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130441032101453 is a prime number
BaseRepresentation
bin11101101010001010101011…
…011000011110111001001101
3122002212001002210201101021212
4131222022223120132321031
5114044121213204221303
61141231441034304205
736322024242544433
oct3552125330367115
9562761083641255
10130441032101453
1138620805037777
1212768419888065
1357a26c8a425a8
14242d545a34d53
15101310d39bad8
hex76a2ab61ee4d

130441032101453 has 2 divisors, whose sum is σ = 130441032101454. Its totient is φ = 130441032101452.

The previous prime is 130441032101431. The next prime is 130441032101459. The reversal of 130441032101453 is 354101230144031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 93690435901924 + 36750596199529 = 9679382^2 + 6062227^2 .

It is a cyclic number.

It is not a de Polignac number, because 130441032101453 - 216 = 130441032035917 is a prime.

It is a super-2 number, since 2×1304410321014532 (a number of 29 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (130441032101459) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65220516050726 + 65220516050727.

It is an arithmetic number, because the mean of its divisors is an integer number (65220516050727).

Almost surely, 2130441032101453 is an apocalyptic number.

It is an amenable number.

130441032101453 is a deficient number, since it is larger than the sum of its proper divisors (1).

130441032101453 is an equidigital number, since it uses as much as digits as its factorization.

130441032101453 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 17280, while the sum is 32.

Adding to 130441032101453 its reverse (354101230144031), we get a palindrome (484542262245484).

The spelling of 130441032101453 in words is "one hundred thirty trillion, four hundred forty-one billion, thirty-two million, one hundred one thousand, four hundred fifty-three".