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 A274119 a(n) = (Product_{i=0..4}(i*n+2) - Product_{i=0..4}(-i*n-1))/(4*n+3). 2
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified September 22 12:52 EDT 2019. Contains 327307 sequences. (Running on oeis4.)