OFFSET
0,2
COMMENTS
Since Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b) ≡ 0 mod (n*k+a+b), then define B(n,k,a,b) = (Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b))/(n*k+a+b), with n*k+a+b <> 0, n >= 0 and k,a,b are integers, such that B(1,n,2,1) = (Product_{i=0..1}(i*n+2) - Product_{i=0..1}(-i*n-1))/(n+3) = A000012(n) with n >= 0;
B(3,n,2,1) = (Product_{i=0..3}(i*n+2) - Product_{i=0..3}(-i*n-1))/(3*n+3) = A100040(n+2) with n >= 0;
B(2*n+1,0,2,1) = (Product_{i=0..2*n+1}(2) - Product_{i=0..2*n+1}(-1))/3 = A002450(n+1) with n >= 0;
B(2*n+1,2,2,1) = (Product_{i=0..2*n+1}(2*i+2) - Product_{i=0..2*n+1}(-2*i-1))/(4*n+5) = A273983(n+1) with n >= 0;
and a(n) is B(2*n+1,1,2,1). - Hong-Chang Wang, Jun 17 2016
LINKS
Hong-Chang Wang, Table of n, a(n) for n = 0..30
Hong-Chang Wang, Definition of the formula B(n,k,a,b)
FORMULA
a(n) = B(2*n+1,1,2,1) = (Product_{i=0..2*n+1}(i+2) - Product_{i=0..2*n+1}(-i-1))/(2*n+4), n >= 0.
a(n) = A062779(n)/(2*(n+1)). - Michel Marcus, Jun 11 2016
a(n) ~ sqrt(Pi)*exp(-2*n)*(48*n*(24*n + 13) + 1177)*4^(n-2)*n^(2*n+1/2)/9. - Ilya Gutkovskiy, Jul 07 2016
EXAMPLE
a(0) = (2*3 - 1*2)/4 = 1.
a(1) = (2*3*4*5 - 1*2*3*4)/6 = 16.
a(2) = (2*3*4*5*6*7 - 1*2*3*4*5*6)/8 = 540.
MATHEMATICA
a[n_] := (Product[i + 2, {i, 0, 2*n + 1}] - Product[-i - 1, {i, 0, 2*n + 1}])/(2*n + 4); Table[a[n], {n, 0, 10}] (* G. C. Greubel, Jun 19 2016 *)
Table[((n+1)(2n+2)!)/(n+2), {n, 0, 30}] (* Harvey P. Dale, Oct 24 2020 *)
PROG
(Python)
# subroutine
def B (n, k, a, b):
pa = pb = 1
for i in range(n+1):
pa *= (i*k+a)
pb *= (-i*k-b)
m = n*k+a+b
p = pa-pb
if m == 0:
return "NaN"
else:
return p/m
# main program
for j in range(101):
print(str(j)+" "+str(B(2*j+1, 1, 2, 1))) # Hong-Chang Wang, Jun 14 2016
(PARI) a(n) = ((2*n+1)!-(2*n)!)/(2*(n+1)) \\ Felix Fröhlich, Jun 11 2016
(PARI) a(n) = (prod(i=0, 2*n+1, i+2)-prod(i=0, 2*n+1, -i-1))/(2*n+4) \\ Felix Fröhlich, Jul 05 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Hong-Chang Wang, Jun 10 2016
STATUS
approved