A273983 -id:A273983 - cp's OEIS frontend

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

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

A274136 a(n) = (n+1)(2n+2)!/(n+2).

Original entry on oeis.org

1, 16, 540, 32256, 3024000, 410572800, 76281004800, 18598035456000, 5762136335155200, 2211729098342400000, 1030334000462807040000, 572721601599913328640000, 374484928188990947328000000, 284562454970932936468070400000

Offset: 0

Hong-Chang Wang, Jun 10 2016

nonn

Sequence is inspired by A273889, A274119 and A273983.
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

			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.

Hong-Chang Wang, Table of n, a(n) for n = 0..30
Hong-Chang Wang, Definition of the formula B(n,k,a,b)

Cf. A000012, A002450, A062779, A100040, A273889, A273983, A274119.

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 *)

a(n) = ((2*n+1)!-(2*n)!)/(2*(n+1)) \\ Felix Fröhlich, Jun 11 2016

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

# 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

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

cp's OEIS Frontend

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

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A274136 a(n) = (n+1)(2n+2)!/(n+2).

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cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

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Crossrefs

Programs

Mathematica

Python

Formula

A274136 a(n) = (n+1)*(2*n+2)!/(n+2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

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

A274136 a(n) = (n+1)(2n+2)!/(n+2).