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4909287731281 is a prime number
BaseRepresentation
bin100011101110000100001…
…0110011111000001010001
3122101022201200222102021221
41013130020112133001101
51120413210124400111
614235143505445041
71014453504125323
oct107341026370121
918338650872257
104909287731281
111623020a3743a
12673552640781
13297c3682850b
1412d87a039613
1587a7da9c371
hex4770859f051

4909287731281 has 2 divisors, whose sum is σ = 4909287731282. Its totient is φ = 4909287731280.

The previous prime is 4909287731263. The next prime is 4909287731287. The reversal of 4909287731281 is 1821377829094.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2499295399056 + 2409992332225 = 1580916^2 + 1552415^2 .

It is a cyclic number.

It is not a de Polignac number, because 4909287731281 - 229 = 4908750860369 is a prime.

It is a super-3 number, since 3×49092877312813 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (4909287731287) 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, 2454643865640 + 2454643865641.

It is an arithmetic number, because the mean of its divisors is an integer number (2454643865641).

Almost surely, 24909287731281 is an apocalyptic number.

It is an amenable number.

4909287731281 is a deficient number, since it is larger than the sum of its proper divisors (1).

4909287731281 is an equidigital number, since it uses as much as digits as its factorization.

4909287731281 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12192768, while the sum is 61.

The spelling of 4909287731281 in words is "four trillion, nine hundred nine billion, two hundred eighty-seven million, seven hundred thirty-one thousand, two hundred eighty-one".