9

From ProofWiki
Jump to navigation Jump to search

Previous  ... Next

Number

$9$ (nine) is:

$3^2$


The $1$st odd prime power:
$9 = 3^2$


The $1$st power of $9$ after the zeroth $1$:
$9 = 9^1$


The larger of the $1$st pair of consecutive powerful numbers:
$8 = 2^3$, $9 = 3^2$


The $2$nd power of $3$ after $(1)$, $3$:
$9 = 3^2$


The sum of the first $2$ cubes:
$9 = 1^3 + 2^3$


The $2$nd square lucky number after $1$:
$1$, $9$, $\ldots$


The $2$nd Kaprekar number after $1$:
$9^2 = 81 \to 8 + 1 = 9$


The $2$nd integer after $1$ whose square has a $\sigma$ value which is itself square:
$\sigma \left({9^2}\right) = 11^2$


The $3$rd square number after $1$, $4$:
$9 = 3^2$
and therefore from Sum of Consecutive Triangular Numbers is Square, the sum of $2$ consecutive triangular numbers:
$9 = 3 + 6$


The $3$rd semiprime after $4$, $6$:
$9 = 3 \times 3$


The $3$rd Cullen number after $1$, $3$:
$9 = 2 \times 2^2 + 1$


The $3$rd square after $1$, $4$ which has no more than $2$ distinct digits


The sum of the first $3$ factorials:
$9 = 1! + 2! + 3!$


The $3$rd of $35$ integers less than $91$ to which $91$ itself is a Fermat pseudoprime:
$3$, $4$, $9$, $\ldots$


The $4$th powerful number after $1$, $4$, $8$


The $4$th lucky number:
$1$, $3$, $7$, $9$, $\ldots$


The $4$th palindromic lucky number:
$1$, $3$, $7$, $9$, $\ldots$


The $4$th subfactorial after $0$, $1$, $2$:
$9 = 4! \left({1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \dfrac 1 {4!} }\right)$


The $5$th trimorphic number after $1$, $4$, $5$, $6$:
$9^3 = 72 \mathbf 9$


The $6$th integer after $0$, $1$, $3$, $5$, $7$ which is palindromic in both decimal and binary:
$9_{10} = 1001_2$


The $6$th positive integer after $2$, $3$, $4$, $7$, $8$ which cannot be expressed as the sum of distinct pentagonal numbers


The $7$th after $1$, $2$, $4$, $5$, $6$, $8$ of the $24$ positive integers which cannot be expressed as the sum of distinct non-pythagorean primes


The $7$th (strictly) positive integer after $1$, $2$, $3$, $4$, $6$, $7$ which cannot be expressed as the sum of exactly $5$ non-zero squares


The $9$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$ such that $5^n$ contains no zero in its decimal representation:
$5^9 = 78 \, 125$


The $9$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$ such that both $2^n$ and $5^n$ have no zeroes:
$2^9 = 512$, $5^9 = 1 \, 953 \, 125$


The $9$th of the trivial $1$-digit pluperfect digital invariants after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$:
$9^1 = 9$


The $9$th of the (trivial $1$-digit) Zuckerman numbers after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$:
$9 = 1 \times 9$


The $9$th of the (trivial $1$-digit) harshad numbers after $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$:
$9 = 1 \times 9$


The $10$th integer $n$ after $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ such that $2^n$ contains no zero in its decimal representation:
$2^9 = 512$


The $10$th integer after $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ which is (trivially) the sum of the increasing powers of its digits taken in order:
$9^1 = 9$


One of the cycle of $5$ numbers (when prepended with zero) to which Kaprekar's process on $2$-digit numbers converges:
$09 \to 81 \to 63 \to 27 \to 45 \to 09$


Every positive integer can be expressed as the sum of at most $9$ positive cubes


The magic constant of a magic cube of order $2$ (if it were to exist), after $1$:
$9 = \displaystyle \dfrac 1 {2^2} \sum_{k \mathop = 1}^{2^3} k = \dfrac {2 \paren {2^3 + 1} } 2$


In ternary:
$100_3 = 9_{10}$


Also see


Previous in Sequence: $1$


Previous in Sequence: $2$


Previous in Sequence: $3$


Previous in Sequence: $4$


Previous in Sequence: $6$


Previous in Sequence: $7$


Previous in Sequence: $8$


Previous in Sequence: $45$


Next in Sequence: $20$ and above


Sources