A062717 Numbers m such that 6*m+1 is a perfect square.
0, 4, 8, 20, 28, 48, 60, 88, 104, 140, 160, 204, 228, 280, 308, 368, 400, 468, 504, 580, 620, 704, 748, 840, 888, 988, 1040, 1148, 1204, 1320, 1380, 1504, 1568, 1700, 1768, 1908, 1980, 2128, 2204, 2360, 2440, 2604, 2688, 2860, 2948, 3128, 3220, 3408, 3504
Offset: 1
Links
- Harry J. Smith, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
Crossrefs
Programs
-
Magma
[(6*n*(n-1) + (2*n-1)*(-1)^n + 1)/4: n in [1..70]]; // Wesley Ivan Hurt, Apr 21 2021
-
Maple
seq(n^2+n+2*ceil(n/2)^2,n=0..48); # Gary Detlefs, Feb 23 2010
-
Mathematica
Select[Range[0, 3999], IntegerQ[Sqrt[6# + 1]] &] (* Harvey P. Dale, Mar 10 2013 *)
-
PARI
je=[]; for(n=0,7000, if(issquare(6*n+1),je=concat(je,n))); je
-
PARI
{ n=0; for (m=0, 10^9, if (issquare(6*m + 1), write("b062717.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 09 2009
-
Python
def A062717(n): return (n*(3*n + 4) + 1 if n&1 else n*(3*n + 2))>>1 # Chai Wah Wu, Jan 31 2023
Formula
G.f.: 4*x^2*(1 + x + x^2) / ( (1+x)^2*(1-x)^3 ).
a(2*k) = k*(6*k+2), a(2*k+1) = 6*k^2 + 10*k + 4. - Mohamed Bouhamida, Nov 06 2007
a(n) = n^2 - n + 2*ceiling((n-1)/2)^2. - Gary Detlefs, Feb 23 2010
From Bruno Berselli, Nov 28 2010: (Start)
a(n) = (6*n*(n-1) + (2*n-1)*(-1)^n + 1)/4.
6*a(n) + 1 = A007310(n)^2. (End)
E.g.f.: (3*x^2*exp(x) - x*exp(-x) + sinh(x))/2. - Ilya Gutkovskiy, Jun 18 2016
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Apr 21 2021
From Amiram Eldar, Mar 11 2022: (Start)
Sum_{n>=2} 1/a(n) = (9-sqrt(3)*Pi)/6.
Sum_{n>=2} (-1)^n/a(n) = 3*(log(3)-1)/2. (End)
Comments