OFFSET
1,2
COMMENTS
Equivalently, numbers n such that oblong(n) + oblong(2*n) is an oblong number, where oblong(n) = A002378(n) = n*(n+1).
Also, x values in the equation A147875(x) = A000217(y) - see Ralf Stephan in Program lines. - Bruno Berselli, May 18 2013
Also, numbers m such that 2*m+1 and 10*m+1 are both squares. - Bruno Berselli, Mar 03 2016
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..798
Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).
FORMULA
G.f.: 12*x*(1+6*x+x^2)/((1-x)*(1-18*x+x^2)(1+18*x+x^2)). [Bruno Berselli, May 18 2013]
a(n) = (1/20)*((3+(-1)^n*sqrt(5))*(2-sqrt(5))^(4*floor(n/2))+(3-(-1)^n*sqrt(5))*(2+sqrt(5))^(4*floor(n/2))-6). [Bruno Berselli, May 18 2013]
a(2*n) = (Fibonacci(6*n-3)^2 + Lucas(6*n-3)*Fibonacci(6*n-1))/2. - Greg Dresden, Sep 24 2023
EXAMPLE
12*13/2 + 24*25/2 = 27*28/2, so 12 is in the sequence.
MATHEMATICA
CoefficientList[Series[12 x (1 + 6 x + x^2)/((1 - x) (1 - 18 x + x^2) (1 + 18 x + x^2)), {x, 0, 20}], x] (* Bruno Berselli, May 18 2013 *)
LinearRecurrence[{1, 322, -322, -1, 1}, {0, 12, 84, 3960, 27144}, 20] (* Harvey P. Dale, Apr 08 2021 *)
PROG
(C)
#include <stdio.h>
#include <math.h>
int main() {
unsigned long long i, s, t;
for (i = 0; i< (1ULL<<31); i++) {
s = 2*i*(2*i+1) + i*(i+1);
t = sqrt(s);
if (s==t*(t+1)) printf("%llu, ", i);
}
return 0;
}
(PARI) for(n=1, 10^9, t=n*(5*n+3)/2; x=sqrtint(2*t); if(t==x*(x+1)/2, print(n))) /* Ralf Stephan, May 17 2013 */
CROSSREFS
Cf. A224419 (numbers n such that triangular(n) + triangular(2*n) is a square).
Cf. A011916 (numbers n such that triangular(2*n) - triangular(n) is a triangular number).
Cf. A225786 (numbers n such that oblong(2*n) + oblong(n) is a square).
Cf. A225839 (triangular numbers of the form triangular(x) + triangular(2*x)).
KEYWORD
nonn,easy
AUTHOR
Alex Ratushnyak, May 16 2013
EXTENSIONS
More terms from Bruno Berselli, May 18 2013
STATUS
approved