

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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,2,1).


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)/((1x)^3*(1+x)) = 3 +1/(1x)^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(n1)2*a(n3)+a(n4) 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^2a; 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*x8*x^2+3*x^3)/((1x)^3*(1+x)) + O(x^60))) \\ Colin Barker, Oct 13 2016


CROSSREFS

Sequence in context: A171188 A145909 A168409 * A006784 A214830 A168456
Adjacent sequences: A135402 A135403 A135404 * A135406 A135407 A135408


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



