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1311513215443 is a prime number
BaseRepresentation
bin10011000101011100010…
…001111000000111010011
311122101020121111101200221
4103011130101320013103
5132441434340343233
62442300103453511
7163516343252224
oct23053421700723
94571217441627
101311513215443
11466234023382
1219221b157297
13968a1453162
144769869124b
15241aea8d42d
hex1315c4781d3

1311513215443 has 2 divisors, whose sum is σ = 1311513215444. Its totient is φ = 1311513215442.

The previous prime is 1311513215431. The next prime is 1311513215447. The reversal of 1311513215443 is 3445123151131.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1311513215443 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 1311513215399 and 1311513215408.

It is not a weakly prime, because it can be changed into another prime (1311513215447) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655756607721 + 655756607722.

It is an arithmetic number, because the mean of its divisors is an integer number (655756607722).

Almost surely, 21311513215443 is an apocalyptic number.

1311513215443 is a deficient number, since it is larger than the sum of its proper divisors (1).

1311513215443 is an equidigital number, since it uses as much as digits as its factorization.

1311513215443 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 21600, while the sum is 34.

Adding to 1311513215443 its reverse (3445123151131), we get a palindrome (4756636366574).

The spelling of 1311513215443 in words is "one trillion, three hundred eleven billion, five hundred thirteen million, two hundred fifteen thousand, four hundred forty-three".