Hexagonal Square Number

Author

Eric W. Weisstein
November 28, 2004

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

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

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

Equation

Resolve[HexagonalNumber[n]==SquareNumber[m]&&m>0&&n>0,{m,n},Integers]

Table[{-(((((3-2 (Sqrt[2])))^(1+2 C[1]))-(((3+2 (Sqrt[2])))^(1+2 C[1])))/(4 (Sqrt[2]))),(1/4) (1+(1/2) ((((3-2 (Sqrt[2])))^(1+2 C[1]))+(((3+2 (Sqrt[2])))^(1+2 C[1]))))}/.C[1]->k,{k,0,4}]//Expand

FullSimplify[HexagonalNumber[(1/4) (1+(1/2) ((((3-2 (Sqrt[2])))^(1+2 k))+(((3+2 (Sqrt[2])))^(1+2k))))],k∈Integers&&k≥0]

FullSimplify[SquareNumber[(((((3-2 (Sqrt[2])))^(1+2k))-(((3+2 (Sqrt[2])))^(1+2 k)))/(4 (Sqrt[2])))],k∈Integers&&k≥0]

FullSimplify[ExpandAll[%],k∈Integers&&k≥0]

Enumeration

Table[(1/32) (-2+(17-12 (Sqrt[2])) (((3-2 (Sqrt[2])))^(4 k))+(((3+2 (Sqrt[2])))^(4 k)) (17+12 (Sqrt[2]))),{k,5}]//Expand

Recurrence

Date: Sat, 11 Sep 2004 19:44:09 -0300
From: "Marcos C S Carreira"

RSolve[{a[1]==1,a[n]==3a[n-1]+(Sqrt[8((a[n-1])^2)+1])},a[n],n]

a[1]:=1
a[n_]:=a[n]=3a[n-1]+(Sqrt[8((a[n-1])^2)+1])

a/@Range[10]