OFFSET
1,1
COMMENTS
a(n) is the position of the row in Pascal's triangle (A007318) in which three consecutive entries appear in the ratio n: n+1: n+2. (Even valid for n = 0 if you allow for a position of -1 to have value 0.) The solution is unique for each n.
The row numbers are given by A060626.
This sequence plus 1 (i.e., a(n) = 2*n^2 + 3*n) is the sequence A014106. - Howard A. Landman, Mar 28 2004
If Y and Z are a 2-blocks of a 2n-set X then, for n>=2, a(n-1) is the number of (2n-2)-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
One might prepend an initial -1: "-1, 4, 13, 26, 43, ..." - Vladimir Joseph Stephan Orlovsky, Oct 25 2008 (This would require too many other changes. - N. J. A. Sloane, Mar 27 2014)
LINKS
Bruno Berselli, Table of n, a(n) for n = 1..1000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011, p. 21. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjic, Two Enumerative Functions. [Broken link]
Leo Tavares, Illustration: Square/Triangular Union
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = n + 4*binomial(2+n, n), with offset 0. - Zerinvary Lajos, May 12 2006
G.f.: x*(4 + x - x^2)/(1 - x)^3. - Vincenzo Librandi, Mar 28 2014
E.g.f.: 1 + exp(x)*(2*x^2 + 5*x - 1). - Stefano Spezia, Jun 16 2024
EXAMPLE
Entries in the ratio 1:2:3 appear in row 14 of Pascal's triangle (A007318) starting at position 4 (1001, 2002, 3003). Entries in the ratio 2:3:4 appear in row 34 of Pascal's triangle starting at position 13 (927983760, 1391975640, 1855967520); and so on (row 62, pos. 26; row 98, pos. 43; ...).
MAPLE
MATHEMATICA
Table[2 n^2 + 3 n - 1, {n, 50}] (* Bruno Berselli, Mar 28 2014 *)
PROG
(Perl) #!/usr/bin/perl $a = 1; while (1) { $k = $a*(2*$a + 3) - 1; print "$k, "; $a ++; }
(Magma) [2*n^2+3*n-1: n in [1..50]]; // Bruno Berselli, Mar 28 2014
(PARI) a(n)=2*n^2+3*n-1 \\ Charles R Greathouse IV, Sep 24 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Howard A. Landman, Mar 08 2004
STATUS
approved