OFFSET
1,2
COMMENTS
Row sums of triangle A131228.
Take the sum of the squares of the first n triangular numbers and divide it by the sum of these n triangular numbers. The sum evenly divides the sum of the squares for the n in this sequence. - J. M. Bergot, May 09 2012
a(n) = the difference between the sum of the terms in antidiagonal(n) and antidiagonal(n-1) in A204008. - J. M. Bergot, Jul 15 2013
LINKS
David Lovler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
FORMULA
From R. J. Mathar, Apr 17 2008: (Start)
a(n) = a(n-2) + 10 = 5*n + ((-1)^n - 7)/2.
G.f.: x*(1 + 6*x + 3*x^2)/((-1+x)^2*(1+x)). (End)
a(n) = 10*n - a(n-1) - 12 with n > 1, a(1)=1. - Vincenzo Librandi, Nov 23 2010
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 3; a(1)=1, a(2)=7, a(3)=11. - Harvey P. Dale, May 20 2012
a(n) = A063097(n) for n >= 2. - Georg Fischer, Nov 02 2018
E.g.f.: 3 + ((10*x - 7)*exp(x) + exp(-x))/2. - David Lovler, Sep 07 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt((5+sqrt(5))/2)*Pi/10 + 3*log(phi)/(2*sqrt(5)), where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023
EXAMPLE
11 = a(3) = 1 mod 10.
11 = a(3) = row 4 sums, triangle A131228: (1 + 3 + 7).
MATHEMATICA
Select[Range[300], MemberQ[{1, 7}, Mod[#, 10]]&] (* or *) LinearRecurrence[ {1, 1, -1}, {1, 7, 11}, 60] (* Harvey P. Dale, May 20 2012 *)
PROG
(PARI) is(n)=n%10==1 || n%10==7 \\ Charles R Greathouse IV, Jul 16 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gary W. Adamson, Jun 20 2007
EXTENSIONS
More terms from R. J. Mathar, Apr 17 2008
STATUS
approved