A358096 - OEIS

A358096

a(n) is the number of ways n can be reached in the algorithm explained in A358094 if the last operation is multiplication.

1, 1, 1, 0, 0, 1, 0, 2, 1, 2, 0, 3, 0, 0, 2, 1, 0, 2, 0, 2, 0, 3, 0, 4, 0, 2, 1, 3, 0, 5, 0, 0, 3, 2, 0, 6, 0, 1, 2, 2, 0, 5, 0, 2, 3, 2, 0, 3, 0, 3, 2, 4, 0, 7, 0, 2, 1, 3, 0, 6, 0, 3, 2, 5, 0, 7, 0, 0, 2, 3, 0, 8, 0, 2, 3, 6, 0, 10, 0, 1

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OFFSET

1,8

LINKS

Yifan Xie, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = A358095(n/2) + A358095(n/3) if n == 0 (mod 6);

a(n) = A358095(n/2) if n == 2 or 4 (mod 6);

a(n) = A358095(n/3) if n == 3 (mod 6);

a(n) = 0 if n == 1 or 5 (mod 6).

EXAMPLE

There are 3 ways to reach 12: (1*3+3)*2=12, (1*2+2)*3=12 and (1+3)*3=12.

PROG

(C++) #include <iostream>

using namespace std; int f(int x, bool y) { if(x<0) return 0; if(x==1) return 1; if(y==0) return f(x-2, 1)+f(x-3, 1); if(y==1) { if(x%6==0) return f(x/2, 0)+f(x/3, 0); if(x%6==1||x%6==5) return 0; if(x%6==2||x%6==4) return f(x/2, 0); if(x%6==3) return f(x/3, 0); } } int n; int main() { cin>>n; cout<<1<<", "; for(int i=2; i<n; i++) cout<<f(i, 1)<<", "; cout<<f(n, 1); return 0; }

CROSSREFS

Cf. A358094, A358095.

Sequence in context: A339898 A273163 A276695 * A071485 A127969 A081733

Adjacent sequences: A358093 A358094 A358095 * A358097 A358098 A358099

KEYWORD

nonn,easy

AUTHOR

Yifan Xie, Nov 01 2022

STATUS

approved