Problems & Puzzles: Puzzles Puzzle 260. Chains of twins There are certain twin primes that when added their sum is equal to the number 'in the middle' of other two larger twin primes.
According with the property described above we say that (5,7) → (11, 13) form a chain of size equal to 2. As a matter of fact this chain is size 2 and not 3 because 11+13 = 24 and 24 is not in the middle of twin primes (23 is prime but 25 is not prime), so the chain really ends here.Are there larger chains? As a matter of fact the solution to this question is the matter of the EIS A068635. This sequence is formed by the smallest prime of the earlier chain of size 1, 2, 3, ... and currently goes like this:
The sixth term was calculated (2002?) by Don Reble and the prime involved is of moderate -and respectable- length (14 digits), so our puzzles are: Q1. Find at least one more term for this sequence. Q2. Find one chain of size 2 using titanic twins. Solution:
J. K. Andersen and Andrew Rupinski discovered that this 'new' puzzle was not a new puzzle at all, but an 'old' puzzle vastly investigated by several well known puzzlers at least in two sites: a) my own Puzzle 119 and b) the 'bi-twins' pages (1 & 2) of Henri Lifchitz. According with the last records described there, both questions (Q1 & Q2) have been responded already: Q1: According with this: The 6th member of the sequence (1228253271*13#*2 -1) was produced by Jack Brennen (1999) 3 years before than Don Reble, The 7th member of the sequence (11228462199623*13# -1)was produced P. Jobling (1999) and the probable (?) 8th member of the sequence (21033215071024191*13# -1)was produced P. Jobling & Carmody (2002) consequently the only interesting questions are: a) Determine if 21033215071024191*13# - 1, the minimal one for the case of size=8 B) Determine the 9th member of the sequence. Q2. Henri Lifchitz has the record (March 2004) of size 2: 853877841*3571# - 1 (1529 digits)
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