hp41programs

Overview

-Ramanujan numbers Tau(n) are defined by

   x.[ (1-x).(1-x²).(1-x³)....... ]²⁴  = Tau(1).x + Tau(2).x² + Tau(3).x³ + ..... + Tau(n).xⁿ + ......

-"TAU" calculates Tau(n) by the recurrence relation:

    (n-1).Tau(n) = SUM_1<=m<=b(n) (-1)^m+1 (2m+1).(n-1-9m(m+1)/2).Tau(n-m(m+1)/2)     where   b(n) = INT ((8n+1)^1/2-1)/2

-However, this algorithm is very sensitive to roundoff errors and results are exact up to n = 43 only.
-Therefore, another approach is necessary to compute Tau(n) for larger arguments.
-"TAU2" employs the formula:

    Tau(n) = n⁴.s(n) - 24 SUM_0<k<n k².(35.k²-52.k.n+18.n²).s(k).s(n-k)     ( n > 1 )  where  s(k) is the sum of the divisors of k.

-Multiple precision is used and Tau(n) may be exactly computed up to n = 13868.
 

1°) Program#1   ( small arguments )
 

-This program calculates and stores Tau(1) , Tau(2) , ...... , Tau(n) into registers R01 , R02 , ....... , Rnn

Data Registers:  R00 = 0 , R01 = Tau(1) , ........... , Rnn = Tau(n)
Flags: /
Subroutines: /

-Synthetic registers M , N , O may be replaced by any unused data registers,
 for instance R97 , R98 , R99  if  n < 97
 
 

 
  ( 106 bytes / SIZE nnn+1, but at least 003 )
 
 

 
Example:    Evaluate Tau(10)

   10  XEQ "TAU"  >>>>  Tau(10) = -115920  in X-register and in R10  ( execution time = 32 seconds )

-We also have   1    , -24 ,  252 , -1472 , 4830 , -6048 , -16744 , 84480 , -113643  ( i-e Tau(1) , .......... , Tau(9) )
    in registers   R01 , R02 , R03 ,   R04  ,  R05 ,    R06 ,     R07  ,    R08  ,     R09       respectively

Notes:

-Without lines 27 and 66  ( FIX 0 and RND ) , Tau(33) would already be wrong !
-This program yields Tau(84) = 6,188,510,373 whereas the correct value is Tau(84) = 6,211,086,336
  and 300 XEQ "TAU" produces a completely meaningless result !
-Obviously, roundoff errors overwhelm the wanted function.
 

2°) Program#2   ( 1 < n < 13869 )
 

-The following program computes Tau(n) by groups of 4 digits in registers R15 R16 R17 R18 R19 R20
 

Data Registers:        R01 = n   ;   R00 and R02 thru R20 are used for temporary data storage and when the program stops:

                                   R03 = Tau(n) = X-register ( approximate value )
                                   R15 to R20 contain the digits of Tau(n) ( exact value )
Flags: /
Subroutines: /

-If you don't have an HP-41CX , replace line 08 by  0  LBL 00  STO IND Y  ISG Y  GTO 00   or add  CLRG  after line 01
 
 

 
   ( 352 bytes / SIZE 021 )
 
 

  15.020  = control number of the exact result ( in registers R15 thru R20 )
  X-output is only an approximate value

Example1:     Calculate  Tau(84)

   84  XEQ "TAU2"  >>>>  Tau(84) =  6211086336 ( 13m07s )    In this case, the approximate value is quite exact!

which is confirmed by   R15 = 0 , R16 = 0 , R17 = 0 , R18 = 62 , R19 = 1108 , R20 = 6336

Example2:    Compute Tau(314)

  314  XEQ "TAU2"  >>>>  Tau(314) = -3.156280211 10¹³  = approximate value ( in 1h07mn20s )

and we have:  R15 = 0 , R16 = 0 , R17 = -31 , R18 = -5628 , R19 = -210 ( which is to be read -0210 ) , R20 = -5744

 whence   Tau(314) = -31562802105744

-Remark that registers R15 to R20 are inferior or equal to zero when Tau(n) < 0

Notes:

-With a good 41-emulator in turbo mode  13868 XEQ "TAU2"  yields:

   R15 = -336 ; R16 = -1948 ; R17 = -0538 ; R18 = -4247 ; R19 = -3535 ; R20 = -1552   ( in about 1h23mn on my PC )

  whence  Tau(13868) = -33619480538424735351552      with a "real" HP-41, execution time is probably of the order of  8 or 9 days!

-For n > 13868 , the calculations involve numbers greater than 10¹⁰ and the results are only approximate,
 but if we use more registers to perform multiplications and additions, this limit can be overcome.