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 Sb . We call it the base sequence. We define a Smarandache Pascal derived sequence Sd as follows:

n

Tn+1 = S rCk .tk+1 , where tk is the kth term of the base sequence.

k=0

Let the terms of the the base sequence Sb be

b1, b2 , b3 , b4 , . . .

Then the Smarandache Pascal derived Sequence Sd

d1, d2 , d3 , d4 , . . . is defined as follows:

d1 = b1

d2 = b1 + b2

d3 = b1 + 2b2 + b3

d4 = b1 + 3b2 + 3b3 + b4

. . .

n

dn+1 = S nCk .bk+1

k=0

 

These derived sequences exhibit interesting properties for some base sequences.

Examples:

{1} Sb 1, 2, 3, 4, . . . ( natural numbers)

Sd 1, 3, 8, 20, 48, 112, 256, . . . ( Smarandache Pascal derived natural number sequence)

The same can be rewritten as

2x2-1 , 3x20, 4x21, 5x22, 6x23, . . .

It can be verified and then proved easily that Tn = 4( Tn-1 - Tn-2 ) for n > 2.

And also that Tn = (n+1) .2n-2

{2} Sb 1, 3, 5, 7, . . . ( odd numbers)

Sd 1, 4, 12, 32, 80, . . .

The first difference 1, 3, 8, 20, 48 , . . . is the same as the Sd for natural numbers.

The sequence Sd can be rewritten as

1.20 , 2.21, 3.22, 4.23, 5.24, . . .

Again we have Tn = 4( Tn-1 - Tn-2 ) for n > 2.Also Tn = n.2n-1.

{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.)

Sb 1, 1, 2, 5, 15, 52, 203, 877, 4140, . . .

We get the derived sequence as follows

Sd 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:

Sb 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 114, 233, . . .

We get the following derived sequence

Sd 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

F2n = S nCk .Fk , where Fk is the kth term of the base Fibonacci sequence.

k=0

Some more interesting properties are given below.

If we take (A) as the base sequence we get the following derived sequence Sdd

Sdd 1, 3, 10, 35, 125, 450, 1625, 5875, 21250, . . .

An interesting observation is ,the first two terms are divisible by 50, the next two terms by 51 , the next two by 52 , the next two by 53 and so on.

T2n T2n-1 0 ( mod 5n )

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 Sd ( 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

F2m+1 = (1/5m) S { 2m+1Cr ( S rCk Fk ) }

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.

F1 , Fd+1 , F2d+1 , F3d+1 , . . . on pascalisation gives the following sequence

 

1, d.F2 , d2.F4 , d3.F6 , d4.F8 , . . ., dn.F2n , . . .

for d = 2 and d = 3.

 

For d = 5 we get the following

Base sequence : F1, F6 , F11, F16, . . .

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

Tn = 20.( Tn-1 - Tn-2 ) ( n > 2)

For d = 8 we get

Base sequence : F1, F9, F17, F25 , . . .

Sb ---- 1, 34, 1597, 75025, . . .

Sd ---- 1, 35, 1666, 79919, . . .

= 1, 35, (35-1). 72 , ( 1666 - 35 ). 72 , . . . etc. which suggests the possibility of

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

Tn = 49.( Tn-1 - Tn-2 ) , (n > 2)

Similarly we have Conjectures:

For d = 10 , Tn = 90.( Tn-1 - Tn-2 ) , (n > 2)

For d = 12 , Tn = 182.( Tn-1 - Tn-2 ) , (n > 2)

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

{5} Smarandache Pascal derived square sequence:

Sb 1, 4, 9, 16, 25, . . .

Sd 1, 5, 18, 56, 160, 432, . . .

Or 1, 5x1, 6x 3, 7x 8, 8x20, 9x48 , . . . , ( Tn = (n+3)tn-1 ) , where tr is the rth term of Pascal derived natural number sequence.

Also one can derive Tn = 2n-2 . ( n+3)(n)/ 2 .

{6} Smarandache Pascal derived cube sequence:

Sb 1, 8, 27, 64, 125

Sd 1, 9, 44, 170, 576, 1792, . . .

We have Tn 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:

Sb 1, 3, 6, 10, 15, 21, . . .

Sd 1, 4, 13, 38, 104, 272, . . .

{8} Smarandache Pascal derived Factorial sequence:

Sb 1, 2, 6, 24, 120, 720, 5040, . . .

Sd 1, 3, 11, 49, 261, 1631, . . .

We can verify that Tn = n.Tn-1 + S Tn-2 + 1.

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

 Smarandache Pascal derived sequence of the kth order.

Consider the natural number sequence again:

Sb 1, 2, 3, 4, 5, . . . The corresponding derived sequence is

Sd 2x2-1 , 3x20, 4x21, 5x22, 6x23, . . . With this as the base sequence we get the derived sequence denoted by Sd2 as

Sdd or Sd2 1, 4, 15, 54, 189, 648, . . . which can be rewritten as

1, 4x30 , 5x31, 6x32 , 7x33 . . .

similarly we get Sd3 as 1, 5x40, 6x41, 7x42 , 8x43 , . . . which suggests the possibility of the terms of Sdk , the kth 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 kth 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.