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Search: id:A115041
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| A115041 |
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y in the smallest x such that x^2 + p = y^n over the set of primes p. |
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+0
1
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| 3, 2, 3, 2, 3, 7, 3, 10, 3, 15, 2, 19, 21, 22, 2, 3, 30, 5, 34, 2, 37, 2, 42, 5, 7, 51, 2, 54, 5, 57, 2, 66, 69, 70, 75, 2, 79, 82, 6, 87, 3, 91, 6, 97, 99, 7, 106, 2, 3, 115, 117, 3, 11, 3, 129, 132, 135, 10, 139, 141, 142, 147, 7, 156, 157, 159, 166, 13, 174, 175, 177, 12, 2
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p.
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EXAMPLE
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5 is the smallest number that when we add its square to prime 2, we get a
perfect power, 3^3. So 3 is the first entry. 498^2 + 997 = 499^2. so 499 is
the last entry in the table.
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PROGRAM
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(PARI) sqplusp(n) = { local(p, x, y, c=0); forprime(p=2, n, for(x=1, n, y=x^2+p; if(ispower(y), print1(power(y)", "); c++; break ) ) ); print(); print(c) } exponent(n) = \ Return the exponent if n is a perfect power { local(x, ln, j, e=0); ln=omega(n); x=factor(n); e=x[1, 2]; for(j=2, ln, if(x[j, 2] < e, e=x[j, 2]) ); return(e) } power(n) = \Return the largest factor for which n is a perfect power { if(ispower(n), return(round(n^(1/exponent(n)))) ) }
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CROSSREFS
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Sequence in context: A097618 A039639 A023509 this_sequence A039641 A093489 A066919
Adjacent sequences: A115038 A115039 A115040 this_sequence A115042 A115043 A115044
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KEYWORD
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easy,nonn
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AUTHOR
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Cino Hilliard (hillcino368(AT)gmail.com), Feb 26 2006
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