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A281445
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Nonnegative k for which (2*k^2 + 1)/11 is an integer.
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1
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4, 7, 15, 18, 26, 29, 37, 40, 48, 51, 59, 62, 70, 73, 81, 84, 92, 95, 103, 106, 114, 117, 125, 128, 136, 139, 147, 150, 158, 161, 169, 172, 180, 183, 191, 194, 202, 205, 213, 216, 224, 227, 235, 238, 246, 249, 257, 260, 268, 271, 279, 282, 290, 293, 301, 304, 312, 315
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OFFSET
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1,1
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COMMENTS
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For prime d < 11, (2*k^2 + 1)/d can provide integers when d = 3 (A186424).
Corresponding values of (2*k^2 + 1)/11 are listed in A179088.
All k == 4 or 7 (mod 11). - Robert Israel, Apr 25 2017
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LINKS
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Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
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FORMULA
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O.g.f.: x*(4 + 3*x + 4*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 4 - 5*exp(-x)/4 - 11*(1 - 2*x)*exp(x)/4.
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = (22*n - 5*(-1)^n - 11)/4. Therefore: a(2*h) = 11*h - 4, a(2*h+1) = 11*h + 4.
If h>0,
h*a(n) + (6*h - 5*(-1)^h - 11)/4 = a(h*n) for odd n; otherwise:
h*a(n) + 4*(h - 1) = a(h*n). Some special cases:
h=2: 2*a(n) - 1 = a(2*n) for odd n, 2*a(n) + 4 = a(2*n) for even n;
h=3: 3*a(n) + 3 = a(3*n) for odd n, 3*a(n) + 8 = a(3*n) for even n;
h=4: 4*a(n) + 2 = a(4*n) for odd n, 4*a(n) + 12 = a(4*n) for even n;
h=5: 5*a(n) + 6 = a(5*n) for odd n, 5*a(n) + 16 = a(5*n) for even n, and so on.
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MAPLE
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seq(seq(11*i+j, j=[4, 7]), i=0..50); # Robert Israel, Apr 25 2017
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MATHEMATICA
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Select[Range[400], IntegerQ[(2*#^2 + 1)/11] &]
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PROG
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(Sage) [k for k in range(400) if ((2*k^2+1)/11).is_integer()]
(MAGMA) &cat [[11*n+4, 11*n+7]: n in [0..30]];
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CROSSREFS
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Cf. A179088.
Cf. A001651: nonnegative k for which (2*k^2 + 1)/3 is an integer.
Sequence in context: A310923 A310924 A310925 * A310926 A310927 A049832
Adjacent sequences: A281442 A281443 A281444 * A281446 A281447 A281448
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KEYWORD
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nonn,easy
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AUTHOR
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Bruno Berselli, Apr 13 2017
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STATUS
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approved
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