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%I #44 Jun 22 2016 23:33:39
%S 11,120,435,1064,2115,3696,5915,8880,12699,17480,23331,30360,38675,
%T 48384,59595,72416,86955,103320,121619,141960,164451,189200,216315,
%U 245904,278075,312936,350595,391160,434739,481440,531371,584640
%N a(n) = (Product_{i=0..4}(i*n+2) - Product_{i=0..4}(-i*n-1))/(4*n+3).
%C Sequence is inspired by A273983. The same argument as in A273889 can be used here to prove the expression evaluates to an integer.
%C 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
%H Hong-Chang Wang, <a href="/A274119/b274119.txt">Table of n, a(n) for n = 0..10000</a>
%H Hong-Chang Wang, <a href="/A274119/a274119_5.png">Definition of the formula B(n,k,a,b)</a>
%H Hong-Chang Wang, <a href="/A274119/a274119_6.png">Proof that (nk+a+b) divides (Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b))</a> - _Hong-Chang Wang_, Jun 17 2016
%F 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
%F Conjectures from _Colin Barker_, Jun 22 2016: (Start)
%F a(n) = 11+42*n+49*n^2+18*n^3.
%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.
%F G.f.: (11+76*x+21*x^2) / (1-x)^4.
%F (End)
%e a(0) = B(4,0,2,1) = (2*2*2*2*2 + 1*1*1*1*1)/3 = 11.
%e a(1) = B(4,1,2,1) = (2*3*4*5*6 + 1*2*3*4*5)/7 = 120.
%e a(2) = B(4,2,2,1) = (2*4*6*8*10 + 1*3*5*7*9)/11 = 435.
%t 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 *)
%o (Python)
%o # subroutine
%o def B (n,k,a,b):
%o pa = pb = 1
%o for i in range(n+1):
%o pa *= (i*k+a)
%o pb *= (-i*k-b)
%o m = n*k+a+b
%o p = pa-pb
%o if m == 0:
%o return "NaN"
%o else:
%o return p/m
%o # main program
%o for j in range(101):
%o print(str(j)+" "+str(B(4,j,2,1))) # _Hong-Chang Wang_, Jun 14 2016
%Y Cf. A008585, A273889, A274117.
%K nonn,hear
%O 0,1
%A _Hong-Chang Wang_, Jun 10 2016
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