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A158460
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Numbers k such that prime(k-1) + 5 is square and equal to prime(k+1) - 1.
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4
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6, 44, 41202, 123125, 141582, 527569, 920270, 975866, 1034000, 1278000, 1504258, 1707305, 1774017, 1863515, 2513332, 2776350, 3315370, 5100781, 5152209, 5746269, 5943102, 7380924, 7891751, 8585974, 10100295, 11022570, 12248841, 13213333, 13654151, 13817964
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OFFSET
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1,1
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COMMENTS
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Conjecture: If the condition holds, prime(k-1) and prime(k) are twin primes of the form 10m+1 and 10m+3. I.e., the last digits are 1 and 3.
This is so because prime(k-1)+5 square => possible ending digits of 0,1,4,5,6,9. To get ending digits for prime(k-1) we subtract 5 to get ending digits 5,6,9,0,1,4. So 1,9 are the only possible endings since 0,4,5,6 => prime(k-1) not prime, impossible. Now by the condition of equality, prime(k-1)+6 = prime(k+1). So if prime(k-1) ends in 9, prime(k-1)+6 ends in 5 => prime(k+1) not prime, impossible. Therefore prime(k-1) ends in 1 and by the condition of prime(k-1) and prime(k) being twin primes, prime(k) ends in 3. This sequence is a variation of the conjecture provided in the link.
The PARI script allows for general investigation of numbers of the form prime(k-1)+a and prime(k+1)-b. The values a=5,7; b=1 consistently yield twin primes when the condition holds. Notice we test for square of the first prime(k-1) retrieval before calling the second prime(k+1). This cuts the search time in half. A much faster and more inclusive program is in the Link.
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LINKS
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Sebastian Martin Ruiz and others, Integers then Equals, digest of 7 messages in primenumbers Yahoo group, Mar 14 - Mar 20, 2009.
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EXAMPLE
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For k=6, prime(6-1) = 11, 11+5 = 16 = prime(6+1)-1 = 17-1 so 6 is the first term in the sequence.
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PROG
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(PARI) integerequal(m, n, a, b) = {local(x, p1, p2); for(x=m, n, p1=prime(x-1); if(issquare(p1+a), p2=prime(x+1); if((p1+a)==(p2-b), print(x", "p1", "prime(x))))); }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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