Given that
k=∞ 1 A*BC*D
Σ  = 
k=0 (3k+1)(3k+2)(3k+3) E
where A, B, C, D and E are, in some order, the number π (pi), the square root of an integer, the natural logarithm of an integer, and two integers, find their values.
Doing the sum until the numbers didn't change resulted in
.1787967688907531
So we want a combination that comes out to about this. The following program assumes the integer whose square root is used is under 23 as is the integer whose natural log is used. The two integers that are used asis are assumed to be under 45. The program tries all possibilities (of course not bothering to put, for example, B = pi, as A = pi already covers this sort of possibility).
DEFDBL AZ
k = 0
denom = 6
DO
term = 1 / denom
ptot = tot
tot = tot + term
k = k + 1
denom = (3 * k + 1) * (3 * k + 2) * (3 * k + 3)
LOOP UNTIL ptot = tot
PRINT tot
pi = ATN(1) * 4
FOR int1 = 1 TO 22
sr = SQR(int1)
FOR int2 = 1 TO 22
lg = LOG(int2)
FOR int3 = 1 TO 44
FOR int4 = 1 TO 44
a = pi
b = sr
c = lg
d = int3: e = int4: GOSUB test
e = lg
d = int3: c = int4: GOSUB test
c = sr
b = lg
d = int3: e = int4: GOSUB test
d = lg
b = int3: e = int4: GOSUB test
e = lg
b = int3: d = int4: GOSUB test
e = sr
b = lg
d = int3: c = int4: GOSUB test
d = lg
b = int3: c = int4: GOSUB test
c = pi
b = sr
a = lg
d = int3: e = int4: GOSUB test
e = lg
d = int3: a = int4: GOSUB test
a = sr
b = lg
d = int3: e = int4: GOSUB test
d = lg
b = int3: e = int4: GOSUB test
e = lg
b = int3: d = int4: GOSUB test
e = sr
b = lg
d = int3: a = int4: GOSUB test
d = lg
b = int3: a = int4: GOSUB test
e = pi
a = sr: b = lg: c = int3: d = int4: GOSUB test
b = int3: c = lg: GOSUB test
a = int4: d = sr: GOSUB test
NEXT
NEXT
NEXT
NEXT
END
test:
IF e <> 0 THEN
v = (a * b  c * d) / e
tst = ABS(tot / v)
IF ABS(tst  1) < 1E08 THEN
PRINT v; a; b; c; d; e
END IF
END IF
RETURN
The following data result:
.178796768891527 3.141592653589793 1.732050807568877 1.09861228866811 3 12
.178796768891527 3.141592653589793 3.464101615137754 1.09861228866811 6 24
.178796768891527 3.141592653589793 3.464101615137754 2.19722457733622 3 24
The value of the 5number formula is shown; we presume the discrepancy from the total is due to rounding errors in the total and in not counting the total infinitely far out. The values of A, B, C, D and E are shown on the rest of each line.
The first line shows A=pi, B=sqrt(3), C=ln(3), D = 3 and E = 12, for
(pi*sqrt(3)  3*ln(3))/12
The second line shows A=pi, B=sqrt(12), C=ln(3), D=6 and E=24, this being the same as the previous with numerator and denominator multiplied by 2.
The third line shows A=pi, B=sqrt(12), C=ln(9), D=3 and E=24, thereby doubling C and halving D from the previous solution.
In each case, of course, A and B can be interchanged as can C and D.

Posted by Charlie
on 20051110 16:09:58 