Interesting sequence

Nice Question. I haven't seen it anywhere.

Here are a few thoughts though.

39 43 50 54 61 65

Difference between the sequences follows this pattern

4 7 4 7 4

Also note, that the 43 in 643

I am guessing that if you do a quick test to the pattern you'll see the pattern happening again

For example, 643 - 4 = 6(39)
643 + 7 = 6(50)

So i think it should be part of the sequence. Ofcourse anyone who can validate or refute that is more than welcome.

i agree with you till the point you say that diff is 4 7 4 7 ... but post that i would like to be devil's advocate ...

if 643 - 4 = 639 and 643 + 7 = 650 is the logic then 143, 243 every such number shud be part of the sequence!

because 143 - 4 = 139 and 143 + 7 = 150
... but that is not the case....

i mean to say that this approach is not fool proof.....can experts pitch in pls??

agreed 100 %

waiting for expert opinion

I think the fastest way to solve this problem is to realize that the sequence moves in larger steps of multiples of 11. Therefore, to discover if the number 643 is in the sequence simply take that number and subtract the first number in the sequence (i.e. 39) from 643. This will leave 604. If 604 is divisible by 11 or leaves a remainder of 4, then the number 643 is in the sequence. The closest multiple of 11 to 604 is 594 with remainder of 10. Therefore, 643 is not in the sequence. You can check this method for other numbers in the sequence to realize that it works. For example, the second number in the sequence is 43. Therefore, 43 - 39 = 4. Whence divided by 11 it leaves a remainder of 4. Therefore, 43 is in the sequence. As another example the second number in the sequence is 50. Therefore, 50-39 = 11, which is a multiple of 11. Therefore, 50 is in the sequence. And just to pick one more number consider the number 72, which would be the next number in the sequence. 72 - 39 = 33, which is a multiple of 11. Therefore 72 is in the sequence.

643 is not in this sequence
- Any term larger than 43 fits the equasion A(n)=A(n-m)+11*(m/2), of which m has to be an even number.

- Given this, the sequence can be written as
n=1, A1=39
n=2, A2=43
n>2, A(n)=A(n-m)+11*(m/2)

-Now the question has transformed into solving "if 643 is devided by 11, and the quotient is an even number, m/2, if the remainder, A(n-m), among 39, 43, 50, 54, 61, or 65."

-As illustrated below
643=11*58+5
643=11*56+27
643=11*54+49
643=11*52+71
643=11*50+93
643=11*48+115
......
and none of 5, 27, 49, 71, 93, 115... is in the sequence.

Hence, 643 is not in the sequence

Thanks a lot!!!

Sorry for noob question, but how do you quickly come with equation that satisfies number greater tan 43?

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I found it ( rather unconventional way), if each of the given numbers in sequence are divided by 11, remains are 4,10,4,10,4,10....so the numbers in these series when divided by 11 should have remainders 4 or 11.
643/11 = 58 + 5 , 5 is the remainder.

so 643 doesn't falls in the sequence.