Indiana · United States
Seymour population
19,478people · latest figure
Seymour has a population of 19,478, making it the 2290th largest place we list in United States. It is in Indiana.
Seymour holds under a tenth of one per cent of United States's 349,035,494 people. New York City, the largest place we list there, is 452 times its size. Twenty years earlier the World Gazetteer recorded 18,730, a rise of 4%.
The 2005 figure comes from the World Gazetteer archive restored on this site. The two sources may draw the settlement's boundary differently, so treat the change as indicative rather than exact. Why figures differ.
Seymour is in Indiana, which has 493 recorded places.
Nearby in size
| Place | Region | Population |
|---|---|---|
| Papillion | Nebraska | 19,510 |
| Montgomery | Illinois | 19,489 |
| Badger | Alaska | 19,482 |
| Lake Shore | Maryland | 19,477 |
| McKeesport | Pennsylvania | 19,453 |
| Weirton Heights | West Virginia | 19,450 |
All places in United States → · Indiana →
Figures are the latest available in GeoNames (CC BY 4.0), which draws on national censuses of varying dates — treat them as recent, not same-year. Country totals are UN World Population Prospects 2024 (medium variant) where the UN lists the country separately; the handful it reports under a parent state keep a GeoNames figure and are marked as such. Sources & method.