OFFSET
1,2
COMMENTS
Triangular numbers belonging to A028387. - Bruno Berselli, Apr 18 2018
Since this sequence lists triangular numbers t such that 4*t + 5 = u^2 and 8*t + 1 = v^2 by definition of triangular numbers, 2*u^2 - 9 = 2*(4*t + 5) - 9 = 8*t + 1 = v^2, that is, sqrt(4*a(n) + 5) is in A075841. - Altug Alkan, Apr 18 2018
LINKS
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).
FORMULA
From Giovanni Resta, Mar 24 2013: (Start)
G.f.: -x*(1 + 20*x + x^2) / ( (x - 1)*(x^2 - 34*x + 1) ).
a(n) = floor(9 * (17 + sqrt(288))^n * (3 - sqrt(8))/32). (End)
a(n) = (A075841(n)^2 - 5)/4. - Altug Alkan, Apr 18 2018
EXAMPLE
Triangular(10) = 55 = 7^2 + 7 - 1, so 55 is in the sequence.
MATHEMATICA
a[1]=1; a[2]=55; a[3]=1891; a[n_] := a[n] = 35*a[n-1] - 35*a[n-2] + a[n-3]; Array[a, 20] (* Giovanni Resta, Mar 24 2013 *)
Table[Floor@(9 (17 + Sqrt@ 288)^n*(3 - Sqrt@ 8)/32), {n, 0, 16}] (* or *)
CoefficientList[Series[-x (1 + 20 x + x^2)/((x - 1) (x^2 - 34 x + 1)), {x, 0, 16}], x] (* Michael De Vlieger, Oct 08 2016 *)
PROG
(C)
#include <stdio.h>
typedef unsigned long long U64;
U64 rootPronic(U64 a) {
U64 sr = 1L<<32, s, b;
while (a < sr*(sr-1)) sr>>=1;
for (b = sr>>1; b; b>>=1) {
s = sr+b;
if (a >= s*(s-1)) sr = s;
}
return sr;
}
int main() {
U64 a, i, t;
for (i=0; i < 1L<<32; ++i) {
a = i*(i+1)/2 + 1;
t = rootPronic(a);
if (a == t*(t-1)) printf("%llu %llu %llu\n", i, t, a-1);
}
return 0;
}
(PARI) Vec(x*(1+20*x+x^2)/(1-35*x+35*x^2-x^3)+O(x^66)) \\ Joerg Arndt, Mar 25 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alex Ratushnyak, Mar 23 2013
STATUS
approved