Base | Representation |
---|---|
bin | 10011000101011101010… |
… | …110001111111011001011 |
3 | 11122101021212012222212121 |
4 | 103011131112033323023 |
5 | 132442003432420023 |
6 | 2442301543552111 |
7 | 163516651511626 |
oct | 23053526177313 |
9 | 4571255188777 |
10 | 1311531138763 |
11 | 46624315543a |
12 | 19222515b637 |
13 | 968a508a281 |
14 | 4769abd8dbd |
15 | 241b142dd5d |
hex | 1315d58fecb |
1311531138763 has 2 divisors, whose sum is σ = 1311531138764. Its totient is φ = 1311531138762.
The previous prime is 1311531138761. The next prime is 1311531138817. The reversal of 1311531138763 is 3678311351131.
It is a weak prime.
It is an emirp because it is prime and its reverse (3678311351131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1311531138763 - 21 = 1311531138761 is a prime.
It is a super-2 number, since 2×13115311387632 (a number of 25 digits) contains 22 as substring.
Together with 1311531138761, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (1311531138761) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655765569381 + 655765569382.
It is an arithmetic number, because the mean of its divisors is an integer number (655765569382).
Almost surely, 21311531138763 is an apocalyptic number.
1311531138763 is a deficient number, since it is larger than the sum of its proper divisors (1).
1311531138763 is an equidigital number, since it uses as much as digits as its factorization.
1311531138763 is an evil number, because the sum of its binary digits is even.
The product of its digits is 136080, while the sum is 43.
Adding to 1311531138763 its reverse (3678311351131), we get a palindrome (4989842489894).
The spelling of 1311531138763 in words is "one trillion, three hundred eleven billion, five hundred thirty-one million, one hundred thirty-eight thousand, seven hundred sixty-three".
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