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1444306497173 is a prime number
BaseRepresentation
bin10101000001000111011…
…000000010111010010101
312010002000012111001111122
4111001013120002322111
5142130414430402143
63023301044413325
7206230155444131
oct25010730027225
95102005431448
101444306497173
11507588401401
121b3abb3a7245
13a6274c41cc7
144dc94b123c1
1527882c60c68
hex15047602e95

1444306497173 has 2 divisors, whose sum is σ = 1444306497174. Its totient is φ = 1444306497172.

The previous prime is 1444306497169. The next prime is 1444306497211. The reversal of 1444306497173 is 3717946034441.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 901753454449 + 542553042724 = 949607^2 + 736582^2 .

It is a cyclic number.

It is not a de Polignac number, because 1444306497173 - 22 = 1444306497169 is a prime.

It is a super-3 number, since 3×14443064971733 (a number of 37 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1444306497103) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 722153248586 + 722153248587.

It is an arithmetic number, because the mean of its divisors is an integer number (722153248587).

Almost surely, 21444306497173 is an apocalyptic number.

It is an amenable number.

1444306497173 is a deficient number, since it is larger than the sum of its proper divisors (1).

1444306497173 is an equidigital number, since it uses as much as digits as its factorization.

1444306497173 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 6096384, while the sum is 53.

The spelling of 1444306497173 in words is "one trillion, four hundred forty-four billion, three hundred six million, four hundred ninety-seven thousand, one hundred seventy-three".