OFFSET
1,1
COMMENTS
The checkpoints, which are the numbers 4, 8, 16, 32, etc., are placed so that the number 2^k is located at the (2^k + k - 5)th position in the sequence, for k>=2; thus:
4 = 2^2 = a(2^2 + 2 - 5) = a( 4 + 2 - 5) = a(1);
8 = 2^3 = a(2^3 + 3 - 5) = a( 8 + 3 - 5) = a(6);
16 = 2^4 = a(2^4 + 4 - 5) = a(16 + 4 - 5) = a(15);
32 = 2^5 = a(2^5 + 5 - 5) = a(32 + 5 - 5) = a(32); etc.
The number of terms that will be placed between successive checkpoints 2^k and 2^(k+1) is (2^(k+1) + (k+1) - 5) - (2^k + k - 5) - 1 = 2^k for each k>=2; i.e., there will be 4 terms placed between 4 and 8, 8 terms placed between 8 and 16, 16 terms placed between 16 and 32, etc
Not every positive integer greater than four will appear in this sequence. If p and q are two consecutive primes with |p-q|>2, then the numbers from (p+2)*3 to (q-1)*3 will not occur in this sequence. No number of the form 2^k*m, where k>1 and m is an odd number not divisible by 3, will occur in this sequence (for example, 20, 28). Also, the numbers of form t^k, where t>3 is an odd prime and k>1 will not occur in this sequence (for example, 25, 49).
No two adjacent terms will share more than one prime factor.
LINKS
Gaurish Korpal, Pulsating graph for first 32 terms
Gaurish Korpal, Comment: 'Be Still My Pulsating Sequence', Quanta Magazine, 14 November 2015
Pradeep Mutalik, Solution: 'Be Still My Pulsating Sequence', Quanta Magazine, 25 November 2015
EXAMPLE
We begin by placing successive powers of 2, starting with 2^2 = 4, with spacing equal to the value of the previous power of 2, in the sequence as checkpoints:
.
.|4 terms|
.|<----->| |<-- 8 terms -->| |<---------- 16 terms --------->|
4,_,_,_,_,8,_,_,_,_,_,_,_,_,16,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,32,...
Then we fill in the remaining locations sequentially, jumping over each checkpoint as we encounter it. Those remaining locations are filled in sequentially, in sets of four terms at a time (i.e., in quadruples). We begin inserting the quadruples of the form {2p, p, 3p, 3p+3}, where p in the j-th quadruple inserted is the j-th odd prime; thus, the first quadruple is {2*3, 3, 3*3, 3*3+3} = {6,3,9,12}, and inserting it gives
4,6,3,9,12,8,_,_,_,_,_,_,_,_,16,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,32,...
Now we jump over the checkpoint 8 and insert the next two quadruples (which have p=5 and p=7, respectively):
4,6,3,9,12,8,10,5,15,18,14,7,21,24,16,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,32,...
Continuing as above, we insert the next 4 quadruples (16 terms) after the checkpoint term 16, the next 8 quadruples (32 terms) after the checkpoint term 32, etc.
CROSSREFS
KEYWORD
nonn
AUTHOR
Gaurish Korpal, Nov 29 2015
STATUS
approved