**PROOF OF THE DEPASCALISATION
THEOREM**

** **Amarnath Murthy,
S.E.(E&T), WLS, Oil and Natural Gas Corporation Ltd., Sabarmati,
Ahmedabad,-380005 INDIA.

In [1] we have defined Pascalisation as follows:

Let b_{1}, b_{2}, . . . be a base sequence. Then the **Smarandache
Pascal derived sequence**

d_{1} , d_{2}, . . . is defined as

d_{1} = b_{1}

d_{2} = b_{1} + b_{2}

d_{3} = b_{1} + 2b_{2} + b_{3}

d_{4} = b_{1} + 3b_{2} + 3b_{3} + b_{4}

. . .

n

d_{n+1} = S ^{n}C_{k}
.b_{k+1}

k=0

Now Given S_{d }the task ahead is to find out the base sequence S_{b}
. We call the process of extracting the base sequence from the Pascal derived
sequence as **Depascalsation.** The interesting observation is that this
again involves the Pascal's triangle , but with a difference.

On expressing b_{k} 's in terms of d_{k}'s We get

b_{1} = d_{1}

b_{2} = -d_{1} + d_{2}

b_{3} = d_{1} - 2d_{2} + d_{3}

b_{4} = -d_{1} + 3d_{2} - 3d_{3} + d_{4}

**. . .**

which suggests the possibility of

n

b_{n+1} = S (-1)^{n+k}.
^{n}C_{k} .d_{k+1}

k=0

This I call as Depascalisation Theorem.

**PROOF: **We shall prove it by induction.

Let the proposition be true for all the numbers l £ k+1. Then we have

b_{k+1} = ^{k}C_{0} (-1)^{k+2} d_{1}
+ ^{k}C_{1} (-1)^{k+1} d_{2} +. . . + ^{k}C_{k}
(-1)^{2}

Also we have

d_{k+2} = ^{k+1}C_{0} b_{1} + ^{k+1}C_{1}
b_{2} + . . . + ^{k+1}C_{r} b_{r+1} + . . . +^{
k+1}C_{k+1} b_{k+2 }, which gives

b_{k+2} = (-1) ^{k+1}C_{0} b_{1} - ^{k+1}C_{1}
b_{2} - . . . - ^{k+1}C_{r} b_{r+1} - . . . + d_{k+2
}

substituting the values of b_{1}, b_{2}, . . .etc. in terms
of d_{1}, d_{2}, . . . , we get the coefficient of d_{1}
as

(-1) ^{k+1}C_{0} + (-^{k+1}C_{1})(-^{1}C_{0})
+(-^{k+1}C_{2})( ^{2}C_{0}) +. . .+ (-1)^{r}
. ^{k+1}C_{r})(^{r}C_{0}) +. . .+ (-1) ^{k+1}(^{k+1}C_{k})(^{k}C_{0})

- ^{k+1}C_{0} + ^{k+1}C_{1}._{ }^{1}C_{0}
- ^{k+1}C_{2 }. ^{2}C_{0} +. . .+ (-1)^{r}
. ^{k+1}C_{r }. ^{r}C_{0} +. . .+ (-1)^{k+1}.^{
k+1}C_{k }._{ }^{k}C_{0}

similarly the coefficient of d_{2} is

^{k+1}C_{1}._{ }^{1}C_{1}
+ ^{k+1}C_{2 }. ^{2}C_{1} +. . .+ (-1)^{r+1}.
^{k+1}C_{r }. ^{r}C_{1} +. . .+ (-1) ^{k+1 }.
^{k+1}C_{k }._{ }^{k}C_{1}

on similar lines we get the coefficient of d_{m+1}
as

^{k+1}C_{m}._{ }^{m}C_{m}
+ ^{k+1}C_{m+1 }. ^{m+1}C_{m} - . . .+ (-1)^{r+m}.
^{k+1}C_{r+m }. ^{r+m}C_{m} +. . .+ (-1) ^{k+m
}. ^{k+1}C_{k }._{ }^{k}C_{m}

k-m

= S (-1)^{h+1}
^{k+1}C_{m+h} . ^{m+h}C_{m}

h=0

(k+1)-m

S (-1)^{h+1}
^{k+1}C_{m+h} . ^{m+h}C_{m} + (-1) ^{k+m}_{
}. ^{k+1}C_{k+1} . ^{k+1}C_{m}
-------------(1)

h=0

Applying theorem {4.2} of reference [2], in (1) we get

= ^{k+1}C_{m }{ 1 + (-1)}^{k+1-m}_{
}+ (-1)^{k+m} . ^{k+1}C_{m}

= (-1)^{k+m} . ^{k+1}C_{m}

which shows that the proposition is true for (k+2) as well. The proposition has already been verified for k+1 = 3 , hence by induction the proof is complete.

In matrix notation if we write

[ b_{1}, b_{2} , . . b_{n}
]_{1xn} * [p_{i,j} ]^{'}_{nxn} = [ d_{1},
d_{2} , . . d_{n} ]_{1xn}

where [p_{i,j} ]^{'}_{nxn}
= the transpose of [p_{i,j} ]_{nxn} and

[p_{i,j} ]_{nxn} is given by p_{i,j}
= ^{i-1}C_{j-1} if i £
j else p_{i,j} = 0

Then we get the following result

If [q_{i,j}]_{nxn} is the transpose
of the inverse of [p_{i,j} ]_{nxn} Then

**q _{i,j} = (-1)^{j+i} .^{i-1}C_{j-1}**

We also have

[ b_{1}, b_{2} , . . b_{n}
] * [q_{i,j} ]^{'}_{nxn} = [ d_{1}, d_{2}
, . . d_{n} ]

where [q_{i,j} ]^{'}_{nxn}
= The Transpose of [q_{i,j} ]_{nxn}

References:

[1] Amarnath Murthy, ' Smarandache Pascal Derived Sequences', SNJ, March ,2000.

[2] Amarnath Murthy, 'More Results and Applications of the Smarandache Star Function., SNJ, VOL.11, No. 1-2-3, 2000.