1211004955507 has 2 divisors, whose sum is σ = 1211004955508. Its totient is φ = 1211004955506.

The previous prime is 1211004955379. The next prime is 1211004955541. The reversal of 1211004955507 is 7055594001121.

It is a strong prime.

It is an emirp because it is prime and its reverse (7055594001121) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1211004955507 - 2⁷ = 1211004955379 is a prime.

It is a super-2 number, since 2×1211004955507² (a number of 25 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 1211004955507.

It is not a weakly prime, because it can be changed into another prime (1211004955577) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 605502477753 + 605502477754.

It is an arithmetic number, because the mean of its divisors is an integer number (605502477754).

Almost surely, 2^{1211004955507} is an apocalyptic number.

1211004955507 is a deficient number, since it is larger than the sum of its proper divisors (1).

1211004955507 is an equidigital number, since it uses as much as digits as its factorization.

1211004955507 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 63000, while the sum is 40.

Adding to 1211004955507 its reverse (7055594001121), we get a palindrome (8266598956628).

The spelling of 1211004955507 in words is "one trillion, two hundred eleven billion, four million, nine hundred fifty-five thousand, five hundred seven".