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131044404122417 is a prime number
BaseRepresentation
bin11101110010111100100111…
…001010000101001100110001
3122011222201111211121211221102
4131302330213022011030301
5114134012414423404132
61142412545042534145
736413434332236402
oct3562744712051461
9564881454554842
10131044404122417
113883368a896555
1212845349967355
13581758517a776
142450823a819a9
151023b74296c62
hex772f27285331

131044404122417 has 2 divisors, whose sum is σ = 131044404122418. Its totient is φ = 131044404122416.

The previous prime is 131044404122389. The next prime is 131044404122419. The reversal of 131044404122417 is 714221404440131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 129578357796001 + 1466046326416 = 11383249^2 + 1210804^2 .

It is a cyclic number.

It is not a de Polignac number, because 131044404122417 - 26 = 131044404122353 is a prime.

Together with 131044404122419, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (131044404122419) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65522202061208 + 65522202061209.

It is an arithmetic number, because the mean of its divisors is an integer number (65522202061209).

Almost surely, 2131044404122417 is an apocalyptic number.

It is an amenable number.

131044404122417 is a deficient number, since it is larger than the sum of its proper divisors (1).

131044404122417 is an equidigital number, since it uses as much as digits as its factorization.

131044404122417 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 86016, while the sum is 38.

Adding to 131044404122417 its reverse (714221404440131), we get a palindrome (845265808562548).

The spelling of 131044404122417 in words is "one hundred thirty-one trillion, forty-four billion, four hundred four million, one hundred twenty-two thousand, four hundred seventeen".