**SMARANDACHE PASCAL DERIVED
SEQUENCES**

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

** **Given a sequence say S_{b} . We call it the base
sequence. We define a **Smarandache Pascal derived sequence S _{d} **as
follows:

n

T_{n+1} = S ^{r}C_{k}
.t_{k+1} , where t_{k} is the k^{th} term of the base
sequence.

k=0

Let the terms of the the base sequence S_{b} be

b_{1, }b_{2 }, b_{3} , b_{4} , . . .

Then the Smarandache Pascal derived Sequence S_{d}

d_{1, }d_{2 }, d_{3} , d_{4} , . . . is
defined as follows:

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

** **

These derived sequences exhibit interesting properties for some base sequences.

Examples:

**{1}** S_{b} ® 1, 2, 3,
4, . . . ( natural numbers)

S_{d} ® 1, 3, 8, 20, 48, 112,
256, . . . ( Smarandache Pascal derived natural number sequence)

The same can be rewritten as

2x2^{-1} , 3x2^{0}, 4x2^{1}, 5x2^{2}, 6x2^{3},
. . .

It can be verified and then proved easily that **T _{n} = 4( T_{n-1}
- T_{n-2} )** for n > 2.

And also that **T _{n} = (n+1) .2^{n-2}**

**{2}** S_{b} ® 1, 3, 5,
7, . . . ( odd numbers)

S_{d} ® 1, 4, 12, 32, 80, .
. .

The first difference 1, 3, 8, 20, 48 , . . . is the same as the S_{d}
for natural numbers.

The sequence S_{d} can be rewritten as

1.2^{0} , 2.2^{1}, 3.2^{2}, 4.2^{3}, 5.2^{4},
. . .

Again we have **T _{n} = 4( T_{n-1} - T_{n-2} )**
for n > 2.Also

**{3}** **Smarandache Pascal Derived Bell Sequence:**

Consider the Smarandache Factor Partitions (SFP) sequence for the square free numbers:

( The same as the **Bell number** sequence.)

S_{b} ® **1**, 1, 2, 5,
15, 52, 203, 877, 4140, . . .

We get the derived sequence as follows

S_{d} ® 1, 2, 5, 15, 52,
203, 877, 4140 , . . .

The **Smarandache Pascal Derived Bell Sequence** comes out to be the
same. We call it **Pascal Self Derived Sequence. **This has been established
in **ref. [1]**

In what follows, we shall see that this Transformation applied to Fibonacci Numbers gives beautiful results.

****{4} Smarandache Pascal derived Fibonacci Sequence:**

Consider the Fibonacci Sequence as the Base Sequence:

S_{b} ® **1, 1, 2, 3, 5, 8,
13, 21, 34, 55, 89, 114, 233, . . .**

We get the following derived sequence

S_{d} ® 1, 2, 5, 13, 34, 89,
233, . . . ---------- **(A)**

It can be noticed that the above sequence is made of the alternate (even numbered terms of the sequence ) Fibonacci numbers.

This gives us the following result on the Fibonacci numbers.

**n**

**F _{2n} = **

**k=0**

Some more interesting properties are given below.

If we take **(A) **as the base sequence we get the following derived
sequence S_{dd}

S_{dd} ® 1, 3, 10, 35, 125,
450, 1625, 5875, 21250, . . .

An interesting observation is ,the first two terms are divisible by 5^{0},
the next two terms by 5^{1} , the next two by 5^{2} , the next
two by 5^{3 }and so on.

**T _{2n} **

On carrying out this division we get the following sequence i.e.

1, 3, 2, 7, 5, 18, 13, 47, 34, 123, 89, . . . ----------- **(B)**

The sequence formed by the odd numbered terms is

1, 2, 5, 13, 34, 89, . . .

which is again nothing but S_{d} ( the base sequence itself.).

Another interesting observation is every even numbered term of **(B)** is
the sum of the two adjacent odd numbered terms. ( 3 = 1+2, 7 = 2 +5, 18 = 5 +
13 etc.)

**CONJECTURE: **Thus we have the possibility of another beautiful result on
the Fibonacci numbers which of-course is yet to be established.

**2m+1 r **

**F _{2m+1} = (1/5^{m}) **

**r =0 k=0**

**Note:** It can be verified that all the above properties hold good for
the **Lucas sequence ( 1 , 3, 4, 7, 11, . . .) **as well.

**Pascalisation of Fibonacci sequence with index in arithmetic progression:**

Consider the following sequence formed by the Fibonacci numbers whose indexes are in A. P.

F_{1 }, F_{d+1} , F_{2d+1} , F_{3d+1} , . .
. on pascalisation gives the following sequence

1, d.F_{2 }, d^{2}.F_{4} , d^{3}.F_{6}
, d^{4}.F_{8} , . . ., d^{n}.F_{2n} , . . .

for d = 2 and d = 3.

For d = 5 we get the following

Base sequence : F_{1}, F_{6} , F_{11}, F_{16},
. . .

1, 13, 233, 4181, 46368, . . .

Derived sequence: 1, 14, 260, 4920, 93200, . . .in which we notice that

260= 20.(14- 1), 4920 = 20.(260 - 14) , 93200 = 4920 - 260 ) etc . which suggests the possibility of

**Conjecture: The terms of the pascal derived sequence for d = 5 are given
by **

**T _{n }= 20.( T_{n-1} - T_{n-2} ) ( n > 2)**

For d = 8 we get

Base sequence : F_{1, }F_{9}, F_{17}, F_{25}
, . . .

S_{b} ---- 1, 34, 1597, 75025, . . .

S_{d} ---- 1, 35, 1666, 79919, . . .

= 1, 35, (35-1). 7^{2} , ( 1666 - 35 ). 7^{2} , . . . etc.
which suggests the possibility of

**Conjecture: The terms of the pascal derived sequence for d = 8 are given
by **

**T _{n }= 49.( T_{n-1} - T_{n-2} ) , (n > 2)**

Similarly we have Conjectures:

For d = 10 , **T _{n }= 90.( T_{n-1} - T_{n-2} ) ,
(n > 2)**

For d = 12 , **T _{n }= 18^{2}.( T_{n-1} - T_{n-2}
) , (n > 2)**

**Note: **There seems to be a direct relation between d and the
coefficient of ( T_{n-1} - T_{n-2} ) (or the common factor) of
each term which is to be explored.

{5} **Smarandache Pascal derived square sequence:**

S_{b} ® 1, 4, 9, 16, 25, . .
.

S_{d} ® 1, 5, 18, 56, 160,
432, . . .

Or 1, 5x**1**, 6x **3**, 7x **8**, 8x**20**, 9x**48 , . . . , (
T _{n} = (n+3)t_{n-1} ) , **where t

Also one can derive **T _{n }= 2^{n-2} . ( n+3)(n)/ 2** .

{6} **Smarandache Pascal derived cube sequence:**

S_{b} ® 1, 8, 27, 64, 125

S_{d} ® 1, 9, 44, 170, 576,
1792, . . .

We have T_{n} º 0 ( mod
(n+1)).

Similarly we have derived sequences for higher powers which can be analyzed for patterns.

{7} **Smarandache Pascal derived Triangular number sequence:**

S_{b} ® 1, 3, 6, 10, 15, 21,
. . .

S_{d} ® 1, 4, 13, 38, 104,
272, . . .

**{8} Smarandache Pascal derived Factorial sequence:**

S_{b} ® 1, 2, 6, 24, 120,
720, 5040, . . .

S_{d} ® 1, 3, 11, 49, 261,
1631, . . .

We can verify that **T _{n} = n.T_{n-1} + **

**Problem: Are there infinitely many primes in the above sequence?**

**Smarandache Pascal derived sequence of the k ^{th }order.**

Consider the natural number sequence again:

S_{b} ® 1, 2, 3, 4, 5, . . .
The corresponding derived sequence is

S_{d} ® 2x2^{-1} ,
3x2^{0}, 4x2^{1}, 5x2^{2}, 6x2^{3}, **. . .**
With this as the base sequence we get the derived sequence denoted by **S _{d2}**
as

S_{dd }or S_{d2} ®
1, 4, 15, 54, 189, 648, **. . .** which can be rewritten as

1, 4x3^{0} , 5x3^{1}, 6x3^{2} , 7x3^{3} **.
. .**

similarly we get S_{d3} as 1, 5x4^{0, }6x4^{1}, 7x4^{2}
, 8x4^{3} , **. . .** which suggests the possibility of the terms of
S_{dk} , the k^{th }order Smarandache Pascal derived natural
number sequence being given by

1, ( k+2) .(k+1)^{0}, (k+3).(k+1)^{1} , (k+4).(k+1)^{2},
. . ., (k+r).(k+1)^{r-2} etc. This can be proved by induction.

** We can take an arithmetic progression with the first term 'a' and
the common difference 'b' as the base sequence and get the derived k ^{th}
order sequences to generalize the above results.**

** **

** Reference:[**1] Amarnath Murthy, ' Generalization of
Partition Function,. Introducing Smarandache Factor Partitions' SNJ, Vol. 11,
No. 1-2-3,2000.