1, 3, 7, 13, 21, ...
What is the 600th member of the series?
What member, above the first, with fewer than 5 digits, is a perfect cube?
What member is a 5-digit palindrome that can also be read as a binary number?
What's the smaller of the two consecutive members that are 1000 apart?
The pth term of the sequence is given by:
T(p) = p(p-1) + 1
So, 600th term of the sequence is 600*599 + 1 = 359401
For the second query, let the jth term of the sequence be a perfect cube. Then:
T(j) = j^2 - j+1 = q^3(say)
Or, 4*q^3-A^2 = 3, where A = (2j-1) and accordingly A must be odd.
Checking for A = 1, 3, 5, ..... in turn, we observe that:
A=1, gives q^3 = 1 which is the first term of the sequence, and thus inadmissible.
A = 37 gives q^3 = 1372/4 = 343 = 7^3. Recalling that :
A = 2j -1, we can now assert that (37+1)/2 = 19th term of the sequence is a perfect cube.
For the third query, the number must be a 5 digit palindrome consisting entirely of 0's and 1's. The possible palindromes satisfying this criteria are 10001, 11011, 10101 and 11111.
Solving for p^2 - p + 1 = 10001, 11011, 10101, 11111 in turn we observe that an integer value of p is achievable only when p^2-p+1 = 10101, giving (2p-1)^2 = 40401, so that p = 101
So the 101th term of the sequence is 10101 which is a 5 digit palindrome such that the said term can also be written as a binary number.
For the final query, let the smaller of the consecutive members be s^2-s+1.
Then, by the problem:
1000 = (s+1)^2 -(s+1) + 1 - (s^2-s+1) = 2s, giving:
s = 500, so that s^2-s+1 = 500*499+1 = 249501.
Hence, 249501 is the required smaller member of the sequence.
Edited on May 9, 2007, 11:51 pm