Properties of the Number 30254

Thirty Thousand Two Hundred Fifty-Four

Basics

Value: 30253 → 30254 → 30255

Parity: even

Prime: No

Previous Prime: 30253

Next Prime: 30259

Digit Sum: 14

Digital Root: 5

Palindrome: No

Factorization: 2 × 7 × 2161

Divisors: 1, 2, 7, 14, 2161, 4322, 15127, 30254

Polygonal

Triangular: No

Square: No

Pentagonal: No

Hexagonal: No

Heptagonal: No

Tetrahedral: No

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Representations

Binary: 111011000101110

Octal: 73056

Duodecimal: 15612

Hexadecimal: 762e

Square: 915304516

Square Root: 173.93677012063895

Classification

Fibonacci: No

Bell Number: No

Factorial Number: No

Regular Number: No

Perfect Number: No

Special

Kaprekar Constant: No

Munchausen Constant: No

Armstrong Number: No

Kaprekar Number: No

Catalan Number: No

Vampire Number: No

Taxicab Number: No

Super Prime: No

Friedman Number: No

Fermat Number: No

Cullen Number: No

OEIS Sequences

Fibonacci sequence beginning 2, 6. A22112

Numbers k such that (k / sum of digits of k) and (k+1 / sum of digits of k+1) are both prime. A85775

Numbers k such that A003415(k) >= A276086(k) and gcd(k, A003415(k)) = gcd(k, A276086(k)), where A003415 is the arithmetic derivative, and A276086 is the primorial base exp-function. A369959

Numbers k such that (A276086(k)/s)^s >= k^s-1 and A276086(k) <= A003415(k), where A003415 is the arithmetic derivative, A276086 is the primorial base exp-function, and s = bigomega(k). A370128

a(n) = round(n*φ²⁰), where φ is the golden ratio, A001622. A4955

Numerators of continued fraction convergents to sqrt(500). A41954

a(n) = gcd(Lucas(n)+1, Fibonacci(n)+1). A115313

Gilda's numbers: numbers k such that if a Fibonacci sequence is formed with first term = a certain absolute value between decimal digits in k (A007953) and second term = sum of decimal digits in k (A040997), then k itself occurs as a term in the sequence. A42947

Sum of Lucas numbers and reflected Lucas numbers (comment to A061084). A75091

Half-indexed Lucas numbers second version L(n)=A000032=Lucas numbers a(0)=2, a(1)=2, a(2)=1, a(3)=2, a(4)=3, a(5)=3, a(2n)=L(n), for n>2: a(2n+1)=L(n)+L(n-3)=2·L(n-1) for n>5: a(n)+a(n+2)=a(n+4) a(2n)=L(n), so a(n)=L(n/2). A127218