# On the generalized constructive set.

Abstract In this paper, we use the elementary methods to study the
properties of the constructive set S, and obtain some interesting
properties for it.

Keywords Generalized constructive set, summation, recurrence equation, characteristic equation.

[section] 1. Introduction and Results

The generalized constructive set S is defined as: numbers formed by digits [d.sub.1], [d.sub.2], ..., [d.sub.m] only, all [d.sub.i] being different each other, 1 [less than or equal to] m [less than or equal to] 9. That is to say,

(1) [d.sub.1], [d.sub.2], ..., [d.sub.m] belongs to S;

(2) If a; b belong to S, then [bar.ab] belongs to S too;

(3) Only elements obtained by rules (1) and (2) applied a finite number of times belongs to S.

For example, the constructive set (of digits 1, 2) is: 1; 2; 11; 12; 21; 22; 111; 112; 121; 122; 211; 212; 221; 222; 1111; 1112; 1121.... And the constructive set (of digits 1, 2, 3) is: 1; 2; 3; 11; 12; 13; 21; 22; 23; 31; 32; 33; 111; 112; 113; 121; 122; 123; 131; 132; 133; 211; 212; 213; 221; 222; 223; 231; 232; 233; 311; 312; 313; 321; 322; 323; 331; 332; 333; 1111;.... In problem 6, 7 and 8 of reference [1], Professor F.Smarandache asked us to study the properties of this sequence. In [2], Gou Su had studied the convergent properties of the series

[[+infinity].summation over (n=1)] 1/[a.sup.[alpha].sub.n,

and proved that the series is convergent if [alpha] > log m, and divergent if [alpha] [less than or equal to] log m, where {[a.sub.n]} denotes the sequence of the constructive set S, formed by digits [d.sub.1]; [d.sub.2], ..., [d.sub.m] only, all [d.sub.i] being different each other, 1 [less than or equal to] m [less than or equal to] 9.

In this paper, we shall use the elementary methods to study the summation [n.summation over (k = 1)] [S.sub.k] and [n.summation over (k = 1)] [T.sub.k], where [S.sub.k] denotes the summation of all k digits numbers in S, [T.sub.k] denotes the summation of each digits of all k digits numbers in S.

That is, we shall prove the following

Theorem 1. For the generalized constructive set S of digits [d.sub.1]; [d.sub.2], ..., [d.sub.m] (1 [less than or equal to] m [less than or equal to] 9), we have

[n.summation over (k = 1)] [S.sub.k] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9 (10 x [(10m).sup.n - 1 / 10m - 1 - [m.sup.n] - 1 / m - 1).

where [S.sub.k] denotes the summation of all k digits numbers in S.

Taking m = 2; [d.sub.1] = 1 and [d.sub.2] = 2 in Theorem 1, we may immediately get

Corollary 1. For the generalized constructive set S of digits 1 and 2, we have

[n.summation over (k = 1)] [S.sub.k] = 1/3 (10 x [20.sup.n] - 1 / 19 - [2.sup.n] + 1).

Taking m = 3, [d.sub.1] = 1, [d.sub.2] = 2 and [d.sub.3] = 3 in Theorem 1, we may immediately get the following:

Corollary 2. For the generalized constructive set S of digits 1; 2 and 3, we have

[n.summation over (k = 1)] [S.sub.k] = 2/3 (10 x [30.sup.n] - 1 / 29 - [3.sup.n]/2 + 1/2).

Theorem 2. For the generalized constructive set S of digits [d.sub.1], [d.sub.2], ..., [d.sub.m] (1 [less than or equal to] m [less than or equal to] 9). we have

[n.summation over (k = 1)] [T.sub.k] = ([d.sub.1] + [d.sub.2] + ... + [d.sub.m]), [nm.sup.n+1] - (n + 1)[m.sup.n] + 1 / [(m - 1).sup.2].

where [T.sub.k] denotes the summation of each digits of all k digits numbers in S.

Taking m = 2, [d.sub.1] = 1 and [d.sub.2] = 2 in Theorem 2, we may immediately get the following:

Corollary 3. For the the generalized constructive set S of digits 1 and 2, we have

[n.summation over (k = 1)] [T.sub.k] = 3n x [2.sup.n+1] - 3(n + 1)[2.sup.n] + 3.

Taking m = 3; [d.sub.1] = 1, [d.sub.2] = 2 and [d.sub.3] = 3 in Theorem 2, we may immediately get

Corollary 4. For the the generalized constructive set S of digits 1; 2 and 3, we have

[n.summation over (k = 1)] [T.sub.k] = 3/2 n x [3.sup.n+1] - 3/2(n + 1)[3.sup.n] + 3/2.

[section] 2. Proof of the theorems

In this section, we shall complete the proof of the theorems. First we prove Theorem 1. Let [S.sub.k] denotes the summation of all k digits numbers in the generalized constructive set S. Note that for k = 1, 2, 3, ..., there are [m.sup.k] numbers of k digits in S. So we have

[S.sub.k] = [10.sup.k-1][m.sup.k-1] ([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[S.sub.k-1]. (1)

Meanwhile, we have

[S.sub.k-1] = [10.sup.k-2][m.sup.k-2]([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[S.sub.k-2]. (2)

Combining (1) and (2), we can get the following recurrence equation

[S.sub.k] - 11m[S.sub.k-1] + 10[m.sup.]2[S.sub.k-2] = 0.

Its characteristic equation

[x.sup.2] - 11mx + 10[m.sup.2] = 0

have two different real solutions

x = m, 10m:

So we let

[S.sub.k] = A x [m.sup.k] + B x [(10m).sup.k].

Note that

[S.sub.0] = 0; [S.sub.1] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m],

we can get

A = - [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9m, B = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9m.

So

[S.sub.k] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9m [((10.m).sup.k] - [m.sup.k]).

Then

[n.summation over (k = 1)] [S.sub.k] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9 (10 x [(10m).sup.n] - 1 / 10m - 1 - [m.sup.n] - 1 / m - 1).

This completes the proof of Theorem 1.

Now we come to prove Theorem 2. Let [T.sub.k] is denotes the summation of each digits of all k digits numbers in the generalized constructive set S.

Similarly, note that for k = 1, 2, 3, ..., there are [m.sup.k] numbers of k digits in S, so we have

[T.sub.k] = [m.sup.k-1]([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[T.sub.k-1] (3)

Meanwhile, we have

[T.sub.k-1] = [m.sup.k-2]([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[T.sub.k-2] (4)

Combining (3) and (4), we can get the following recurrence equation

[T.sub.k] - 2m[T.sub.k-1] + [m.sup.2][T.sub.k-2] = 0,

its characteristic equation

[x.sup.2] - 2mx + [m.sup.2] = 0

have two solutions

[x.sub.1] = [x.sub.2] = m.

So we let

[T.sub.k] = A x [m.sup.k] + k x B x [m.sup.k].

Note that

[T.sub.0] = 0; [T.sub.1] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m].

We may immediately deduce that

A = 0; B = [d.sub.1] + [d.sub.2] + ... + [d.sub.m]/m,

So

[T.sub.k] = ([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) x [km.sup.k-1].

Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

This completes the proof of Theorem 2.

References

[1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, 1993.

[2] Gou Su, On the generalized constructive set, Research on Smarandache problems in number theory, Hexis, 2005, 53-55.

Qianli Yang

Department of Mathematics, Weinan Teacher's College Weinan, Shaanxi, P.R.China

Keywords Generalized constructive set, summation, recurrence equation, characteristic equation.

[section] 1. Introduction and Results

The generalized constructive set S is defined as: numbers formed by digits [d.sub.1], [d.sub.2], ..., [d.sub.m] only, all [d.sub.i] being different each other, 1 [less than or equal to] m [less than or equal to] 9. That is to say,

(1) [d.sub.1], [d.sub.2], ..., [d.sub.m] belongs to S;

(2) If a; b belong to S, then [bar.ab] belongs to S too;

(3) Only elements obtained by rules (1) and (2) applied a finite number of times belongs to S.

For example, the constructive set (of digits 1, 2) is: 1; 2; 11; 12; 21; 22; 111; 112; 121; 122; 211; 212; 221; 222; 1111; 1112; 1121.... And the constructive set (of digits 1, 2, 3) is: 1; 2; 3; 11; 12; 13; 21; 22; 23; 31; 32; 33; 111; 112; 113; 121; 122; 123; 131; 132; 133; 211; 212; 213; 221; 222; 223; 231; 232; 233; 311; 312; 313; 321; 322; 323; 331; 332; 333; 1111;.... In problem 6, 7 and 8 of reference [1], Professor F.Smarandache asked us to study the properties of this sequence. In [2], Gou Su had studied the convergent properties of the series

[[+infinity].summation over (n=1)] 1/[a.sup.[alpha].sub.n,

and proved that the series is convergent if [alpha] > log m, and divergent if [alpha] [less than or equal to] log m, where {[a.sub.n]} denotes the sequence of the constructive set S, formed by digits [d.sub.1]; [d.sub.2], ..., [d.sub.m] only, all [d.sub.i] being different each other, 1 [less than or equal to] m [less than or equal to] 9.

In this paper, we shall use the elementary methods to study the summation [n.summation over (k = 1)] [S.sub.k] and [n.summation over (k = 1)] [T.sub.k], where [S.sub.k] denotes the summation of all k digits numbers in S, [T.sub.k] denotes the summation of each digits of all k digits numbers in S.

That is, we shall prove the following

Theorem 1. For the generalized constructive set S of digits [d.sub.1]; [d.sub.2], ..., [d.sub.m] (1 [less than or equal to] m [less than or equal to] 9), we have

[n.summation over (k = 1)] [S.sub.k] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9 (10 x [(10m).sup.n - 1 / 10m - 1 - [m.sup.n] - 1 / m - 1).

where [S.sub.k] denotes the summation of all k digits numbers in S.

Taking m = 2; [d.sub.1] = 1 and [d.sub.2] = 2 in Theorem 1, we may immediately get

Corollary 1. For the generalized constructive set S of digits 1 and 2, we have

[n.summation over (k = 1)] [S.sub.k] = 1/3 (10 x [20.sup.n] - 1 / 19 - [2.sup.n] + 1).

Taking m = 3, [d.sub.1] = 1, [d.sub.2] = 2 and [d.sub.3] = 3 in Theorem 1, we may immediately get the following:

Corollary 2. For the generalized constructive set S of digits 1; 2 and 3, we have

[n.summation over (k = 1)] [S.sub.k] = 2/3 (10 x [30.sup.n] - 1 / 29 - [3.sup.n]/2 + 1/2).

Theorem 2. For the generalized constructive set S of digits [d.sub.1], [d.sub.2], ..., [d.sub.m] (1 [less than or equal to] m [less than or equal to] 9). we have

[n.summation over (k = 1)] [T.sub.k] = ([d.sub.1] + [d.sub.2] + ... + [d.sub.m]), [nm.sup.n+1] - (n + 1)[m.sup.n] + 1 / [(m - 1).sup.2].

where [T.sub.k] denotes the summation of each digits of all k digits numbers in S.

Taking m = 2, [d.sub.1] = 1 and [d.sub.2] = 2 in Theorem 2, we may immediately get the following:

Corollary 3. For the the generalized constructive set S of digits 1 and 2, we have

[n.summation over (k = 1)] [T.sub.k] = 3n x [2.sup.n+1] - 3(n + 1)[2.sup.n] + 3.

Taking m = 3; [d.sub.1] = 1, [d.sub.2] = 2 and [d.sub.3] = 3 in Theorem 2, we may immediately get

Corollary 4. For the the generalized constructive set S of digits 1; 2 and 3, we have

[n.summation over (k = 1)] [T.sub.k] = 3/2 n x [3.sup.n+1] - 3/2(n + 1)[3.sup.n] + 3/2.

[section] 2. Proof of the theorems

In this section, we shall complete the proof of the theorems. First we prove Theorem 1. Let [S.sub.k] denotes the summation of all k digits numbers in the generalized constructive set S. Note that for k = 1, 2, 3, ..., there are [m.sup.k] numbers of k digits in S. So we have

[S.sub.k] = [10.sup.k-1][m.sup.k-1] ([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[S.sub.k-1]. (1)

Meanwhile, we have

[S.sub.k-1] = [10.sup.k-2][m.sup.k-2]([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[S.sub.k-2]. (2)

Combining (1) and (2), we can get the following recurrence equation

[S.sub.k] - 11m[S.sub.k-1] + 10[m.sup.]2[S.sub.k-2] = 0.

Its characteristic equation

[x.sup.2] - 11mx + 10[m.sup.2] = 0

have two different real solutions

x = m, 10m:

So we let

[S.sub.k] = A x [m.sup.k] + B x [(10m).sup.k].

Note that

[S.sub.0] = 0; [S.sub.1] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m],

we can get

A = - [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9m, B = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9m.

So

[S.sub.k] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9m [((10.m).sup.k] - [m.sup.k]).

Then

[n.summation over (k = 1)] [S.sub.k] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m] / 9 (10 x [(10m).sup.n] - 1 / 10m - 1 - [m.sup.n] - 1 / m - 1).

This completes the proof of Theorem 1.

Now we come to prove Theorem 2. Let [T.sub.k] is denotes the summation of each digits of all k digits numbers in the generalized constructive set S.

Similarly, note that for k = 1, 2, 3, ..., there are [m.sup.k] numbers of k digits in S, so we have

[T.sub.k] = [m.sup.k-1]([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[T.sub.k-1] (3)

Meanwhile, we have

[T.sub.k-1] = [m.sup.k-2]([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) + m[T.sub.k-2] (4)

Combining (3) and (4), we can get the following recurrence equation

[T.sub.k] - 2m[T.sub.k-1] + [m.sup.2][T.sub.k-2] = 0,

its characteristic equation

[x.sup.2] - 2mx + [m.sup.2] = 0

have two solutions

[x.sub.1] = [x.sub.2] = m.

So we let

[T.sub.k] = A x [m.sup.k] + k x B x [m.sup.k].

Note that

[T.sub.0] = 0; [T.sub.1] = [d.sub.1] + [d.sub.2] + ... + [d.sub.m].

We may immediately deduce that

A = 0; B = [d.sub.1] + [d.sub.2] + ... + [d.sub.m]/m,

So

[T.sub.k] = ([d.sub.1] + [d.sub.2] + ... + [d.sub.m]) x [km.sup.k-1].

Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

This completes the proof of Theorem 2.

References

[1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, 1993.

[2] Gou Su, On the generalized constructive set, Research on Smarandache problems in number theory, Hexis, 2005, 53-55.

Qianli Yang

Department of Mathematics, Weinan Teacher's College Weinan, Shaanxi, P.R.China

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Author: | Yang, Qianli |
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Publication: | Scientia Magna |

Date: | Jun 1, 2006 |

Words: | 1336 |

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