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First differences of A030299.
5

%I #31 Nov 09 2021 07:25:53

%S 11,9,102,9,81,18,81,9,913,9,81,18,81,9,702,9,171,27,72,18,693,18,72,

%T 27,171,9,702,9,81,18,81,9,8024,9,81,18,81,9,702,9,171,27,72,18,693,

%U 18,72,27,171,9,702,9,81,18,81,9,5913,9,81,18,81,9,1602,9,261

%N First differences of A030299.

%C From _M. F. Hasler_, Jan 12 2013: (Start)

%C Note [updated Mar 03 2013]: The definition of sequence A030299 has been slightly modified in Jan. 2013, and as a consequence the following properties remain valid beyond the first A007489(9)-1 = 409112 terms, which had not been the case before, when A030299 had been defined through concatenation of the lexicographically ordered permutations, which in case of elements >= 10 broke up the nice mathematical properties (esp. of the sequence A219664 = 9*A217626 cited below).

%C This sequence taken modulo 9 is zero except (possibly) at indices where a run of permutations ends in A030299. (These indices are given by A007489(n), n>0.) There it equals (mod 9) the "n" of the following run. E.g., a(1)=2 (mod 9), and A030299(1+1)=12 is the start of the run for n=2; a(3)=3 (mod 9) and A030299(3+1)=123 is the start of the run for n=3, a(9)=4 (mod 9) and A030299(9+1)=1234 is the start of the run for n=4, etc.

%C The subsequence between these indices (A007489(n)+1,...,A007489(n+1)-1), always starts with the same terms, listed in A219664 = 9*A217626 (= A209280 = A107346 where the latter are defined). (End)

%H Antti Karttunen, <a href="/A220664/b220664.txt">Table of n, a(n) for n = 1..5912</a>

%F a(n) = A030299(n+1) - A030299(n).

%F a(n) = A219664(n-A007489(k)), for A007489(k) < n < A007489(k+1). - _M. F. Hasler_, Jan 13 2013

%e A030299 starts (1, 12, 21, 123, 132, 213, 231, 312, ...), the first differences thereof yield (11, 9, 102, 9, 81, 18, 81, ...).

%p (l-> seq(l[j]-l[j-1], j=2..nops(l)))([seq(map(x-> parse(cat(x[])),

%p combinat[permute](n))[], n=0..5)])[]; # _Alois P. Heinz_, Nov 09 2021

%o (Scheme): (define (A220664 n) (- (A030299 (+ 1 n)) (A030299 n)))

%o (PARI) {A030299=concat( vector( 5,k, vecsort( vector( (#k=vector(k, j, 10^j)~\10)!, i, numtoperm(#k, i-1)*k )))); A220664=vecextract(A030299,"^1")-vecextract(A030299,"^-1")} \\ _M. F. Hasler_, Jan 12 2013

%o (Python)

%o from itertools import permutations

%o def pmap(s, m): return sum(s[i-1]*10**(m-i) for i in range(1, len(s)+1))

%o def agen():

%o m = 1

%o while True:

%o for s in permutations(range(1, m+1)): yield pmap(s, m)

%o m += 1

%o def aupton(terms):

%o alst, g = [], agen()

%o t = next(g)

%o while len(alst) < terms:

%o t, prevt = next(g), t

%o alst += [t - prevt]

%o return alst

%o print(aupton(65)) # _Michael S. Branicky_, Nov 09 2021

%Y The repeating part is given by A219664, equal to A107346 for indices < 5!.

%K nonn,base

%O 1,1

%A _Antti Karttunen_, Dec 17 2012