OFFSET
0,1
COMMENTS
All terms are multiples of 3.
Difference table of c(n):
1/3, 1/6, 2/15, 7/60, 2/21,...
-1/6, -1/30, -1/60, -1/84, -1/105,...
2/15, 1/60, 1/210, 1/420, 1/630,...
-7/60, -1/84, -1/420, -1/1260, -1/2520,... .
This is an autosequence of the second kind; the inverse binomial transform is the signed sequence. The main diagonal is the first upper diagonal multiplied by 2.
Denominators of the main diagonal: A051133(n+1).
Denominators of the first upper diagonal; A000911(n).
Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/A027642(n).
Are the numerators of the main diagonal (-1)^n? If yes, what is the value of 1/3 - 1/30 + 1/210,... or 1 - 1/10 + 1/70 - 1/420, ... , from A002802(n)?
Is a(n+40) - a(n) divisible by 10?
No: a(5) = 21 but a(45) = 12972. # Robert Israel, Jul 17 2023
Are the common divisors to A014206(n) and A007531(n+3) of period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2?
Reduce c(n) = f(n) = b(n)/a(n) = 1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, ... .
Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind
1, 1/2, 1/3, 1/4, 1/5, 1/6, ...
0, 1/6, 1/6, 3/20, 2/15, 5/42, ...
1/3, 1/6, 2/15, 7/60, 11/105, 2/21, ...
0, 1/10, 1/10, 13/140, 3/35, 5/63, ...
1/5, 1/10, 3/35, 11/140, 23/315, 43/630, ...
0, 1/14, 1/14, 17/252, 4/63, ...
This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.
Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).
First row: see A003506, Leibniz's Harmonic Triangle.
a(n) is the denominator of the third row f(n).
The first column is 1, 0, 1/3, 0, 1/5, 0, 1/7, 0, ... . Numerators: A093178(n+1). This incites, considering tan(1), to introduce before the first row
Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .
LINKS
Robert Israel, Table of n, a(n) for n = 0..10000
FORMULA
The sum of the difference table main diagonal is 1/3 - 1/30 + 1/210 - ... = 10*A086466-4 = 4*(sqrt(5)*log(phi)-1) = 0.3040894... - Jean-François Alcover, Apr 22 2014
a(n) = (n+1)*(n+2)*(n+3)/gcd(4*n - 4, n^2 + n + 2), where gcd(4*n - 4, n^2 + n + 2) is periodic with period 16. - Robert Israel, Jul 17 2023
MAPLE
seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))), n=0..1000);
MATHEMATICA
Denominator[Table[(n^2+n+2)/Times@@(n+{1, 2, 3}), {n, 0, 50}]] (* Harvey P. Dale, Mar 27 2015 *)
PROG
(PARI) for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014
CROSSREFS
KEYWORD
AUTHOR
Paul Curtz, Apr 18 2014
EXTENSIONS
More terms from Colin Barker, Apr 18 2014
STATUS
approved