OFFSET
1,2
COMMENTS
Solves the following arithmetic problem. Let (3) q#(i/j)=(q+h) be defined thus: Given q, an unknown fraction, q=qd/d, of integer value (initially) and h equal to any desired integer value and j equal to any desired integer value and i equal to j(q+h)+h.
Let '#' denote the repeated addition of the denominator j to the denominator of q and the repeated addition of the numerator i to the numerator of q, each addition recursively replacing the prior q fractions numerator and denominator respectively. This results in a series of fractions of mostly non-integer values.
After the first three iterations for any case would result in the following fractions: 1) (qd+i) / (d+j) 2) ((qd+i)+i) / ((d+j)+j) 3) (((qd+i)+i)+i) / (((d+j)+j)+j)
The question is, what is the smallest initial denominator, d (of q=qd/d) that in the course of the repeated additions will result in fractions of integer value, where every integer, from the initial (q+1)...(q+h) will be formed?
For example, let q = 4, h = 5, so (q+h)=9 and j = 1, so that for i we have i = 14. So in terms of q#(i/j)=(q+h) we have 4#(14/1)=9.
In this sample, since h=5 and j=1 we get from (2) above that d(5,1) = 252 as the solution. Applying this solution, we see then that the initial numerator of q, or q*d, becomes 4*252 = 1008 and the initial denominator is 252. Alternatively, in terms of q#(i/j)=(q+h) we have (1008/252)#(14/1)=9. We see that the repeated additions yield:
1) 1008 +14 and 252 +1 ~ 1022/253
2) 1022 +14 and 253 +1 ~ 1036/254
...
27) 1372 +14 and 277 +1 ~ 1386/279
28) 1386 +14 and 278 +1 ~ 1400/280 =5
...
63) 1876 +14 and 314 +1 ~ 1890/315 =6
...
108) 2506 +14 and 359 +1 ~ 2520/360 =7
...
168) 3346 +14 and 419 +1 ~ 3360/420 =8
...
252) 4522 +14 and 503 +1 ~ 4536/504 =9
Note that h=5 and j=1 were chosen so that d(5,1) = a(5) of the sequence. Also note that by the definition of q#(i/j)=(q+h) all answers are only a function of h and j.
Note also that q=qd/d can also be expressed as 0=0d/d and that any q#(i/j)=(q+h) can be expressed as 0#(i/j)=h [after adjustments are made to i]. For instance 4#(14/1)=9 is the equivalent of 0#(10/1)=5 in terms of d(h,j).
Interestingly: For any q#(i/j)=p and r#(s/j)=t then (q+r)#((i+s)/j)=(p+t). Also for any q#(i/j)=p then (qr)#((ir)/j)=(pr).
The sequence A025558 can be calculated from this formula when h = j, in otherwords using the sequence of d(1, 1)...d(n, n). i.e. a(7) = 735 = d(7,7) = lcm(49,21,35,7,21,7,1) a(8) = 2240 = d(8,8) = lcm(64,28,16,10,32,416,1)
Denominator of the sum of all elements of n X n Hilbert Matrix M[i,j] with alternate signs. M[i,j] = 1/(i+j-1)(i,j = 1..n). - Alexander Adamchuk, Apr 11 2006
LINKS
Eric Weisstein's World of Mathematics, Hilbert Matrix.
Eric Weisstein's World of Mathematics, Harmonic Number.
FORMULA
a(n) = Denominator[Sum[Sum[(-1)^(i+j)*1/(i+j-1),{i,1,n}],{j,1,n}]]. - Alexander Adamchuk, Apr 11 2006
EXAMPLE
a(5) = lcm(9,4,7,3) = 252
a(7) = lcm(13,6,11,5,9,4,1) = 25740
a(10)= lcm(19,9,17,4,3,7,13,3,11,1) = 11639628
a(14)= lcm(27,13,25,6,23,11,3,5,19,9,17,4,15,1) = 2868336900
n=2: HilbertMatrix[n,n]
1 1/2
1/2 1/3
so a(2) = Denominator[(1 - 1/2 - 1/2 + 1/3)] = Denominator[1/3] = 3.
The n X n Hilbert matrix begins:
1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 ...
1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 ...
1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 ...
1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 ...
1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
MATHEMATICA
Denominator[Table[Sum[Sum[(-1)^(i+j)*1/(i+j-1), {i, 1, n}], {j, 1, n}], {n, 1, 40}]] (* Alexander Adamchuk, Apr 11 2006 *)
CROSSREFS
KEYWORD
nonn,uned
AUTHOR
Scott C. Macfarlan (scottmacfarlan(AT)covance.com), Mar 01 2004
STATUS
approved