# [seqfan] Re: English number words modulo themselves

Eric Angelini Eric.Angelini at kntv.be
Sat Jun 22 07:30:19 CEST 2013

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Propulsé d'un aPhone
Many thanks, Maximilian!
And you're right, of course,
loops.
Best,
É.

Le 22 juin 2013 à 01:08, "Maximilian Hasler" <maximilian.hasler at gmail.com> a écrit :

> On Fri, Jun 21, 2013 at 6:01 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>>
>> Replace each letter (of the English number word N) by its rank
>> in the alphabet. Sum the ranks.
>> Divide the sum by the number N.
>
> EA21(N)={w=select(t->t>64,Vec(Vecsmall(English(N)))); sum(i=1,#w,w[i]%32)%N}
> /* for English(), see A052360 */
>
> vector(99,n,EA21(n))
> = [0, 0, 2, 0, 2, 4, 2, 1, 6, 9, 8, 3, 8, 6, 5, 0, 7, 1, 10, 7, 15,
> 11, 2, 23, 24, 3, 10, 16, 4, 10, 10, 30, 24, 24, 2, 8, 17, 35, 25, 4,
> 36, 16, 11, 12, 36, 44, 8, 37, 28, 16, 49, 20, 16, 18, 53, 6, 17, 57,
> 49, 37, 9, 31, 27, 29, 9, 17, 28, 10, 1, 40, 2, 24, 20, 22, 2, 10, 21,
> 3, 73, 74, 27, 50, 47, 50, 31, 40, 52, 35, 27, 87, 30, 53, 50, 53, 34,
> 43, 55, 38, 30]
>>
>> [Are loops possible? Number N producing the remainder M, which,
>> in turn, produces the remainder N?]
>
> I don't think so because  f(N) = (...) % N  <  N,
> (where % = remainder operator)
> so the sequence of iterations is strictly decreasing
> (and stops at 0).
>
> Maximilian
>
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