55

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Number

$55$ (fifty-five) is:

$5 \times 11$


The $1$st of the $3$ repdigit numbers which are also triangular


The $1$st of the $4$ cubic recurring digital invariants:
$55 \to 250 \to 133 \to 55$


The $2$nd square pyramorphic number after $1$:
$55 = \displaystyle \sum_{k \mathop = 1}^5 k^2 = \dfrac {5 \paren {5 + 1} \paren {2 \times 5 + 1} } 6$


The $2$nd after $1$ of the $4$ square pyramidal numbers which are also triangular


The $4$th Kaprekar number after $1$, $9$, $45$:
$55^2 = 3025 \to 30 + 25 = 55$


The $5$th heptagonal number after $1$, $7$, $18$, $34$:
$55 = 1 + 7 + 11 + 16 + 21 = \dfrac {5 \paren {5 \times 5 - 3} } 2$


The $5$th square pyramidal number after $1$, $5$, $14$, $30$:
$55 = 1 + 4 + 9 + 14 + 25 = \dfrac {5 \paren {5 + 1} \paren {2 \times 5 + 1} } 6$


The $5$th and last after $0$, $1$, $3$, $21$ of the $5$ Fibonacci numbers which are also triangular


The $5$th palindromic triangular number after $0$, $1$, $3$, $6$
$55 = \dfrac {10 \times \paren {10 + 1} } 2$
The $1$st of those which can be split into two palindromic halves


The $10$th Fibonacci number, after $1$, $1$, $2$, $3$, $5$, $8$, $13$, $21$, $34$:
$55 = 21 + 34$


The $10$th triangular number after $1$, $3$, $6$, $10$, $15$, $21$, $28$, $36$, $45$:
$55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = \dfrac {10 \times \paren {10 + 1} } 2$


The $19$th semiprime after $4$, $6$, $9$, $10$, $14$, $15$, $21$, $22$, $25$, $26$, $33$, $34$, $35$, $38$, $39$, $46$, $49$, $51$:
$55 = 5 \times 11$


The $21$st of $35$ integers less than $91$ to which $91$ itself is a Fermat pseudoprime:
$3$, $4$, $9$, $10$, $12$, $16$, $17$, $22$, $23$, $25$, $27$, $29$, $30$, $36$, $38$, $40$, $43$, $48$, $51$, $53$, $55$, $\ldots$


The $24$th and last after $1$, $2$, $4$, $5$, $6$, $8$, $9$, $12$, $13$, $15$, $16$, $17$, $20$, $24$, $25$, $27$, $28$, $32$, $35$, $26$, $39$, $48$, $51$ of the $24$ positive integers which cannot be expressed as the sum of distinct non-pythagorean primes


The $26$th odd positive integer that cannot be expressed as the sum of exactly $4$ distinct non-zero square numbers all of which are coprime
$1$, $3$, $5$, $7$, $\ldots$, $35$, $37$, $41$, $43$, $45$, $47$, $49$, $53$, $55$, $\ldots$


The $33$rd and last positive integer after $2$, $3$, $4$, $7$, $8$, $\ldots$, $33$, $37$, $38$, $42$, $43$, $44$, $45$, $46$, $49$, $50$, $54$ which cannot be expressed as the sum of distinct pentagonal numbers


There are exactly $55$ sets of $4$ integers $\set {a, b, c, d}$ such that all integers can be written in the form:
$n = a x^2 + b y^2 + c z^2 + d w^2$
for integer $x, y, z, w$


Also see



Sources