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A366457
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a(n) = n^2 + 58.
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1
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58, 59, 62, 67, 74, 83, 94, 107, 122, 139, 158, 179, 202, 227, 254, 283, 314, 347, 382, 419, 458, 499, 542, 587, 634, 683, 734, 787, 842, 899, 958, 1019, 1082, 1147, 1214, 1283, 1354, 1427, 1502, 1579, 1658, 1739, 1822, 1907, 1994, 2083, 2174, 2267, 2362, 2459, 2558, 2659, 2762, 2867, 2974, 3083
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OFFSET
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0,1
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COMMENTS
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Leonhard Euler observed that the polynomial n^2 + n + 41 takes distinct prime values for the 40 consecutive integers n = 0 to 39. Legendre showed that the first 29 terms of 2*n^2 + 29 (n = 0 to 28) are primes.
For even n = 2*m we have a(n) = 2*(2*m^2 + 29). It follows that a(n) is double a prime for the 29 even values of n in the integer interval [0, 57]. Calculation shows that a(n) takes distinct prime values for the 29 odd values of n in the interval [0, 57], except for a(29) = 29*31, a(33) = 31*37, a(41) = 37*47 and a(53) = 47*61. See the example section below.
The polynomial n^2 + 232 has similar properties. See A048988.
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LINKS
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FORMULA
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G.f.: (59*x^2 - 115*x + 58)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 58, a(1) = 59 and a(2) = 62.
Sum_{n>=0} 1/a(n) = (1+sqrt(58)*Pi*coth(sqrt(58)*Pi))/116 = 0.2148763... - R. J. Mathar, Apr 24 2024
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EXAMPLE
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The sequence terms factorized for 0 <= n <= 57:
[2*29, 59, 2*31, 67, 2*37, 83, 2*47, 107, 2*61, 139, 2*79, 179, 2*101, 227, 2*127, 283, 2*157, 347, 2*191, 419, 2*229, 499, 2*271, 587, 2*317, 683, 2*367, 787, 2*421, (29*31), 2*479, 1019, 2*541, (31*37), 2*607, 1283, 2*677, 1427, 2*751, 1579, 2*829, (37*47), 2*911, 1907, 2*997, 2083, 2*1087, 2267, 2*1181, 2459, 2*1279, 2659, 2*1381, (47*61), 2*1487, 3083, 2*1597, 3307].
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MAPLE
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seq(n^2 + 58, n = 0..50);
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MATHEMATICA
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Table[n^2 + 58, {n, 0, 50}]
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PROG
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(PARI) vector(50, n, n^2 + 58)
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CROSSREFS
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KEYWORD
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nonn,easy,changed
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AUTHOR
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STATUS
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approved
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