[seqfan] Re: Three infinite sequences

[Brad Klee]

Hi Eric,

It would be nice to also obtain the following finite counts:

Number of unique terms in addition sequences without zeros.

Number of unique terms in multiplication sequence without ones.

And then the counts that treat numbers with permuted digits as the same.

It seems that the maximum unpadded values in your sequences are:

9,111,111,111

And 

8,222

For Mult. Series brute force could probably answer easily.

Maybe consider a variant where a block equals to sum/ multiple of remaining digits. For example:

5127, 5+7=12
8864, 8*8=64

Or a partitioning scheme such as

6657, 6+6=5+7
45252, 4*5=2*2*5

These sequences would have infinite number of terms even after reduction of padding by 1s and 0s. I wouldn't be surprised if intersection sequences worked out infinitely infinite as well. 

Cheers,

Brad

> On May 30, 2015, at 5:42 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
> 
> 
> Hello SeqFans,
> 
> S gathers the integers that have a digit
> which is the sum of all other digits (of the said integer):
> 
> S= 11,22,33,44,55,66,77,88,99,101,110,
> 112,121,123,132,134,143,145,154,156,
> 165,167,176,178,187,189,198,211,...
> S is infinite: you can write as many zeroes
> as you want after 101...
> ------------
> P gathers the integers that have a digit
> which is the product of all other digits (of the said integer):
> 
> P=11,22,33,44,55,66,77,88,99,100,111,
> 122,133,144,155,166,177,188,199,200,
> 212,221,224,242,300,313,331,400,414,...
> P is infinite: you can write as many "1"s
> after 11...
> ------------
> INT is the intersection between S and P:
> 
> INT=11,22,33,44,55,66,77,88,99,224,
> 242,422,1001,1010,1100,1236,1263,1326,1362,1623,
> 1632,2136,2163,2316,2361,2613,2631,...
> 
> INT is infinite: you can write as many
> zeroes as you want after 1001...
> 
> Best,
> É.
> 
> 
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