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A027862
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Primes of the form j^2 + (j+1)^2.
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41
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5, 13, 41, 61, 113, 181, 313, 421, 613, 761, 1013, 1201, 1301, 1741, 1861, 2113, 2381, 2521, 3121, 3613, 4513, 5101, 7321, 8581, 9661, 9941, 10513, 12641, 13613, 14281, 14621, 15313, 16381, 19013, 19801, 20201, 21013, 21841, 23981, 24421, 26681
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OFFSET
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1,1
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COMMENTS
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Also, primes of the form 4*k+1 which are the hypotenuse of one and only one right triangle with integral legs. - Cino Hilliard, Mar 16 2003
Equivalently, primes of the form (m^2+1)/2 (take m=2*j+1). These primes a(n) have nontrivial solutions of x^2 == 1 (Modd a(n)) given by x=x(n)=A002731(n). For Modd n see a comment on A203571. See also A206549 for such solutions for primes of the form 4*k+1, given in A002144.
E.g., a(3)=41, A002731(3)=9, 9^2=81, floor(81/41)=1 (odd),
-81 = -2*41 + 1 == 1 (mod 41), hence 9^2 == 1 (Modd 41). - Wolfdieter Lang, Feb 24 2012
Also primes of the form 4*k+1 that are the smallest side length of one and only one integer Soddyian triangle (see A230812). - Frank M Jackson, Mar 13 2014
Also, primes of the form (m^2+1)/2. - Zak Seidov, May 01 2014
Primes in the main diagonal of A000027 when represented as an array read by antidiagonals. - Clark Kimberling, Mar 12 2023
The diophantine equation x^2 + … + (x + r)^2 = p may be rewritten to A*x^2 + B*x + C = p, where A = (r + 1), B = r*(r + 1), C = r*(r + 1)*(2*r + 1)/6. If gcd(A, B, C) > 1 no solution for a prime p exists. The gcd(A, B, C) = 1 holds only for r = 1, 2, 5 (gcd is the greatest common divisor). For r = 1 we have x^2 + (x + 1)^2 = p, thus for x from A027861 we calculate primes p from A027862. For r = 2 we have x^2 + (x + 1)^2 + (x + 2)^2 = p, thus for x from A027863 we calculate primes p from A027864. For r = 5 we have x^2 + … + (x + 5)^2 = p, thus for x from A027866 we calculate primes p from A027867. - Ctibor O. Zizka, Oct 04 2023
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REFERENCES
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D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc. Boston, MA, 1976, p. 271.
Morris Kline, Mathematical Thought from Ancient to Modern Times, 1972. pp. 275.
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LINKS
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FORMULA
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EXAMPLE
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13 is in the sequence because it is prime and 13 = 2^2 + 3^2. - Michael B. Porter, Aug 27 2016
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MATHEMATICA
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Select[Table[n^2+(n+1)^2, {n, 200}], PrimeQ] (* Harvey P. Dale, Aug 22 2012 *)
Select[Total/@Partition[Range[200]^2, 2, 1], PrimeQ] (* Harvey P. Dale, Apr 20 2016 *)
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PROG
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(PARI) je=[]; for(n=1, 500, if(isprime(n^2+(n+1)^2), je=concat(je, n^2+(n+1)^2))); je
(PARI) fermat(n) = { for(x=1, n, y=2*x*(x+1)+1; if(isprime(y), print1(y" ")) ) }
(Magma) [ a: n in [0..150] | IsPrime(a) where a is n^2+(n+1)^2 ]; // Vincenzo Librandi, Dec 18 2010
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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