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 b₁, b₂, . . . be a base sequence. Then the Smarandache Pascal derived sequence

d₁ , d₂, . . . is defined as

d₁ = b₁

d₂ = b₁ + b₂

d₃ = b₁ + 2b₂ + b₃

d₄ = b₁ + 3b₂ + 3b₃ + b₄

. . .

n

d_n+1 = S ⁿC_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

b₁ = d₁

b₂ = -d₁ + d₂

b₃ = d₁ - 2d₂ + d₃

b₄ = -d₁ + 3d₂ - 3d₃ + d₄

. . .

which suggests the possibility of

n

b_n+1 = S (-1)^n+k. ⁿC_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 = ^kC₀ (-1)^k+2 d₁ + ^kC₁ (-1)^k+1 d₂ +. . . + ^kC_k (-1)²

Also we have

d_k+2 = ^k+1C₀ b₁ + ^k+1C₁ b₂ + . . . + ^k+1C_r b_r+1 + . . . + ^k+1C_k+1 b_k+2 , which gives

b_k+2 = (-1) ^k+1C₀ b₁ - ^k+1C₁ b₂ - . . . - ^k+1C_r b_r+1 - . . . + d_k+2

substituting the values of b₁, b₂, . . .etc. in terms of d₁, d₂, . . . , we get the coefficient of d₁ as

(-1) ^k+1C₀ + (-^k+1C₁)(-¹C₀) +(-^k+1C₂)( ²C₀) +. . .+ (-1)^r . ^k+1C_r)(^rC₀) +. . .+ (-1) ^k+1(^k+1C_k)(^kC₀)

- ^k+1C₀ + ^k+1C₁. ¹C₀ - ^k+1C₂ . ²C₀ +. . .+ (-1)^r . ^k+1C_r . ^rC₀ +. . .+ (-1)^k+1. ^k+1C_k . ^kC₀

similarly the coefficient of d₂ is

^k+1C₁. ¹C₁ + ^k+1C₂ . ²C₁ +. . .+ (-1)^r+1. ^k+1C_r . ^rC₁ +. . .+ (-1) ^k+1 . ^k+1C_k . ^kC₁

on similar lines we get the coefficient of d_m+1 as

^k+1C_m. ^mC_m + ^k+1C_m+1 . ^m+1C_m - . . .+ (-1)^r+m. ^k+1C_r+m . ^r+mC_m +. . .+ (-1) ^k+m . ^k+1C_k . ^kC_m

k-m

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

h=0

(k+1)-m

S (-1)^{h+1 k+1}C_m+h . ^m+hC_m + (-1) ^k+m . ^k+1C_k+1 . ^k+1C_m -------------(1)

h=0

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

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

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

[ b₁, b₂ , . . b_n ]_1xn * [p_i,j ]^'_nxn = [ d₁, d₂ , . . 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-1C_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-1C_j-1

We also have

[ b₁, b₂ , . . b_n ] * [q_i,j ]^'_nxn = [ d₁, d₂ , . . 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.