On the basis of the above definition of the U.S., we can make its exact mathematical computation as follows:

The first thing to be calculated is the number of seeds which are filled in Xo. This number can be calculated if we know-

1. The dimensions of a single mustard (*sarṣapa*) seed.

2. The number of such mustard seeds contained in the right circular cylinder of diameter 1,00,00 *yojanas *and height 1000 *yojanas*.

3. The number of seeds contained in the conical heap.

Now, we proceed to calculate these three numbers. We find that *sarṣapa *is the name of a linear measure which is equal to the diameter of one mustard seed (which is also called *sarṣapa*) and which has a perfectly spherical shape. According to one tradition found in the *Trilokasāra*,^{[1]} we get

8 *sarṣapas *= 1 *yava*

8 *yavas *= 1 *aṅgula*.

1 *aṅgula *= 64 *sarṣapas*

Now, as we have already seen the diameters of the continents of oceans are measured by '*paramāṇāngula*'; we get, therefore, from the table of space-points (see supra p. 334-336).

1 *yojana *= (768000 x 500) *utṣedha- aṅgula*

1 *yojana *= (768000 x 500 x 64) *sarṣapas*

= 24576000000 *sarṣapas*

radius of the circular base of the cylinder

= 50000 *yojanas*

= 12288 x 1011 *sarṣapas*

*yojanas*

= 24576 x 109 *sarṣapas*

*sarṣapa*= 4/3 π (radius)

^{3}

radius of the seed of *sarṣapa *= ½ *sarṣapa*;

*sarṣapa*

= 4/3 π (½)^{3} cubic *sarṣapas*

= π/6 cubic *sarṣapas*

Now, volume of right circular cylinder

= π x (radius)^{2} x height................. (1)

Now, we should know the height of the right circular conical heap above the cylinder.^{[2]} According to the *Tilokasāra*,^{[3]} if the seeds of *tila*, *sarṣapa*, *valla*, *arhada*, *caṇā *(gram), *atasi*, *kulattha*, *rajāmaṣa *etc. are filled in a right circular cylinder in such a way that a conical heap is formed, then

height of the conical heap = circumfernce of the circular base of the cylinder /11

Volume of the right circular cone

= 1/3 π (radius)^{2} x height

^{[4]} = [1/3 π (radius)^{2}] x

= 1/3 π (radius)^{2} x

= 2/33 π^{2} (radius)^{3}........................................ (2)

Now, putting the values of radius and height in equation no. 1 above, we get the

Volume of right circular cylinder

= π x (12288 x 10^{11})

^{2}x (24576 x 10

^{9}) cubic-

*sarṣapas*

Volume of one seed of *sarṣapa *is π/6 cubic *sarṣapas*. Therefore, the number of *sarṣapa *seeds in the cylinder

= 6 x (12288 x 10^{11})^{2} x (24576 x 10^{9})

= 222,651,104,624,64 x 10^{31}

= 2.226 x 1044 approximately................... (3)

By substituting the value of radius in equation no. 2, we get the volume of right circular conical heap

= 12/33 x π x (12288 x 10^{11})^{3}

= 2.119 x 1045 approximately^{[5]}................... (4)

The number of *sarṣapas *in the conical heap

= 2.341 x 1045 approximately.^{[6]}

Thus, the total number of seeds in the cylinderical container of 10000 *yojanas *diameter and 1000 *yojanas *height together with the conical heap is 2.341 x 1045.

We have seen that

(1) Volume of the cylindrical container is

= πr^{2}h

where r is the radius of the circular base and h is the height of the cylinder

(2) Volume of the conical heap is

= 2/33 π^{2}r^{2}

(radius of the circular base of the cone is also r).

(3) Volume of a single seed is π/6

Thus, the total number of seeds in the cylindrical container with the conical heap above is

Now we know that h, the height of the cylindrical container, is the same in all containers A, B, C and X_{o}, X_{1}, X_{2}...... etc. but the r, the radius is different in them.

Value of h is

= 24576 x 109 *sarṣapa*

The *anavasthita *cylinders are made again and again with changed diameters. Let us denote

the original anavasthita cylinder as X_{o}

firstly made new anavasthita cylinder as X_{1}

secondly made new anavasthita cylinder as X_{2}

and so on.

Let the

radius of Xo be denoted as r_{o}

radius of X1 be denoted as r_{1}

radius of X2 be denoted as r2

and so on.

Radius of A, B and C is also r_{o}, which is

= 12288 x 10^{11 }*sarṣapas*.

Let k_{o}, k_{1}, k_{2}...... denote the total number of seeds in X_{o}, X_{1},X_{2}...... respectively.

Thus,

k_{o} = 6r_{o}^{2}h + 4/11 πr_{o}^{3 }.................. (0)

= 2.341 x 10^{45} approximately.

Now, by dropping one seed each from X_{o} into the rings of continents and oceans, the last seed will be dropped in the k_{o }^{th }ring with radius

r_{1}, = 2^{(k}^{o}^{-1)}r_{o} = 2^{(k}^{o}^{-2)} x 2r_{o}

Now, as 2r_{o} = 1 lakh *yojanas*

r_{1} = 2^{(k}^{o}^{-2)} lakh *yojanas*

When X_{o} becomes emptied, one seed is dropped in the cylinder A.

Total number of seeds contained in X_{1} is

X_{1} = 6r_{1}^{2}h + 4/11 πr_{1}3.................. (1)

^{[7]} = 6 [2^{2(ko-1)}r_{o}^{2}]h + 4/11 π[2^{3(ko-1)}r_{o}^{3}]

Now, by dropping one seed each in the rings ahead the last seed will be dropped in

(k_{o}+k_{1})^{th} ring.

The radius of this ring will be

r_{2} = 2^{(ko+k1-2)} lakh *yojanas*

= 2^{(ko+k1-1)}r_{o}

In this way, on emptying the X_{1}, second seed will be dropped in the cylinder A.

Total number of seeds in X2 will be

k^{2} = 6r_{2 }^{2} h + 4/11 πr_{2}^{3}.................. (2)

In the above manner, the last seed of X_{2} will be dropped in (k_{o}+k_{1}+k_{2})^{th} ring.

Now, third seed will be dropped in the cylinder A and X_{3 }will be made with radius r_{3}.

Total number of seeds in X_{3} will be

k_{3} = 6r_{3}^{2}h + 4/11 πr_{3} ^{3}.................. (3)

This process will continue till A is filled completely with a conical heap above. It means that when ko seeds will be dropped in A, it will be so filled. This will happen when the cylinder

X_{ko-1 }will be emptied. Its last seed will be dropped in the

Now the first seed will be dropped in the cylinder B, and a new *anavasthita *cylinder X_{ko} will be made. Its radius will be

Now, again to fill the cylinder A which is already emptied, the whole process as above will be repeated. It means that when the cylinder X_{2ko}-1 will be emptied, cylinder A will be filled second time and then the second seed will be dropped in the cylinder B. Thus, by repeating the above process ko times, the cylinder B will be completely filled. This will happen when the cylinder X_{2k0}-1 will be emptied. Its last seed will be dropped in the

Now the first seed will be dropped in the cylinder C, and a new *anavasthita *cylinder will be made. Its radius will be and the total number of seeds contained in it will be The values of can be found from the original equations.

Now, by repeating the process again ko times, the cylinder C will be completely filled. This will happen when the cylinder will be emptied. Its last seed will be dropped in the

Now all the three cylinders A, B and C are completely filled. A new *anavasthita *cylinder will be made. The radius will be given by

The number *Utkṛṣṭa Saṃkhyāta *(**U. S.**)

Now, we can see that although the value of kko 3 is definitely defined by the equation (o) to K_{0}^{2}, yet it will not be possible to compute it mathematically in terms of digits. However, by substituting approximate values of k_{o} etc. in the above equations, we can get a glimpse of the gigentic value of U. S.

By simplifying the equations (1) to (k_{ko 2}) as shown in foot note no. 1 on p. 362, and by neglecting the second term therein, we get

And

Now, by putting the values of k_{1}, k_{2} in these equations, we get,

By putting k′ in place of k_{o}-1 of and expanding the right side of equation we get

Now, the value of the last term of the right side is the highest and values of other terms are not less than one; hence, by neglecting other terms, we get,

By substituting

Now by putting m in place of 2^{n} in the above relation we get,

Now, the last term in the above relation will have the highest value. It will be

Thus, by neglecting other terms, we get

here m = 2^{n} and

We have already found out that k_{o}>10^{45}. By using logarithm,^{[8]} we get,

Thus the value of is greater than

It means that the value of is greater than the number obtained by raising

to itself 10^{133} times.^{[9]}

*Trilokasāra *with the *bhāṣā-ṭikā *(commentary in vernacular language) by Pt. Todarmalji, *śloka *18. However, in the *Lalitavistara *of Buddhist tradition, we get:

7 *sarṣapas *= 1 *yava*

7 *yavas *= 1 *aṅgulī-parva*.

Dr. Laxmichand Jain, in his *Prastāvanā *(Preface) to '*Jambūdvīpapṇṇattisaṃgaho*', has written an essay titled '*Tiloyapaṇṇatti kā Gaṇita*' (mathematics of *Tilloyapaṇṇattī*), in which he writes: "Here, we do not know how much and why this height of the cone is taken."

*Tilasarisavavallāḍhai caṇaatasikulattharāyamāsādi pariṇāhekkārasamo vaho jadi gayaṇgo rāsī-Trilokasāra*, vese 23

*Trilokasāra*in place of 1/3 π (radius)

^{2}we get [circumference]

^{2}But this is only approximately true, because if we take π=3, then

volume = 1/3 π (radius)^{2}

= 1/3 x 3 x (radius)2

= (radius)2

And also {{46ee6124bc36f9fa54d7f388c89c54e2722776f1}}= (radius)^{2}

^{2}= (circumference/6)

^{2}But, as we have already seen the correct value of π is not 3, but 3.1415..... (See, Appendix II, supra p. 346).

This value is found with the help of logarithm. The value of log π is taken as.4971. If we want to get the exact value of the number of seeds of *sarṣapa *contained in the conical heap, we have to take the value of π up to 32 decimal points. Here, however, we have taken this value up to 4 decimal points.

*Trilokasāra*, the total number of seeds of

*sarṣapa*contained in the cylindrical container and the conical heap is found as follows:

1. Value of π is taken as 3.

2. The volume of the sphere is taken as 9/2 (radius)^{3}.

3. The volume of the right circular cone of the height Oce/11 is

{{c924d6fcbde8f5b76dfc6d0f91e11956b11f312c}}

4. Volume of cylinderical container is taken as [3 x (radius)^{2} x (height)]

*Trilokasāra*, this value (of the no. of seeds in the cylindrical container) is given to be 197,912,092,999,680,000,000,000,000,000,000,000,000,000,000 and no. of seeds in the conical heap is given as 179, 920, 084, 545, 163, 636, 363, 636, 363, 636, 363, 636, 363, 636, 363, 336, 6

^{4}/11. Total no. of seeds required is given as 199, 711, 293, 845, 131, 636, 363, 636, 363, 636, 363, 636, 363, 636, 363, 636, 363, 6

^{4}/11 = 1.997 x 10

^{45}approximately

*iloyapaṇṇatti kā Gaṇita*has taken {{c60cf5068f14b9c3712c9908a8468efc668b73c5}} But this seems to be erroneous; maybe he has neglected the number of seeds contained in the conical heap.

^{2 }= 2x2 = 4 By raising twice 2 to the power 2, we get {{fef048771658aef5d63ffa20aee1bdfd7aa57fbf}} {{8b7f96b036220421182552ed62bde5323ac29fc4}} Then, again it would be beyond imagination to gauge the value of {{3beaba32da88eee4cbdb39f941af53a26770ee6b}} Where {{5a76f0c25ee4e406244343f19e3d31c9790fa158}} and ultimately the immense value of k or

*Utkṛṣṭa saṃkhyāta*can also be imagined.