861
Jump to navigation
Jump to search
Number
$861$ (eight hundred and sixty-one) is:
- $3 \times 7 \times 41$
- The $1$st term of the $2$nd sequence of $3$ consecutive triangular numbers all of which are sphenic:
| \(\ds \quad \ \ \) | \(\ds T_{41}\) | \(=\) | \(\ds 861\) | \(\ds = 3 \times 7 \times 41\) | which is sphenic | |||||||||
| \(\ds T_{42}\) | \(=\) | \(\ds 903\) | \(\ds = 3 \times 7 \times 43\) | which is sphenic | ||||||||||
| \(\ds T_{43}\) | \(=\) | \(\ds 946\) | \(\ds = 2 \times 1 \times 43\) | which is sphenic |
- The $21$st hexagonal number after $1$, $6$, $15$, $28$, $45$, $66$, $91$, $\ldots$, $378$, $435$, $496$, $561$, $630$, $703$, $780$:
- $861 = \ds \sum_{k \mathop = 1}^{21} \paren {4 k - 3} = 21 \paren {2 \times 21 - 1}$
- The $41$st triangular number after $1$, $3$, $6$, $10$, $15$, $\ldots$, $325$, $351$, $378$, $406$, $435$, $465$, $496$, $528$, $561$, $595$, $630$, $666$, $703$, $741$, $780$, $820$:
- $861 = \ds \sum_{k \mathop = 1}^{41} k = \dfrac {41 \times \paren {41 + 1} } 2$
- The $43$rd Smith number after $4$, $22$, $27$, $58$, $\ldots$, $648$, $654$, $663$, $666$, $690$, $706$, $728$, $729$, $762$, $778$, $825$, $852$:
- $8 + 6 + 1 = 3 + 7 + 4 + 1 = 15$