perplexus.info :: Sequences : Generalizing II

Generalizing II (Posted on 2015-11-20)

1, 3, 4, 9, 10, 12, 13, 27 ... is a sequence of all possible sums of distinct positive powers of three in an ascending order.

What is the n'th (e.g. 666th) number in the above sequence?

No Solution Yet Submitted by Ady TZIDON

Rating: 4.0000 (1 votes)

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a way to calculate

| Comment 1 of 2

DefDbl A-Z

Dim crlf$, ttl

Private Sub Form_Load()

Form1.Visible = True

Text1.Text = ""

crlf = Chr$(13) + Chr$(10)

Open "genlii.txt" For Output As #2

ttl = 0

addIn 0

Close 2

Text1.Text = Text1.Text & crlf & " done"

End Sub

Sub addIn(wh)

DoEvents

pwr = Int(3 ^ wh + 0.5)

For addend = 0 To pwr Step pwr

ttl = ttl + addend

If ttl > 0 Then

Print #2, mform(ttl, "############0")

' Text1.Text = Text1.Text & ttl & crlf

End If

If wh < 15 Then

addIn wh + 1

End If

ttl = ttl - addend

End Sub

Function mform$(x, t$)

a$ = Format$(x, t$)

If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$

mform$ = a$

End Function

After sorting the output, the following are the first 666 lines, to show the exemplified case:

This is Sloane's A005836

That site gives

a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0

but surely this must be on the basis that a(2)=1 and a(1)=0 also, making a(3)=3*a(2)=3, a(4)=a(3)+1=4, a(5)=3*a(3)=9, a(6)=a(5)+1=10, a(7)=3*a(4)=12, etc.

which could be used to build the above table, or recursively as follows:

The example given in the puzzle statement, the 666th number in the sequence, is a(667) -- only on the basis of that formula -- and by the above table is 21,981.

The recursion:

a(667)=3*a(334)

a(334)=a(333)+1

a(333)=3*a(167)

a(167)=3*a(84)

a(84)=a(83)+1

a(83)=3*a(42)

a(42)=a(41)+1

a(41)=3*a(21)

a(21)=3*a(11)

a(11)=3*a(6)

a(6)=a(5)+1

a(5)=3*a(3)

a(3)=3*a(2)

On the basis that a(2)=1 we can fill this in from bottom to top:

a(667)=3*a(334)       21981  -- last calculated
a(334)=a(333)+1        7327          ^
a(333)=3*a(167)        7326          |
a(167)=3*a(84)         2442          |
a(84)=a(83)+1           814          |
a(83)=3*a(42)           813          |
a(42)=a(41)+1           271          |
a(41)=3*a(21)           270          | 
a(21)=3*a(11)            90               |
a(11)=3*a(6)             30          |   
a(6)=a(5)+1              10          |
a(5)=3*a(3)               9          ^
a(3)=3*a(2)               3  -- first calculated

Posted by Charlie on 2015-11-20 14:45:19

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