1421131054333 has 2 divisors, whose sum is σ = 1421131054334. Its totient is φ = 1421131054332.

The previous prime is 1421131054319. The next prime is 1421131054351. The reversal of 1421131054333 is 3334501311241.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1217010525124 + 204120529209 = 1103182^2 + 451797^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1421131054333 is a prime.

It is a super-2 number, since 2×1421131054333² (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1421131054295 and 1421131054304.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1421131054303) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 710565527166 + 710565527167.

It is an arithmetic number, because the mean of its divisors is an integer number (710565527167).

Almost surely, 2^{1421131054333} is an apocalyptic number.

It is an amenable number.

1421131054333 is a deficient number, since it is larger than the sum of its proper divisors (1).

1421131054333 is an equidigital number, since it uses as much as digits as its factorization.

1421131054333 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 12960, while the sum is 31.

Adding to 1421131054333 its reverse (3334501311241), we get a palindrome (4755632365574).

The spelling of 1421131054333 in words is "one trillion, four hundred twenty-one billion, one hundred thirty-one million, fifty-four thousand, three hundred thirty-three".