OFFSET
1,2
COMMENTS
a(n) = 21322314 for n > 12. - Reinhard Zumkeller, Jan 25 2014
The digits of each term a(n) are a permutation of those of the corresponding term A063850(n). - Chayim Lowen, Jul 16 2015
REFERENCES
C. Fleenor, "A litteral sequence", Solution to Problem 2562, Journal of Recreational Mathematics, vol. 31 No. 4 pp. 307 2002-3 Baywood NY.
Problem in J. Recreational Math., 30 (4) (1999-2000), p. 309.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
V. Bronstein and A. S. Fraenkel, On a curious property of counting sequences, Amer. Math. Monthly, 101 (1994), 560-563.
Onno M. Cain and Sela T. Enin, Inventory Loops (i.e. Counting Sequences) have Pre-period 2 max S_1 + 60, arXiv:2004.00209 [math.NT], 2020.
X. Gourdon and B. Salvy, Effective asymptotics of linear recurrences with rational coefficients, Discrete Mathematics, vol. 153, no. 1-3, 1996, pages 145-163.
James Henle, Is (some) mathematics poetry?, Journal of Humanistic Mathematics 1:1 (2011), pp. 94-100.
Madras Math's Amazing Number Facts, Fact No. 13
Madras Math, Descriptive Number
Trevor Scheopner, The Cyclic Nature (and Other Intriguing Properties) of Descriptive Numbers, Princeton Undergraduate Mathematics Journal, Issue 1, Article 4.
L. J. Upton, Letter to N. J. A. Sloane, Jan 8 1991.
Index entries for linear recurrences with constant coefficients, signature (1).
FORMULA
a(n+1) = A047842(a(n)). - M. F. Hasler, Feb 25 2018
G.f.: x*(1 + 10*x + 10*x^2 + 1091*x^3 + 2000*x^4 + 208101*x^5 + 101000*x^6 - 99990*x^7 - 98010*x^8 + 31007101*x^9 + 10001000*x^10 - 9900990*x^11 - 9899010*x^12) / (1 - x). - Colin Barker, Aug 23 2018
EXAMPLE
The term after 312213 is obtained by saying "Two 1's, two 2's, two 3's", which gives 21-22-23, i.e., 212223.
MATHEMATICA
RunLengthEncode[x_List] := (Through[{Length, First}[ #1]] &) /@ Split[ Sort[x]]; LookAndSay[n_, d_:1] := NestList[ Flatten[ RunLengthEncode[ # ]] &, {d}, n - 1]; F[n_] := LookAndSay[n, 1][[n]]; Table[ FromDigits[ F[n]], {n, 25}] (* Robert G. Wilson v, Jan 22 2004 *)
a[1] = 1; a[n_] := a[n] = FromDigits[Reverse /@ Sort[Tally[a[n-1] // IntegerDigits], #1[[1]] < #2[[1]]&] // Flatten]; Array[a, 26] (* Jean-François Alcover, Jan 25 2016 *)
PROG
(Haskell)
import Data.List (group, sort, transpose)
a005151 n = a005151_list !! (n-1)
a005151_list = 1 : f [1] :: [Integer] where
f xs = (read $ concatMap show ys) : f ys where
ys = concat $ transpose [map length zss, map head zss]
zss = group $ sort xs
-- Reinhard Zumkeller, Jan 25 2014
(PARI) say(n) = {digs = digits(n); d = vecsort(digs, , 8); s = ""; for (k=1, #d, nbk = #select(x->x==d[k], digs); s = concat(s, Str(nbk)); s = concat(s, d[k]); ); eval(s); }
lista(nn) = {print1(n = 1, ", "); for (k=1, nn, m = say(n); print1(m, ", "); n = m; ); } \\ Michel Marcus, Feb 12 2016
(PARI) a(n, show_all=1, a=1)={for(i=2, n, show_all&&print1(a", "); a=A047842(a)); a} \\ M. F. Hasler, Feb 25 2018
(PARI) Vec(x*(1 + 10*x + 10*x^2 + 1091*x^3 + 2000*x^4 + 208101*x^5 + 101000*x^6 - 99990*x^7 - 98010*x^8 + 31007101*x^9 + 10001000*x^10 - 9900990*x^11 - 9899010*x^12) / (1 - x) + O(x^40)) \\ Colin Barker, Aug 23 2018
(Python)
from itertools import accumulate, groupby, repeat
def summarize(n, _):
return int("".join(str(len(list(g)))+k for k, g in groupby(sorted(str(n)))))
def aupton(nn): return list(accumulate(repeat(1, nn+1), summarize))
print(aupton(25)) # Michael S. Branicky, Jan 11 2021
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved