ROBERT J. ANCONA MD – NPI #1346342680
Pediatrics

A pediatrician trained to care for children in the diagnosis, treatment and prevention of infectious diseases. This specialist can apply specific knowledge to affect a better outcome for pediatric infections with complicated courses, underlying diseases that predispose to unusual or severe infections, unclear diagnoses, uncommon diseases and complex or investigational treatments.

ROBERT ANCONA is a physician located in BALTIMORE, MD. NPPES has assigned the NPI number 1346342680 to ROBERT ANCONA on September 05, 2006. It is a Type-1 NPI, indicating this NPI number is associated with an individual. The primary taxonomy selected by this provider is 2080P0208X from the Health Care Provider Taxonomy code set, which is classified as Pediatrics, specializing in Pediatric Infectious Diseases

The NPI profile was previously updated about 19 years ago on Jul 08, 2007. Use the NPI data found here to bill health insurance companies, identify providers enrolled in Medicare and Medicaid services or other HIPAA compliant transactions. See the complete NPI profile for ROBERT ANCONA below.

NPI Profile for
ROBERT J. ANCONA

NPI Number
1346342680
Enumeration Date

(about 20 years ago)
Entity Type
Type-1  Individual (Male)
Legal Name
ROBERT J. ANCONA
Credentials
MD
Primary location
2 HAMILL ROAD
SUITE 405
BALTIMORE, MD 21210
Phone: (410) 323-1144 Fax: (410) 323-6161
Mailing address
Same as primary location
Sole Proprietor
No
Updated
Taxonomy Code(s)

A taxonomy code is a code that describes the health care service provider's type, classification, and the area of specialization. The primary specialty for this provider is indicated as (Primary) below.

Taxonomy Classification / Specialization State License
2080P0208X
- Pediatrics / Pediatric Infectious Diseases (Primary)
MD D0023664

The taxonomy codes are selected by the provider at the time of NPI registration. Selection of a taxonomy code does not replace any credentialing or validation process that the provider requesting the code should complete.