1. We need to count functions $f:A\to B$ where $A=\{1,2,3,4,5\},\qquad B=\{1,2,3,4,5,6\}$ and the condition is $f(1)+f(2)=f(4)-1.$

2. Rearrange the condition: $f(4)=f(1)+f(2)+1.$ So once $f(1)$ and $f(2)$ are chosen, $f(4)$ is fixed.

But since $f(4)\in B=\{1,2,3,4,5,6\}$, we must have $1\le f(1)+f(2)+1\le 6.$ Because $f(1),f(2)\ge 1$, the lower bound is automatic. Hence we only need $f(1)+f(2)\le 5.$

3. Count ordered pairs $(f(1),f(2))$ with values in $\{1,2,3,4,5,6\}$ such that $f(1)+f(2)\le 5.$

Possible sums:

• Sum $2$: $(1,1)$ → $1$ pair
• Sum $3$: $(1,2),(2,1)$ → $2$ pairs
• Sum $4$: $(1,3),(2,2),(3,1)$ → $3$ pairs
• Sum $5$: $(1,4),(2,3),(3,2),(4,1)$ → $4$ pairs

Total number of valid ordered pairs: $1+2+3+4=10.$

4. For each such valid choice of $f(1),f(2)$, the value of $f(4)$ is uniquely determined by $f(4)=f(1)+f(2)+1.$

5. The values $f(3)$ and $f(5)$ are unrestricted, each can be any of the $6$ elements of $B$. So number of choices for these is $6\times 6=36.$

6. Therefore total number of functions is $10\times 36=360.$

Thus the required integer is $\boxed{360}.$