A328059 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.

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, 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, 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

Offset

0,4

Links

S. Brunner, Table of n, a(n) for n = 0..10000

S. Brunner, Link for python program which shows calculation steps for each number

S. Brunner, List for n = 0..5600 with calculation steps for each n

Example

a(10)=2 because you need 2 calculation steps to get to 10.
The numbers for the first calculation are: a(n-2)=a(8)=2, a(n-1)=a(9)=3.
We need to add these 2 numbers: 2+3=5 (1st step).
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).
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.
The numbers we need to use {a(n-5) to a(n)} are: 5,3,5,4,3,5.
1. 5+3=8 | 2. 8*5=40 | 3. 40-4=36 | 4. 36+3=39 | 5. 39-5=34.

Prog

(Python)
def A(z):
    a=[0, 1, 1]
    n=3
    an=2
    while n<=z:
        list1=[a[n-an]]
        nn=an-2
        while nn>0:
            l=len(list1)
            i=0
            while i<l:
                v=list1[0]
                list1.remove(v)
                list1.append(v+a[n-nn-1])
                if v-n<300:
                    list1.append(v*a[n-nn-1])
                list1.append(v-a[n-nn-1])
                i=i+1
            nn=nn-1
        c=list1.count(n-an-a[n-1])+list1.count(n/an-a[n-1])+list1.count(n+an-a[n-1])
        c=c+list1.count((n-an)/a[n-1])+list1.count((n/an)/a[n-1])+list1.count((n+an)/a[n-1])
        c=c+list1.count(n-an+a[n-1])+list1.count(n/an+a[n-1])+list1.count(n+an+a[n-1])
        if c==0:
            an=an+1
        else:
            a.append(an)
            print(n, an)
            n=n+1
            an=2
    return a

Crossrefs

Sequence in context: A131287 A102640 A332581 * A123674 A363219 A371531

Adjacent sequences: A328056 A328057 A328058 * A328060 A328061 A328062

Keyword

nonn

Author

S. Brunner, Oct 03 2019

Status

approved