Amy the algebraist has an abacus with five rows, each of which contains nine beads. Used conventionally, this can represent any whole number between 0 and 99999. In order to represent the number \[a_1+10 a_2 +100 a_3 + 1000 a_4 + 10000 a_5\] one should put the abacus in position \((a_1,a_2,a_3,a_4,a_5)\), meaning that on the right hand side there are \(a_1\) beads from the top row, \(a_2\) beads from the second row, \(a_3\) beads from the third row, \(a_4\) beads from the 4th row, and \(a_5\) beads from the fifth row, with all other beads being on the left hand side. Here \(a_1, a_2, a_3, a_4\) and \(a_5\) are each whole numbers between \(0\) and \(9\).

Being an algebraist, Amy has to do lots of calculations involving the golden mean \(\phi=\frac{1+\sqrt{5}}{2}\). To help, she uses her abacus but now puts it in position \((a_1,a_2,a_3,a_4,a_5)\) to denote the number \[ a_1+(\phi\times a_2)+(\phi^2\times a_3)+(\phi^3\times a_4) +(\phi^4\times a_5). \] Some numbers now have multiple representations, because \(\phi^2=\phi+1\). For example, positions \((2,1,0,0,0)\) and \((1,0,1,0,0)\) both represent the number \(2+\phi\).

How many ways are there of representing \(4+7\phi\) on Amy's abacus?

MathsBombe Competition 2020 is organised by the The Department of Mathematics at The University of Manchester.

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