A241269 - OEIS

%I #46 Jul 17 2023 17:17:44

%S 3,6,15,60,105,21,126,360,495,330,429,1092,1365,420,1020,2448,2907,

%T 1710,1995,4620,5313,759,3450,7800,8775,4914,5481,12180,13485,3720,

%U 8184,17952,19635,10710,11655,25308,27417,3705,15990,34440,37023,19866,21285,45540

%N Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).

%C All terms are multiples of 3.

%C Difference table of c(n):

%C    1/3,     1/6,    2/15,    7/60,    2/21,...

%C   -1/6,    -1/30,  -1/60,   -1/84,   -1/105,...

%C    2/15,    1/60,   1/210,   1/420,   1/630,...

%C   -7/60,   -1/84,  -1/420,  -1/1260, -1/2520,... .

%C 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.

%C Denominators of the main diagonal: A051133(n+1).

%C Denominators of the first upper diagonal; A000911(n).

%C c(n) is a companion to A026741(n)/A045896(n).

%C Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/A027642(n).

%C 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)?

%C Is a(n+40) - a(n) divisible by 10?

%C No: a(5) = 21 but a(45) = 12972. # _Robert Israel_, Jul 17 2023

%C 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?

%C 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, ... .

%C Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind

%C   1,    1/2,  1/3,    1/4,     1/5,     1/6,   ...

%C   0,    1/6,  1/6,    3/20,    2/15,    5/42,  ...

%C   1/3,  1/6,  2/15,   7/60,   11/105,   2/21,  ...

%C   0,    1/10, 1/10,  13/140,   3/35,    5/63,  ...

%C   1/5,  1/10, 3/35,  11/140,  23/315,  43/630, ...

%C   0,    1/14, 1/14,  17/252,   4/63,   ...

%C This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.

%C Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).

%C First row: see A003506, Leibniz's Harmonic Triangle.

%C Second row: A026741/A045896.

%C a(n) is the denominator of the third row f(n).

%C 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

%C Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .

%H Robert Israel, <a href="/A241269/b241269.txt">Table of n, a(n) for n = 0..10000</a>

%F c(n) = A014206(n)/A007531(n+3).

%F 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

%F 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

%p seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))),n=0..1000);

%t Denominator[Table[(n^2+n+2)/Times@@(n+{1,2,3}),{n,0,50}]] (* _Harvey P. Dale_, Mar 27 2015 *)

%o (PARI) for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ _Colin Barker_, Apr 18 2014

%K nonn,frac,look

%O 0,1

%A _Paul Curtz_, Apr 18 2014

%E More terms from _Colin Barker_, Apr 18 2014