[seqfan] Re: English number words modulo themselves
Maximilian Hasler
maximilian.hasler at gmail.com
Sat Jun 22 01:07:36 CEST 2013
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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