Можем да опитаме да използваме календарна таблица тук, която включва всички възможни начални дати на месеца, които може да се появят в очаквания резултат:
with calendar as (
select '2017-09-01'::date as dt union all
select '2017-10-01'::date union all
select '2017-11-01'::date union all
select '2017-12-01'::date union all
select '2018-01-01'::date union all
select '2018-02-01'::date union all
select '2018-03-01'::date union all
select '2018-04-01'::date union all
select '2018-05-01'::date union all
select '2018-06-01'::date union all
select '2018-07-01'::date union all
select '2018-08-01'::date
)
select
t.id as subscription_id,
c.dt,
t.amount_monthly
from calendar c
inner join your_table t
on c.dt >= t.start_date and
c.dt < t.start_date + (t.month_count::text || ' month')::interval
order by
t.id,
c.dt;