This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Example a(2)=11+1=12, a(3)=1112+12=1124.
#!/usr/bin/perl -l use strict; use warnings; use bigint; $=1; { print my $o=$; s/((\d)\2*)/(length $1).$2/ge; $+=$o; <>, redo; } __END_
The sequence starting at 7 is 7 (prime), 17 (prime), 1117 (prime), and 3117 (composite), so a(3) = 7.
from sympy import isprime, nextprime
from itertools import count, groupby, islice
def LS(n):
return int(''.join(str(len(list(g)))+k for k, g in groupby(str(n))))
def f(n): return 0 if not isprime(n) else 1 + f(LS(n))
def agen(startn=0, startk=1):
n, adict = startn, {i:-1 for i in range(startn)}
for k in count(startk):
fk = f(k)
if fk not in adict: adict[fk] = k
while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 7))) # Michael S. Branicky, Jul 27 2022
1.139366173592639521191348935319958...
A005150(7) = 13112221 is prime and a(7) = 13112221. A005150(8) = 1113213211 = 11 * 19 * 1831 * 2909, hence a(8) = 2909.
The 11 partitions of 6 generate 7 Look-and-Say partitions as follows: 6 -> 111111 51 -> 111111 42 -> 111111 411 -> 21111 33 -> 222 321 -> 111111 3111 -> 3111 222 -> 33 2211 -> 222 21111 -> 411 111111 -> 6, so that a(6) counts these 7 partitions: 111111, 21111, 222, 3111, 33, 411, 6.
LS[part_List] := Reverse[Sort[Flatten[Map[Table[#[[2]], {#[[1]]}] &, Tally[part]]]]]; LS[n_Integer] := #[[Reverse[Ordering[PadRight[#]]]]] &[DeleteDuplicates[Map[LS, IntegerPartitions[n]]]]; TableForm[t = Map[LS[#] &, Range[10]]](*A239454,array*)
Flatten[t](*A239454,sequence*)
Map[Length[LS[#]] &, Range[25]](*A239455*)
(* Peter J. C. Moses, Mar 18 2014 *)
disjointFamilies[y_]:=Select[Tuples[IntegerPartitions/@Length/@Split[y]],UnsameQ@@Join@@#&];
Table[Length[Select[IntegerPartitions[n],Length[disjointFamilies[#]]>0&]],{n,0,10}] (* Gus Wiseman, Aug 11 2025 *)
23 has "one 2, one 3", so a(23) = 1213.
-- see Watkins link, p. 3.
import Data.List (unfoldr, group); import Data.Tuple (swap)
a045918 0 = 10
a045918 n = foldl (\v d -> 10 * v + d) 0 $ say $ reverse $ unfoldr
(\x -> if x == 0 then Nothing else Just $ swap $ divMod x 10) n
where say = concat . map code . group
code xs = [toInteger $ length xs, head xs]
-- Reinhard Zumkeller, Aug 09 2012
LS:=n-> if n>9 then LS(op(convert(n,base,10))) else for i from 2 to nargs do if args[i] <> n then RETURN(( LS( args[i..nargs] )*10^length(i-1) + i-1)*10 + n ) fi od: 10*nargs + n fi; # M. F. Hasler, Nov 14 2006
LookAndSayA[n_] := FromDigits@ Flatten@ IntegerDigits@ Flatten[ Through[{Length, First}[#]] & /@ Split@ IntegerDigits@ n] (* Robert G. Wilson v, Jan 27 2012 *)
A045918(a)={my(c=1);for(j=2,#a=Vec(Str(a)),if(a[j-1]==a[j],a[j-1]="";c++,a[j-1]=Str(c,a[j-1]);c=1));a[#a]=Str(c,a[#a]);eval(concat(a))} \\ M. F. Hasler, Jan 27 2012
from re import finditer def A045918(n): return int(''.join([str(len(m.group(0)))+m.group(0)[0] for m in finditer(r'(\d)\1*',str(n))])) # Chai Wah Wu, Dec 03 2014
from itertools import groupby
def LS(n): return int(''.join(str(len(list(g)))+k for k, g in groupby(str(n))))
print([LS(n) for n in range(48)]) # Michael S. Branicky, Jul 27 2022
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.
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
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 *)
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
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
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
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
E.g. the term after 3112 is obtained by saying "one 3, two 1's, one 2", which gives 132112.
a006751 = foldl1 (\v d -> 10 * v + d) . map toInteger . a088203_row -- Reinhard Zumkeller, Aug 09 2012
RunLengthEncode[ x_List ] := (Through[ {First, Length}[ #1 ] ] &) /@ Split[ x ]; LookAndSay[ n_, d_:1 ] := NestList[ Flatten[ Reverse /@ RunLengthEncode[ # ] ] &, {d}, n - 1 ]; F[ n_ ] := LookAndSay[ n, 2 ][ [ n ] ]; Table[ FromDigits[ F[ n ] ], {n, 11} ] (* Zerinvary Lajos, Mar 21 2007 *)
# This outputs the first n elements of the sequence, where n is given on the command line.
$s = 2;
for (2..shift @ARGV) {
print "$s, ";
$s =~ s/(.)\1*/(length $&).$1/eg;
}
print "$s\n";
## Arne 'Timwi' Heizmann (timwi(AT)gmx.net), Mar 12 2008
l=[2]
n=s=1
y=''
while n<21:
x=str(l[n - 1]) + ' '
for i in range(len(x) - 1):
if x[i]==x[i + 1]: s+=1
else:
y+=str(s)+str(x[i])
s=1
x=''
n+=1
l.append(int(y))
y=''
s=1
print(l) # Indranil Ghosh, Jul 05 2017
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