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13044000000001 = 35713652758331
BaseRepresentation
bin1011110111010000101011…
…1111001000100000000001
31201011222211210021010020011
42331310022333020200001
53202203103000000001
643424155245144521
72514253323415645
oct275641277104001
951158753233204
1013044000000001
114179a24a41a63
12156802125b141
1373807565c389
14331493946c25
151794873e3d51
hexbdd0afc8801

13044000000001 has 4 divisors (see below), whose sum is σ = 13047652761904. Its totient is φ = 13040347238100.

The previous prime is 13043999999969. The next prime is 13044000000049. The reversal of 13044000000001 is 10000000044031.

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.

It is a cyclic number.

It is not a de Polignac number, because 13044000000001 - 25 = 13043999999969 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (13044000000061) 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, 1826375595 + ... + 1826382736.

It is an arithmetic number, because the mean of its divisors is an integer number (3261913190476).

Almost surely, 213044000000001 is an apocalyptic number.

It is an amenable number.

13044000000001 is a deficient number, since it is larger than the sum of its proper divisors (3652761903).

13044000000001 is an equidigital number, since it uses as much as digits as its factorization.

13044000000001 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3652761902.

The product of its (nonzero) digits is 48, while the sum is 13.

Adding to 13044000000001 its reverse (10000000044031), we get a palindrome (23044000044032).

The spelling of 13044000000001 in words is "thirteen trillion, forty-four billion, one", and thus it is an aban number.

Divisors: 1 3571 3652758331 13044000000001