Base | Representation |
---|---|
bin | 10011000111000111100… |
… | …011111011100011110001 |
3 | 11122112220001022112012211 |
4 | 103013013203323203301 |
5 | 133004131124012012 |
6 | 2443154441452121 |
7 | 163612060151041 |
oct | 23070743734361 |
9 | 4575801275184 |
10 | 1313313110257 |
11 | 466a78015a55 |
12 | 192641aa1041 |
13 | 96aca307385 |
14 | 477c973b721 |
15 | 24267ac56a7 |
hex | 131c78fb8f1 |
1313313110257 has 2 divisors, whose sum is σ = 1313313110258. Its totient is φ = 1313313110256.
The previous prime is 1313313110177. The next prime is 1313313110269. The reversal of 1313313110257 is 7520113133131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 995317494336 + 317995615921 = 997656^2 + 563911^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1313313110257 is a prime.
It is not a weakly prime, because it can be changed into another prime (1313313110657) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 656656555128 + 656656555129.
It is an arithmetic number, because the mean of its divisors is an integer number (656656555129).
Almost surely, 21313313110257 is an apocalyptic number.
It is an amenable number.
1313313110257 is a deficient number, since it is larger than the sum of its proper divisors (1).
1313313110257 is an equidigital number, since it uses as much as digits as its factorization.
1313313110257 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5670, while the sum is 31.
Adding to 1313313110257 its reverse (7520113133131), we get a palindrome (8833426243388).
The spelling of 1313313110257 in words is "one trillion, three hundred thirteen billion, three hundred thirteen million, one hundred ten thousand, two hundred fifty-seven".
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