1421130001 has 4 divisors (see below), whose sum is σ = 1422031600. Its totient is φ = 1420228404.

The previous prime is 1421129971. The next prime is 1421130013. The reversal of 1421130001 is 1000311241.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also an emirpimes, since its reverse is a distinct semiprime: 1000311241 = 53 ⋅18873797.

It is a cyclic number.

It is not a de Polignac number, because 1421130001 - 2¹⁵ = 1421097233 is a prime.

It is a super-2 number, since 2×1421130001² = 4039220959484520002, which contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (1421130031) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 448431 + ... + 451588.

It is an arithmetic number, because the mean of its divisors is an integer number (355507900).

Almost surely, 2^1421130001 is an apocalyptic number.

It is an amenable number.

1421130001 is a deficient number, since it is larger than the sum of its proper divisors (901599).

1421130001 is an equidigital number, since it uses as much as digits as its factorization.

1421130001 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 901598.

The product of its (nonzero) digits is 24, while the sum is 13.

The square root of 1421130001 is about 37697.8779376240. The cubic root of 1421130001 is about 1124.2889336382.

Adding to 1421130001 its reverse (1000311241), we get a palindrome (2421441242).

The spelling of 1421130001 in words is "one billion, four hundred twenty-one million, one hundred thirty thousand, one".