I can calculate power I need that is 1300*24*3600*7 = 786.246MW

Where does the 3600 come from in that equation? There are 24 hours in a day, not 3600. There are 3600 seconds in an hour, but that's not relevant unless you're calculating joules.

When you multiply a power figure (in watts or kilowatts) by a period of time measured in hours, the units become Wh (watt-hours) or kWh (kilowatt-hours), i.e. quantities of energy, not power. Alternatively you can express energy in joules. One joule is one watt for one second, so 1 Wh (one watt-hour) is 3600 joules.

You have a computer that draws 1.3 kW. If you run it for 24 hours, it will use (1.3 × 24) = 31.2 kWh, or 112.32 MJ (megajoules). If you run it for seven full days, it will use (7 × 31.2) = 218.4 kWh, or 786.24 MJ.

Battery capacities are specified in amp-hours. Multiply the battery voltage (in volts) by its capacity (in amp-hours) to get watt-hours.

Assuming you have a 24V battery and you need 218,400 Wh, it will need to have a capacity of 9100 Ah. You might be able to achieve this with a number of truck batteries in parallel; I don't know.

I couldn't understand your comment about charging for six hours. Those calculations assume no recharging for a week.

All of those numbers are far more approximate than they look. There are inefficiencies and approximations involved.

I would first try to reduce your computer's power consumption. Many fast laptops consume less than 100 watts. Next, you can reduce losses by using a computer power supply that can run directly from the battery, instead of using an inverter to step the battery voltage up to AC mains voltage.