Pentagonal Square Triangular Number

Author

Eric W. Weisstein
September 4, 2003

This notebook downloaded from http://mathworld.wolfram.com/notebooks/IntegerSequences/PentagonalSquareTriangularNumber.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/PentagonalSquareTriangularNumber.html.

©2005 Wolfram Research, Inc. except for portions noted otherwise

Initialization

<<Figurate`

Analytic Solution

PentagonalNumber[l]==SquareNumber[m]==TriangularNumber[n]

Reduce[PentagonalNumber[n]==SquareNumber[k]==TriangularNumber[m]&&(k|m|n)∈Integers&&k>0&&m>0&&n>0,{k,m,n},Integers]

FindInstance[PentagonalNumber[n]==SquareNumber[k]==TriangularNumber[m]&&(k|m|n)∈Integers&&k>1&&m>1&&n>1,{k,m,n},Integers]

Search

Reduce[PentagonalNumber[n]==TriangularNumber[k]&&(k|n)∈Integers&&k>0&&n>0,{k,n},Integers]/.C[1]:>x

Triangular

FullSimplify[(1/12) (-6+3 (((7-4 (Sqrt[3])))^x)+(Sqrt[3]) (((7-4 (Sqrt[3])))^x)+3 (((7+4 (Sqrt[3])))^x)-(Sqrt[3]) (((7+4 (Sqrt[3])))^x))]

FullSimplify[TriangularNumber[(1/12) (-6+(((7-4 (Sqrt[3])))^x) (3+(Sqrt[3]))-(-3+(Sqrt[3])) (((7+4 (Sqrt[3])))^x))]]

Square

FullSimplify[(1/6) (1+(1/2) (-(((7-4 (Sqrt[3])))^x)-(Sqrt[3]) (((7-4 (Sqrt[3])))^x)-(((7+4 (Sqrt[3])))^x)+(Sqrt[3]) (((7+4 (Sqrt[3])))^x)))]

FullSimplify[PentagonalNumber[(1/12) (2-(((7-4 (Sqrt[3])))^x) (1+(Sqrt[3]))+(-1+(Sqrt[3])) (((7+4 (Sqrt[3])))^x))]]

Pentagonal Square

PentagonalSquare[x_]:=(1/48) (-4+(((7-4 (Sqrt[3])))^(2 x)) (2+(Sqrt[3]))-(-2+(Sqrt[3])) (((7+4 (Sqrt[3])))^(2 x)))

Table[PentagonalSquare[x],{x,10}]//Expand

Select[Table[PentagonalSquare[x],{x,200}]//Expand,SquareQ]//Timing

PentagonalSquare[9690]//N