A274117 -id:A274117 - cp's OEIS frontend

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

Hong-Chang Wang, Jun 10 2016

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

			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.

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

Cf. A008585, A273889, A273983, A274117.

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 *)

# 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

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
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)
E.g.f.: exp(x)*(11 + 109*x + 103*x^2 + 18*x^3). - Stefano Spezia, Aug 07 2024

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula