A274119 a(n) = (Product_{i=0..4}(i*n+2) - Product_{i=0..4}(-i*n-1))/(4*n+3).

11, 120, 435, 1064, 2115, 3696, 5915, 8880, 12699, 17480, 23331, 30360, 38675, 48384, 59595, 72416, 86955, 103320, 121619, 141960, 164451, 189200, 216315, 245904, 278075, 312936, 350595, 391160, 434739, 481440, 531371, 584640

Offset

0,1

Comments

Sequence is inspired by A273983. The same argument as in A273889 can be used here to prove the expression evaluates to an integer.

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(2*n,2,2,1) = (Product_{i=0..2*n}(2*i+2) - Product_{i=0..2*n}(-2*i-1))/ (4*n+3) = A273889(n+1), n >= 0; B(2*n,3,2,1) = (Product_{i=0..2*n}(3*i+2) - Product_{i=0..2*n}(-3*i-1))/(6*n+3) = A274117(n+1), n >= 0; B(2,n,2,1) = (Product_{i=0..2}(i*n+2) - Product_{i=0..2}(-i*n-1))/(2*n+3) = A008585(n+1), n >= 0; and a(n) is B(4,n,2,1). - Hong-Chang Wang, Jun 17 2016

Links

Hong-Chang Wang, Table of n, a(n) for n = 0..10000

Hong-Chang Wang, Definition of the formula B(n,k,a,b)

Hong-Chang Wang, Proof that (nk+a+b) divides (Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b)) - Hong-Chang Wang, Jun 17 2016

Formula

a(n) = B(4,n,2,1) = (Product_{i=0..4}(i*n+2) - Product_{i=0..4}(-i*n-1))/(4*n+3), n >= 0. - Hong-Chang Wang, Jun 14 2016

Conjectures from Colin Barker, Jun 22 2016: (Start)

a(n) = 11+42*n+49*n^2+18*n^3.

a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.

G.f.: (11+76*x+21*x^2) / (1-x)^4.

(End)

Example

a(0) = B(4,0,2,1) = (2*2*2*2*2 + 1*1*1*1*1)/3 = 11.
a(1) = B(4,1,2,1) = (2*3*4*5*6 + 1*2*3*4*5)/7 = 120.
a(2) = B(4,2,2,1) = (2*4*6*8*10 + 1*3*5*7*9)/11 = 435.

Mathematica

B[n_, k_] := (Product[k (i - 1) + 1, {i, 2 n - 1}] + Product[k (i - 1) + 2, {i, 2 n - 1}])/(2 k (n - 1) + 3); Table[B[3, n], {n, 0, 31}] (* Michael De Vlieger, Jun 10 2016 *)

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(4, j, 2, 1)))  # Hong-Chang Wang, Jun 14 2016

Crossrefs

Cf. A008585, A273889, A274117.

Sequence in context: A060499 A164828 A060498 * A171316 A081122 A004190

Adjacent sequences: A274116 A274117 A274118 * A274120 A274121 A274122

Keyword

nonn,hear

Author

Hong-Chang Wang, Jun 10 2016

Status

approved