A328059 - OEIS

%I #27 Dec 20 2019 23:38:45

%S 0,1,1,2,2,4,2,3,2,3,2,3,2,3,4,4,2,3,3,4,4,4,6,5,5,5,4,6,4,5,3,5,4,3,

%T 5,5,4,3,4,5,2,4,3,6,4,5,5,7,4,5,5,6,7,5,3,5,4,3,6,5,2,7,6,7,5,5,4,6,

%U 6,7,5,6,6,6,8,5,7,7,6,5,5,7,5,7,3,4,5,6,5,6,6,7,7,7,4,8,6,7,7,6,5

%N a(n) is the number of additions, subtractions and/or multiplications needed to calculate n using numbers a(n-a(n)) to a(n) in order and in single step calculations.

%H S. Brunner, <a href="/A328059/b328059.txt">Table of n, a(n) for n = 0..10000</a>

%H S. Brunner, <a href="https://pastebin.com/BDzgnvhP">Link for python program which shows calculation steps for each number</a>

%H S. Brunner, <a href="https://pastebin.com/T3umyseT">List for n = 0..5600 with calculation steps for each n</a>

%e a(10)=2 because you need 2 calculation steps to get to 10.

%e The numbers for the first calculation are: a(n-2)=a(8)=2, a(n-1)=a(9)=3.

%e We need to add these 2 numbers: 2+3=5 (1st step).

%e To get 10 as result we need to multiply 5 from step 1 with a(n) itself, so we get 5*2=10 (2nd step).

%e For n=34 we need 5 calculation steps, because there is no solution with 4 or fewer steps, but we pretend we tested these cases before.

%e The numbers we need to use {a(n-5) to a(n)} are: 5,3,5,4,3,5.

%e 1. 5+3=8 | 2. 8*5=40 | 3. 40-4=36 | 4. 36+3=39 | 5. 39-5=34.

%o (Python)

%o def A(z):

%o     a=[0,1,1]

%o     n=3

%o     an=2

%o     while n<=z:

%o         list1=[a[n-an]]

%o         nn=an-2

%o         while nn>0:

%o             l=len(list1)

%o             i=0

%o             while i<l:

%o                 v=list1[0]

%o                 list1.remove(v)

%o                 list1.append(v+a[n-nn-1])

%o                 if v-n<300:

%o                     list1.append(v*a[n-nn-1])

%o                 list1.append(v-a[n-nn-1])

%o                 i=i+1

%o             nn=nn-1

%o         c=list1.count(n-an-a[n-1])+list1.count(n/an-a[n-1])+list1.count(n+an-a[n-1])

%o         c=c+list1.count((n-an)/a[n-1])+list1.count((n/an)/a[n-1])+list1.count((n+an)/a[n-1])

%o         c=c+list1.count(n-an+a[n-1])+list1.count(n/an+a[n-1])+list1.count(n+an+a[n-1])

%o         if c==0:

%o             an=an+1

%o         else:

%o             a.append(an)

%o             print(n,an)

%o             n=n+1

%o             an=2

%o     return a

%K nonn

%O 0,4

%A _S. Brunner_, Oct 03 2019