OFFSET
1,2
COMMENTS
Apparently a(n) = A062318(n+2) - 1.
The terms beginning with a(2) are the row numbers in Pascal's Triangle where every 3rd element in those rows is divisible by 3 and none of the other elements in those rows are divisible by 3. - Thomas M. Green, Apr 03 2013
REFERENCES
Thomas M. Green, Prime Patterns in Pascal's Triangle, paper in review process, 2013.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
Index entries for linear recurrences with constant coefficients, signature (1,3,-3).
FORMULA
a(n) = A038754(n+1) - 2.
a(n) = 3*a(n-2) + 4 for n > 2; a(1) = 1, a(2) = 4.
a(n) = (5 - (-1)^n)*3^(1/4*(2*n - 1 + (-1)^n))/2 - 2.
G.f.: x*(1 + 3*x)/((1 - x)*(1 - 3*x^2)).
E.g.f.: 2*(cosh(sqrt(3)*x) - cosh(x)) + sqrt(3)*sinh(sqrt(3)*x) - 2*sinh(x). - Stefano Spezia, Dec 31 2022
EXAMPLE
For n = 3, a(3) = 7. The binomial coefficients of the 7th row of Pascal's Triangle are 1 7 21 35 35 21 7 1 and every 3rd element is a multiple of 3. - Thomas M. Green, Apr 03 2013
MATHEMATICA
Accumulate[Transpose[NestList[{Last[#], 3*First[#]}&, {1, 3}, 40]][[1]]] (* Harvey P. Dale, Feb 17 2012 *)
PROG
(Magma) T:=[ n le 2 select 2*n-1 else 3*Self(n-2): n in [1..33] ]; [ n eq 1 select T[1] else Self(n-1)+T[n]: n in [1..#T]];
(PARI) a(n) = (2+n%2)*3^(n\2)-2 \\ Charles R Greathouse IV, Jul 15 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Aug 10 2009
EXTENSIONS
Incorrect formula removed by Stefano Spezia, Dec 31 2022
STATUS
approved