|
|
A055808
|
|
a(n) and floor(a(n)/4) are both squares; i.e., squares that remain squares when written in base 4 and last digit is removed.
|
|
19
|
|
|
0, 1, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Let A(x) = (1 + k*x + (k-1)*x^2). Then the coefficients of (A(x))^2 sum to 4*k^2, where k = (n - 1). Examples: if k = 3 we have (1 + 3*x + 2*x^2)^2 = (1 + 6*x + 13x^2 + 12*x^3 + 4*x^4), and ( 1 + 6 + 13 + 12 + 4) = 36. If k = 4 we have (1 + 4*x + 3*x^2)^2 = (1 + 8*x + 22*x^2 + 24*x^3 + 9*x^4), and (1 + 8 + 22 + 24 + 9) = 64 = a(5). - Gary W. Adamson, Aug 02 2015
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 4*(-1+n)^2 for n>1; a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>4; G.f.: x*(x^3-7*x^2-x-1) / (x-1)^3. - Colin Barker, Sep 15 2014
|
|
EXAMPLE
|
36 is in the sequence because 36 = 6^2 = 210 base 3 and 21 base 4 = 9 = 3^2.
|
|
MATHEMATICA
|
Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {4, 16, 36}, 50]] (* Vincenzo Librandi, Aug 03 2015 *)
|
|
PROG
|
(PARI) concat(0, Vec(x*(x^3-7*x^2-x-1)/(x-1)^3 + O(x^100))) \\ Colin Barker, Sep 15 2014
(PARI) is_ok(n)=issquare(n) && issquare(floor(n/4));
first(m)=my(v=vector(m), r=0); for(i=1, m, while(!is_ok(r), r++); v[i]=r; r++; ); v; /* Anders Hellström, Aug 08 2015 */
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|