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Suppose I know the sequence is of the $a n^4 + b n^3 + c n^2 + d n$, with positive integers $a,b,c,d$, what is the easiest way to get them from the first few coefficients?

Here are the sequences I'm looking at. Mathematica's FindSequenceFunction works for the first one, but not for the rest

{15,48,105,192,315}
{15, 64, 183, 432, 895}
{15,80,273,720,1595}
{15, 144, 603, 1728, 3975}

Obtained from moments of Gaussian matrix products like $\operatorname{Tr}(AAAA^TA^TA^T)$ (background)

Clear[a, A, n];
getMoment[n_, expr_] := (
   traceExpr = Tr[expr] /. A -> Array[a, {n, n}];
   distSpec = # \[Distributed] NormalDistribution[0, 1] & /@ 
     Variables[traceExpr];
   Expectation[traceExpr, distSpec]
   );
exprs = {A . A . A . A . A . A, 
   A\[Transpose] . A\[Transpose] . A . A\[Transpose] . A . A, 
   A . A . A . A\[Transpose] . A\[Transpose] . A\[Transpose], 
   A . A\[Transpose] . A . A\[Transpose] . A . A\[Transpose]};
getMomentSequence[expr_] := Table[getMoment[n, expr], {n, 1, 5}];
table = Table[
   With[{seq = getMomentSequence[expr]},
    {TraditionalForm@expr, TraditionalForm[seq], 
     Expand[FindSequenceFunction[seq][n]]}],
   {expr, exprs}];
TableForm[table]
Yaroslav Bulatov
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  • quarticCoeffs[ll_List /; Length[ll] == 5] := Module[{x}, CoefficientList[InterpolatingPolynomial[ll, x], x]] seems to work for this. – Daniel Lichtblau Dec 04 '23 at 16:45

2 Answers2

3
seq={{15,48,105,192,315},{15,64,183,432,895},{15,80,273,720,1595},{15,144,603,1728,3975}};

form=a n^4+b n^3+c n^2+d n;

form/.First@Solve[Table[form,{n,5}]==#]&/@seq

{8 n + 6 n^2 + n^3, 4 n + 10 n^2 + n^4, 4 n + 8 n^2 + n^3 + 2 n^4, 4 n^2 + 6 n^3 + 5 n^4}
azerbajdzan
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2

form is the product { n^4, n^3, n^2, n} . {a, b, c, d}

First we calculate matrix

m=Table[{ n^4, n^3, n^2, n},{n,1,5}];

abcd=Map[LinearSolve[m, #] &, seq] ({{0, 1, 6, 8}, {1, 0, 10, 4}, {2, 1, 8, 4}, {5, 6, 4, 0}})

check result

MapThread[0 == Norm[m . #2 - #1] &, {seq, abcd}]
(*{True, True, True, True}*)
Ulrich Neumann
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