NEW FAMILY OF SEQUENCES

THE GENERATION OF NEW SEQUENCES (Part H)

Picture of a square

Introduction

Recently a new method for the generation of squares of squares were produced in Part IA through Part IVE. In addition, a new interleaved sequence of numbers was developed from this work and the sequence awarded the Sloane number A178218. OEIS has also published four other numbers based on this particular type of sequence. These numbers are A214345, A214493, A214393 and A214405.

This site will show that these four sequences as well as an infinitude of others are generated from the square sequence method developed in Part IA and ending with Part IVE .

Some Background

The previous parts showed that a tuple (a1,b1,c1) can be converted into a different tuple (a,b,c) basically a transformation of the type (a1,b1,c1) ⇒ (a,b,c). In addition, the initial tuples start out with the tuples (1,b1,1) or (1,1,c1) in which the b1s and the c1s have the following values:

b1 =  k
b1 = −k
c1 =  k
c1 = −k

where k is any natural integer from 1 to ∞ used in calculating f and the denominator d = 2(2b1c1 − 1 ) in the equation:

f = [2e2n2 + (4c1 − 4b1) en +(1 − 2b12 + c12)] / {2(2b1 − c1 − 1)}

This equation is critical is that it is the initial starting point for generating the interleaved sequences.

Table and General Sequence

The tables listed below although produced according to the methods of Parts IA through IVE, were actually computed and outputted by a computer program. Four sequences are apparent as highlighted in color. The one in green, the one in teal, the one in pink and the one in white. For example each green tuple ends with a number which is repeated in the next green tuple. Likewise for the teal, the pink and white tuples.

  
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Table I
a1 b1 c1
11-49
111-29
121-9
13111
14131
15151
16171
17191
181111
191131
1101151
1111171
1121191
1131211
1141231
  
f = S/d
24
6
-8
-18
-24
-26
-24
-18
-8
6
24
46
72
102
136
Table II
a b c
2525-25
717-23
-713-17
-1713-7
-23177
-252525
-233747
-175373
-773103
797137
25125175
47157217
73193263
103233313
137277367
  
Δ
0
240
120
-120
-240
0
840
2520
5280
9360
15000
22440
31920
43680
57969
For the tuple (1,1, −49)   f is calculated to be:
f = [2e2n2 + (− 196 − 4)en + 2400]/2×50 = [2e2n2 − 200en + 2400]/100
Setting e = 10 and g = 20 affords f = 2n2 − 20n + 24
Substituting this value of f in
(f + 1, f + 10n + 1, f + 20n − 49) as was shown in Part IA
gives the general equations for the complete sequence

a = (2n2 − 20n + 24 + 1 ) = (2n2 − 20n + 25 )
b = (2n2 − 20n + 24 + 10n + 1 ) = (2n2 − 10n + 25)
c = (2n2 − 20n + 24 + 20n − 49 ) = (2n2 − 25)

Separation of Sequences

To separate out the four sequences and generate the equations for each, the following method was found. All we need are two equations to generate the interleaved sequences in this case we wil take a and b. Since two sequences may be produced from table II above, a factor fs = 5 will be used to convert a ⇒ new a and b ⇒ new b. This is accomplished by substituting the value of 5n for n in the equations for a and b, to give the equations for the first sequence:

a = (50n2 − 100n + 25 )
b = (50n2 − 50n + 25)

Using these equations (or copying directly from Table II) we obtain for the first sequence:

25, 25, -25, 25, 25, 125, 175, 325, 425, 625, 775, 1025, 1225, 1525, 1775, 2125, 2425, 2825, 3175, 3625, 4025, 4525, 4975, 5525, 6025, 6625, 7175, 7825, 8425

which on dividing each entry by 25 affords the oeis Sloane sequence A178218:
1, 1, -1, 1, 1, 5, 7, 13, 17, 25, 31, 41, 49, 61, 71, 85, 97, 113, 127, 145, 161, 181, 199, 221, 241, 265, 287, 313, 337


To obtain the equations for the second sequence we use the second line of table II (1,11,-29) and solve for f.

For the tuple (1,11,-29)   f is calculated to be:
f = [2e2n2 + (−116 − 44)en + 600]/2×50 = [2e2n2 − 160en]+ 600/100
and by setting e = 10 affords the general equation f = 2n2 − 16n + 6
Substituting this value of f in
(f + 1, f + 10n + 11, f + 20n − 29) as was shown in Part IA
gives the general equations for the complete sequence starting at point (1,11,-29)

a = (2n2 − 16n + 6 + 1 ) = (2n2 − 16n + 7 )
b = 2(n2 − 16n + 6 + 10n + 11 ) = (2n2 − 6n + 17)
c = (2n2 − 16n + 6 + 20n − 29) = (2n2 + 4n − 23)

Substituting the value of 4n for n in the equations for a and b as was done above, affords the equations for the second sequence:

a = (50n2 − 80n + 7 )
b = (50n2 − 30n + 17)

Using these equations (or copying directly from Table II) we obtain for the second sequence identical to the Sloane No. A214393 except for the first three numbers which are also part of the sequence:

7, 17, -23, 37, 47, 157, 217, 377, 487, 697, 857, 1117, 1327, 1637, 1897, 2257, 2567, 2977, 3337, 3797, 4207, 4717, 5177, 5737, 6247, 6857, 7417, 8077, 8687, 9397, 10057, 10817, 11527, 12337, 13097, 13957, 14767, 15677, 16537, 17497, 18407, 19417, 20377, 21437, 22447, 23557, 24617, 25777
The Generating Function (G.f.) for this sequence is (7 + 3x − 57x2 + 97x3) /(1+x)(1−x)3.
Note that a G.f. as defined by Wikepedia is a formal power series in one indeterminate, whose coefficients encode information about a sequence of numbers an that is indexed by the natural numbers, i.e., the coefficients of the equation are each of the terms in the sequence being looked at which in this case are :
7 + 17x − 23x2 + 37x3 + 47x4+ 157x5 + 217x6 + 377x7 + 487x8 + 697x9 + 857x10 ...+ mxN where m and N approach ∞ and are obtained by dividing the numerator by the denominator above.


To obtain the equations for the third sequence we use the third line of table II (1,21,-9) and solve for f.

For the tuple (1,21,-9)   f is calculated to be:
f = [2e2n2 + (− 36 − 84)en − 800]/2×50 = [2e2n2 − 120en − 800]/100
and by setting e = 10 affords the general equation f = 2n2 − 12n − 8
Substituting this value of f in
(f + 1, f + 10n + 21, f + 20n − 9) as was shown in Part IA
gives the general equations for the complete sequence starting at point (1,21,-9)

a = (2n2 − 12n − 8 + 1 ) = (2n2 − 12n − 7 )
b = (2n2 − 12n − 8 + 10n + 21 ) = (2n2 − 2n + 13)
c = (2n2 − 12n − 8 + 20n − 9) = (2n2 + 8n − 17)

Substituting the value of 4n for n in the equations for a and b as was done above, affords the equations for the third sequence:

a = (50n2 − 60n − 7 )
b = (50n2 − 10n + 13)

Using these equations (or copying directly from Table II) we obtain for the third sequence:

-7, 13, -17, 53, 73, 193, 263, 433, 553, 773, 943, 1213, 1433, 1753, 2023, 2393, 2713, 3133, 3503, 3973, 4393, 4913, 5383, 5953, 6473, 7093, 7663, 8333, 8953, 9673, 10343, 11113, 11833, 12653, 13423, 14293, 15113, 16033, 16903, 17873, 18793, 19813, 20783, 21853, 22873, 23993, 25063, 26233
The Generating Function (G.f.) for this sequence is ( −7 + 27x − 43x2 + 73x3 /(1+x)(1−x)3 where the coefficients are:
−7 + 13x − 17x2 + 53x3 + 73x4+ 193x5 + 263x6 + 433x7 + 553x8 + 773x9 + 943x10 ...+ mxN where m and N approach ∞.


To obtain the equations for the fourth sequence we use the fourth line of table II (1,31,11) and solve for f.

For the tuple (1,31,11)   f is calculated to be:
f = [2e2n2 + (44 − 124)en − 1800]/2×50 = [2e2n2 − 80en − 1800]/100
and by setting e = 10 affords the general equation f = 2n2 − 8n − 18
Substituting this value of f in
(f + 1, f + 10n + 31, f + 20n + 11) as was shown in Part IA
gives the general equations for the complete sequence starting at point (1,31,11)

a = (2n2 − 8n − 18 + 1 ) = (2n2 − 8n − 17)
b = (2n2 − 8n − 18 + 10n + 31 ) = (2n2 + 2n + 13)
c = (2n2 − 8n − 18 + 20n + 11) = (2n2 + 12n − 7)

Substituting the value of 4n for n in the equations for a and b as was done above, affords the equations for the third sequence:

a = (50n2 − 40n − 17 )
b = (50n2 + 10n + 13)

Using these equations (or copying directly from Table II) we obtain for the fourth sequence:

-17, 13, -7, 73, 103, 233, 313, 493, 623, 853, 1033, 1313, 1543, 1873, 2153, 2533, 2863, 3293, 3673, 4153, 4583, 5113, 5593, 6173, 6703, 7333, 7913, 8593, 9223, 9953, 10633, 11413, 12143, 12973, 13753, 14633, 15463, 16393, 17273, 18253, 19183, 20213, 21193, 22273, 23303, 24433, 25513, 26693
The Generating Function (G.f.) for this sequence is ( −17 + 47x − 33x2 + 53x3 /(1+x)(1−x)3 where the coefficients are:
−17 + 13x − 7x2 + 73x3 + 103x4+ 233x5 + 313x6 + 493x7 + 623x8 + 853x9 + 1033x10 ...+ mxN where m and N approach ∞.


To obtain the equations for the fifth sequence we use the fifth line of table II (1,41,31) and solve for f.

For the tuple (1,41,31)   f is calculated to be:
f = [2e2n2 + (124 − 164)en − 2400]/2×50 = [2e2n2 − 40en − 2400]/100
and by setting e = 10 affords the general equation f = 2n2 − 4n − 24
Substituting this value of f in
(f + 1, f + 10n + 41, f + 20n + 31) as was shown in Part IA
gives the general equations for the complete sequence starting at point (1,41,31)

a = (2n2 − 4n − 24 + 1 ) = (2n2 − 4n − 23)
b = (2n2 − 4n − 24 + 10n + 41 ) = (2n2 + 6n + 17)
c = (2n2 − 4n − 24 + 20n + 31) = (2n2 + 16n + 7)

Substituting the value of 4n for n in the equations for a and b as was done above, affords the equations for the third sequence:

a = (50n2 − 20n − 23 )
b = (50n2 + 30n + 17)

Using these equations (or copying directly from Table II) we obtain for the fourth sequence:

-23, 17, 7, 97, 137, 277, 367, 557, 697, 937, 1127, 1417, 1657, 1997, 2287, 2677, 3017, 3457, 3847, 4337, 4777, 5317, 5807, 6397, 6937, 7577, 8167, 8857, 9497, 10237, 10927, 11717, 12457, 13297, 14087, 14977, 15817, 16757, 17647, 18637, 19577, 20617, 21607, 22697, 23737, 24877, 25967, 27157
The Generating Function (G.f.) for this sequence is ( −23 + 63x − 27x2 + 37x3 /(1+x)(1−x)3 where the coefficients are:
−23 + 17x + 7x2 + 97x3 + 137x4+ 277x5 + 367x6 + 557x7 + 697x8 + 937x9 + 1127x10 ...+ mxN where m and N approach ∞.


So in effect what we have done is to produce four new sequences via a completely new route as well as the original sequence with the Sloane number A178218. In addition, all the sequences are related via the general sequence formula (before factoring in fs) and thus constitute a family of interleaved sequences.

This concludes Part H. Go back to homepage.


Copyright © 2012 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com