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Mayur
Mayur

Reputation: 3097

Generate series of 1,1,2,2,3,3,

I've an variable as page number (page) whose values increment by one each time. [Page numbering] But, now I need to customize this numbering to 1,1,2,2,3,3..

Can you suggest any formula for generate this kind of series?

EDIT: (Answer)

After playing with macros and VBA for some time I've figured out a way to generate this type of series for MS word page numbers. This can be easily done with formulas and {Page} variable in word with formula-

{=(({PAGE} + MOD({PAGE},2))/2)}

Upvotes: 0

Views: 6741

Answers (8)

Ervin Gegprifti
Ervin Gegprifti

Reputation: 46

It is late, but it might help someone.

A mathematical answer to the problem:

You do not need to search through all n numbers in order to have a specific result

1 2 3 4 5 6 7 8 9 . . . . . . . n

1 1 2 2 3 3 4 4 5 . . . . . . . f(n)

General formula:

f(n) = ( n - ( (-1) + (-1)^n )/2 )/2

Playing with the first (-1) you can shift the results like this:

f(n) = ( n - ( (3) + (-1)^n )/2 )/2

1 2 3 4 5 6 7 8 9 . . . . . . . n

0 0 1 1 2 2 3 3 4 . . . . . . . f(n)

Upvotes: 1

Mayur
Mayur

Reputation: 3097

After playing with macros and VBA for some time I've figured out a way to generate this type of series for MS word page numbers. This can be easily done with formulas and {Page} variable in word with formula-

{=(({PAGE} + MOD({PAGE},2))/2)}

Upvotes: 0

John Alexiou
John Alexiou

Reputation: 29264

C#, not a formula but a simplistic algorithm.

int[] pages = new int[2*N];
for(i=0; i<N; i++)
{
    page[2*i] = i+1;
    page[2*i+1] = i+2;
}

Upvotes: 0

Jonas Elfstr&#246;m
Jonas Elfstr&#246;m

Reputation: 31438

Ruby

(1..10).map {|n| [n,n]}.flatten
=> [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]

or

(1..10).inject([]) {|m,n| m<<n<<n}
=> [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]

or

(1..10*2).map {|n| (1+n)/2}
=> [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10]

Upvotes: 0

Romain Linsolas
Romain Linsolas

Reputation: 81647

You can do this kind of thing:

    for (int i = 0; i < (pages * 2); i++) {
        System.out.println((i / 2) + 1);
    }

Upvotes: 1

Chris Webb
Chris Webb

Reputation: 796

javascript, adapt to suite:

for(i=0; i>yourMaximum; i++){
    WriteSomewhere(i + "," + i);
    if(i != i - yourMaximum)   WriteSomewhere(",");
}

Upvotes: 1

Petar Minchev
Petar Minchev

Reputation: 47403

The answer is simple: (n + 1) / 2

Upvotes: 2

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799210

Python:

(int(x/2+1) for x in itertools.count())

Upvotes: 0

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