Search a number
-
+
130053201349 is a prime number
BaseRepresentation
bin111100100011111000…
…1100101110111000101
3110102200122001011011011
41321013301211313011
54112322104420344
6135425015242221
712252560654221
oct1710761456705
9412618034134
10130053201349
1150178707095
1221256679371
13c357c11116
14641a597781
1535b2871d34
hex1e47c65dc5

130053201349 has 2 divisors, whose sum is σ = 130053201350. Its totient is φ = 130053201348.

The previous prime is 130053201337. The next prime is 130053201353. The reversal of 130053201349 is 943102350031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 107734932900 + 22318268449 = 328230^2 + 149393^2 .

It is an emirp because it is prime and its reverse (943102350031) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-130053201349 is a prime.

It is a super-2 number, since 2×1300532013492 (a number of 23 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 130053201349.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (130053201149) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65026600674 + 65026600675.

It is an arithmetic number, because the mean of its divisors is an integer number (65026600675).

Almost surely, 2130053201349 is an apocalyptic number.

It is an amenable number.

130053201349 is a deficient number, since it is larger than the sum of its proper divisors (1).

130053201349 is an equidigital number, since it uses as much as digits as its factorization.

130053201349 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 9720, while the sum is 31.

The spelling of 130053201349 in words is "one hundred thirty billion, fifty-three million, two hundred one thousand, three hundred forty-nine".