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131401322421217 is a prime number
BaseRepresentation
bin11101111000001001000001…
…001001010101101111100001
3122020020211210002101001010021
4131320021001021111233201
5114210334402014434332
61143244533440433441
736451301156564122
oct3570110111255741
9566224702331107
10131401322421217
1138960a95977112
12128a2559018881
135842126147504
142463c02284049
15102d0b3876c97
hex778241255be1

131401322421217 has 2 divisors, whose sum is σ = 131401322421218. Its totient is φ = 131401322421216.

The previous prime is 131401322421191. The next prime is 131401322421277. The reversal of 131401322421217 is 712124223104131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 93309108592896 + 38092213828321 = 9659664^2 + 6171889^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-131401322421217 is a prime.

It is a super-2 number, since 2×1314013224212172 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (131401322421277) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65700661210608 + 65700661210609.

It is an arithmetic number, because the mean of its divisors is an integer number (65700661210609).

Almost surely, 2131401322421217 is an apocalyptic number.

It is an amenable number.

131401322421217 is a deficient number, since it is larger than the sum of its proper divisors (1).

131401322421217 is an equidigital number, since it uses as much as digits as its factorization.

131401322421217 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 16128, while the sum is 34.

Adding to 131401322421217 its reverse (712124223104131), we get a palindrome (843525545525348).

The spelling of 131401322421217 in words is "one hundred thirty-one trillion, four hundred one billion, three hundred twenty-two million, four hundred twenty-one thousand, two hundred seventeen".