143121001011349 has 2 divisors, whose sum is σ = 143121001011350. Its totient is φ = 143121001011348.

The previous prime is 143121001011301. The next prime is 143121001011397. The reversal of 143121001011349 is 943110100121341.

It is a balanced prime because it is at equal distance from previous prime (143121001011301) and next prime (143121001011397).

It can be written as a sum of positive squares in only one way, i.e., 131970697622500 + 11150303388849 = 11487850^2 + 3339207^2 .

It is a cyclic number.

It is not a de Polignac number, because 143121001011349 - 2¹³ = 143121001003157 is a prime.

It is a super-3 number, since 3×143121001011349³ (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a self number, because there is not a number n which added to its sum of digits gives 143121001011349.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (143121001011749) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71560500505674 + 71560500505675.

It is an arithmetic number, because the mean of its divisors is an integer number (71560500505675).

Almost surely, 2^{143121001011349} is an apocalyptic number.

It is an amenable number.

143121001011349 is a deficient number, since it is larger than the sum of its proper divisors (1).

143121001011349 is an equidigital number, since it uses as much as digits as its factorization.

143121001011349 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2592, while the sum is 31.

The spelling of 143121001011349 in words is "one hundred forty-three trillion, one hundred twenty-one billion, one million, eleven thousand, three hundred forty-nine".