130040114261 has 2 divisors, whose sum is σ = 130040114262. Its totient is φ = 130040114260.

The previous prime is 130040114219. The next prime is 130040114287. The reversal of 130040114261 is 162411040031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 83111124100 + 46928990161 = 288290^2 + 216631^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-130040114261 is a prime.

It is a super-2 number, since 2×130040114261² (a number of 23 digits) contains 22 as substring.

It is a Sophie Germain prime.

It is a Curzon number.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (130040114761) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65020057130 + 65020057131.

It is an arithmetic number, because the mean of its divisors is an integer number (65020057131).

Almost surely, 2^130040114261 is an apocalyptic number.

It is an amenable number.

130040114261 is a deficient number, since it is larger than the sum of its proper divisors (1).

130040114261 is an equidigital number, since it uses as much as digits as its factorization.

130040114261 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 576, while the sum is 23.

Adding to 130040114261 its reverse (162411040031), we get a palindrome (292451154292).

The spelling of 130040114261 in words is "one hundred thirty billion, forty million, one hundred fourteen thousand, two hundred sixty-one".