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A135405
Sequence where the sum of each pair of consecutive elements is a square.
1
0, 1, 8, 8, 17, 19, 30, 34, 47, 53, 68, 76, 93, 103, 122, 134, 155, 169, 192, 208, 233, 251, 278, 298, 327, 349, 380, 404, 437, 463, 498, 526, 563, 593, 632, 664, 705, 739, 782, 818, 863, 901, 948, 988, 1037, 1079, 1130, 1174, 1227, 1273, 1328, 1376, 1433, 1483
OFFSET
0,3
COMMENTS
This covers squares of all consecutively increasing integers with the exception of 2.
It is actually possible to cover all nonnegative integers by using the given formula starting with n=-2, thus giving terms 2, -2, 3, 1, 8, 8, 17, 19, 30, etc. - Vladimir Joseph Stephan Orlovsky, Feb 12 2015
FORMULA
a(n) = (n+2)*(n+1)/2 + 2*(-1)^n for n>0.
From R. J. Mathar, Dec 12 2007: (Start)
O.g.f.: x*(1 +6*x -8*x^2 +3*x^3)/((1-x)^3*(1+x)) = -3 +1/(1-x)^3 + 2/(1+x).
a(n) = A000217(n+1) + 2*(-1)^n if n>0.
(End)
E.g.f.: -3 + 2*exp(-x) + (1/2)*(2 + 4*x + x^2)*exp(x). - G. C. Greubel, Oct 12 2016
From Colin Barker, Oct 13 2016: (Start)
a(n) = (-4*(-1)^n+n+n^2)/2 for n>1.
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n>4.
(End)
EXAMPLE
a(1) = 1 because 0 + 1 = 1^2.
a(2) = 8 because 1 + 8 = 9 = 3^2.
a(3) = 8 because 8 + 8 = 16 = 4^2.
MATHEMATICA
a=1; lst={0, a}; Do[a=n^2-a; AppendTo[lst, a], {n, 3, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 17 2008 *)
Table[(n+2)*(n+1)/2 + 2*(-1)^n, {n, 0, 25}] (* G. C. Greubel, Oct 12 2016 *)
PROG
(Magma) [0] cat [(n+2)*(n+1)/2+2*(-1)^n: n in [1..60]]; // Vincenzo Librandi, Feb 14 2015
(PARI) concat(0, Vec(x*(1+6*x-8*x^2+3*x^3)/((1-x)^3*(1+x)) + O(x^60))) \\ Colin Barker, Oct 13 2016
CROSSREFS
Sequence in context: A171188 A145909 A168409 * A006784 A214830 A168456
KEYWORD
nonn,easy
AUTHOR
Alexander R. Povolotsky, Dec 11 2007, Apr 02 2008
EXTENSIONS
More terms from Vladimir Joseph Stephan Orlovsky, Dec 17 2008
STATUS
approved