1535400131 has 2 divisors, whose sum is σ = 1535400132. Its totient is φ = 1535400130.

The previous prime is 1535400107. The next prime is 1535400133. The reversal of 1535400131 is 1310045351.

It is a strong prime.

It is an emirp because it is prime and its reverse (1310045351) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1535400131 - 2¹⁸ = 1535137987 is a prime.

It is a super-2 number, since 2×1535400131² = 4714907124549634322, which contains 22 as substring.

Together with 1535400133, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 1535400097 and 1535400106.

It is not a weakly prime, because it can be changed into another prime (1535400133) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 767700065 + 767700066.

It is an arithmetic number, because the mean of its divisors is an integer number (767700066).

Almost surely, 2^1535400131 is an apocalyptic number.

1535400131 is a deficient number, since it is larger than the sum of its proper divisors (1).

1535400131 is an equidigital number, since it uses as much as digits as its factorization.

1535400131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 900, while the sum is 23.

The square root of 1535400131 is about 39184.1821530066. The cubic root of 1535400131 is about 1153.6494351594.

Adding to 1535400131 its reverse (1310045351), we get a palindrome (2845445482).

The spelling of 1535400131 in words is "one billion, five hundred thirty-five million, four hundred thousand, one hundred thirty-one".