$75 = 3 \times 5^2$
According to A036378 there are $75$ primes between $2^9 = 512$ and $2^{10} = 1024$
The $75$ primes are:
| ________ | ________ | ________ |
|---|---|---|
| $1) 521$ | $2) 523$ | $3) 541$ |
| $4) 547$ | $5) 557$ | $6) 563$ |
| $7) 569$ | $8) 571$ | $9) 577$ |
| $10) 587$ | $11) 593$ | $12) 599$ |
| $13) 601$ | $14) 607$ | $15) 613$ |
| $16) 617$ | $17) 619$ | $18) 631$ |
| $19) 641$ | $20) 643$ | $21) 647$ |
| $22) 653$ | $23) 659$ | $24) 661$ |
| $25) 673$ | $26) 677$ | $27) 683$ |
| $28) 691$ | $29) 701$ | $30) 709$ |
| $31) 719$ | $32) 727$ | $33) 733$ |
| $34) 739$ | $35) 743$ | $36) 751$ |
| $37) 757$ | $38) 761$ | $39) 769$ |
| $40) 773$ | $41) 787$ | $42) 797$ |
| $43) 809$ | $44) 811$ | $45) 821$ |
| $46) 823$ | $47) 827$ | $48) 829$ |
| $49) 839$ | $50) 853$ | $51) 857$ |
| $52) 859$ | $53) 863$ | $54) 877$ |
| $55) 881$ | $56) 883$ | $57) 887$ |
| $58) 907$ | $59) 911$ | $60) 919$ |
| $61) 929$ | $62) 937$ | $63) 941$ |
| $64) 947$ | $65) 953$ | $66) 967$ |
| $67) 971$ | $68) 977$ | $69) 983$ |
| $70) 991$ | $71) 997$ | $72) 1009$ |
| $73) 1013$ | $74) 1019$ | $75) 1021$ |
Consider writing down a sequence using the following rules
1) Write down the first odd non-negative integer, $1$
2) Write down the next two even numbers, $2, 4$
3) Write down the next three odd numbers, $5, 7, 9$
4) Write down the next four even numbers $10, 12, 14, 16$
5) Write down the next five odd numbers $17, 19, 21, 23, 25$
6) Well, you get the idea.
The first $47$ members of this sequence looks like this:
$1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, 19, 21, 23, 25, 26, 28, 30, 32, 34, 36, 37, 39, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 77, 79, 81$
As you can see $75$ is the $42^{nd}$ member, of what is known as the Connell Sequence, see A001614.
The formula for this sequence is
$a(n) = 2n - \lfloor \frac {1 + \sqrt {8n - 7} } {2} \rfloor $ where $\lfloor \rfloor$ indicates the floor function.
Thus we can calculate the $75^{th}$ member of this function:
$a(75) = (2 \times 75) - \lfloor \frac {1 + \sqrt {(8 \times 75) - 7} } {2} \rfloor $
$a(75) = 150 - \lfloor \frac {1 + \sqrt {600 - 7} } {2} \rfloor $
$a(75) = 150 - \lfloor \frac {1 + \sqrt {593} } {2} \rfloor $
$a(75) = 150 - \lfloor \frac {1 + 24.35159 } {2} \rfloor $
$a(75) = 150 - \lfloor \frac {25.35159 } {2} \rfloor $
$a(75) = 150 - \lfloor 12.67579 \rfloor$
$a(75) = 150 - 12$
$\underline {\underline {a(75) = 138}}$
As an aside note that the number at the end of each sub-sequence above is the square of the index of the sub-sequence, i.e the $5^{th}$ sub-sequence ends in $25$.
No comments:
Post a Comment