a)/(b - c)), (20 + a)/(
 b - (12 + a)/(b - (6 + a)/(b - c))), (30 + a)/(
 b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))), (42 + a)/(
 b - (30 + a)/(b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c))))), (
 56 + a)/(b - (42 + a)/(
  b - (30 + a)/(
   b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))))), (72 + a)/(
 b - (56 + a)/(
  b - (42 + a)/(
   b - (30 + a)/(
    b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c))))))), (90 + a)/(
 b - (72 + a)/(
  b - (56 + a)/(
   b - (42 + a)/(
    b - (30 + a)/(
     b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))))))), (
 110 + a)/(
 b - (90 + a)/(
  b - (72 + a)/(
   b - (56 + a)/(
    b - (42 + a)/(
     b - (30 + a)/(
      b - (20 + a)/(b - (12 + a)/(b - (6 + a)/(b - c)))))))))}

I suspect the continued fraction is probably a hypergeometric function of some sort.

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    $\begingroup$ Some of the constants look suspicous: A002378 Oblong (or promic, pronic, or heteromecic) numbers: a(n) = n*(n+1). (Formerly M1581 N0616) oeis.org/… $\endgroup$ – flinty May 14 at 21:05
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    $\begingroup$ ... in fact it's simpler than that - the constants in the fraction from bottom to top appear in Table[(n - 4) (n - 5), {n, 7, 50}] $\endgroup$ – flinty May 14 at 21:21
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I wrote the recursion directly by solving for x[k] in terms of x[k-1]:

Clear[x];
x[k_] := x[k] = (a + k + k^2)/(b - x[k - 1])
x[1] = c; 

This clearly gives a "continued fraction-like" answer:

x[#] & /@ Range[10]

enter image description here

Observing that the coefficients in each stack are the same:

coef={6, 12, 20, 30, 42, 56, 72, 90, 110}

I tried:

FindSequenceFunction[{6, 12, 20, 30, 42, 56, 72, 90, 110}]

2 + 3 #1 + #1^2 &

Of course this is not a proof of anything, but likely this pattern continues and might lead you to a relatively simple general form.

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