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3954066044953 is a prime number
BaseRepresentation
bin111001100010100000101…
…100111011100000011001
3112000000010122212000101211
4321202200230323200121
51004240411401414303
612224250045543121
7555446304555613
oct71424054734031
915000118760354
103954066044953
111294a01222352
1253a3a8874aa1
13228b3656298b
14d9540a35db3
156ccc358466d
hex398a0b3b819

3954066044953 has 2 divisors, whose sum is σ = 3954066044954. Its totient is φ = 3954066044952.

The previous prime is 3954066044837. The next prime is 3954066045031. The reversal of 3954066044953 is 3594406604593.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 3616508530944 + 337557514009 = 1901712^2 + 580997^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-3954066044953 is a prime.

It is a super-3 number, since 3×39540660449533 (a number of 39 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (3954066045953) 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, 1977033022476 + 1977033022477.

It is an arithmetic number, because the mean of its divisors is an integer number (1977033022477).

Almost surely, 23954066044953 is an apocalyptic number.

It is an amenable number.

3954066044953 is a deficient number, since it is larger than the sum of its proper divisors (1).

3954066044953 is an equidigital number, since it uses as much as digits as its factorization.

3954066044953 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 41990400, while the sum is 58.

The spelling of 3954066044953 in words is "three trillion, nine hundred fifty-four billion, sixty-six million, forty-four thousand, nine hundred fifty-three".