A036059
The summarize Fibonacci sequence: summarize the previous two terms!.
Original entry on oeis.org
1, 1, 21, 1221, 3231, 233231, 533221, 15534221, 3514334231, 3534533241, 3544832231, 183544733221, 28172544634231, 2827162554535241, 2827265554337241, 2837267544338231, 3847264544637221, 3847362564636221, 2837662564536221, 2827863534537221, 3837564524538221
Offset: 0
a(20) = 3837564524538221;
a(21) = 4837265534637221;
a(22+16*k) = 3837365544636221, k >= 0;
a(36) = a(20+16) = 3837265554834221 <> a(20);
a(37) = a(21+16) = 3837266544735221 <> a(21);
a(38) = a(22+16) = 3837365544636221 = a(22). - _Reinhard Zumkeller_, Aug 10 2014
-
import Data.List (sort, group)
a036059 n = a036059_list !! n
a036059_list = map (read . concatMap show) fss :: [Integer] where
fss = [1] : [1] : zipWith h (tail fss) fss where
h vs ws = concatMap (\us -> [length us, head us]) $
group $ reverse $ sort $ vs ++ ws
-- Reinhard Zumkeller, Aug 10 2014
-
a:= proc(n) option remember; `if`(n<2, 1, (p-> parse(cat(seq((c->
`if`(c=0, [][], [c, 9-i][]))(coeff(p, x, 9-i)), i=0..9))))(
add(x^i, i=map(x-> convert(x, base, 10)[], [a(n-1),a(n-2)]))))
end:
seq(a(n), n=0..20); # Alois P. Heinz, Jun 18 2022
-
a[0] = a[1] = 1; a[n_] := a[n] = FromDigits @ Flatten @ Reverse @ Select[ Transpose @ { DigitCount[a[n-1]] + DigitCount[a[n-2]], Append[ Range[9], 0]}, #[[1]] > 0&];
Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Dec 30 2017 *)
-
def aupton(nn):
alst = [1, 1]
for n in range(2, nn+1):
prev2, anstr = sorted(str(alst[-2]) + str(alst[-1])), ""
for d in sorted(set(prev2), reverse=True):
anstr += str(prev2.count(d)) + d
alst.append(int(anstr))
return alst
print(aupton(20)) # Michael S. Branicky, Feb 02 2021
A036058
Summarize digits of preceding number, by decreasing digit value. Start with a(0) = 0.
Original entry on oeis.org
0, 10, 1110, 3110, 132110, 13123110, 23124110, 1413223110, 1423224110, 2413323110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110
Offset: 0
The third term is 1110 because the second term contains one 1 and one 0.
-
a(n)=if(n>9,1433223110,[0,10,1110,3110,132110,13123110,23124110,1413223110, 1423224110,2413323110][n+1]) \\ Charles R Greathouse IV, Jul 24 2012
-
a(n,a=0)={for(k=1,n,a==(a=A244112(a))&&break);a} \\ M. F. Hasler, Feb 25 2018
A300191
Look and Say the digits of the last three terms, by increasing digit value. Start with a(1)=1, a(2)=2 and a(3)=3.
Original entry on oeis.org
1, 2, 3, 111213, 412223, 51423314, 7152433415, 515253543517, 613263548527, 5142634495263718, 515263549546372819, 516263648566373839, 515283648596374849, 414283649596376859, 3132837475106377869, 10413283746576677869, 20513283644596976859, 30514283545596976859
Offset: 1
1, 2, 3 => there are one '1', one '2' and one '3': 111213;
2, 3, 111213 => there are four '1', two '2' and two '3': 412223;
3, 111213, 412223 => there are five '1', four '2', three '3', one '4': 51423314.
-
P:=proc(q) local a,b,c,d,k,n,y,x; y:=array(1..3); x:=array(0..9);
y[1]:=1; y[2]:=2; y[3]:=3;
print(1); print(2); print(3); a:=[]; b:=[1]; c:=[2];
for n from 1 to q do for k from 0 to 9 do x[k]:=0; od;
a:=b; b:=c; c:=convert(y[3],base,10);
for k from 1 to nops(a) do x[a[k]]:=x[a[k]]+1; od;
for k from 1 to nops(b) do x[b[k]]:=x[b[k]]+1; od;
for k from 1 to nops(c) do x[c[k]]:=x[c[k]]+1; od;
y[1]:=y[2]; y[2]:=y[3]; y[3]:=0;
for k from 0 to 9 do if x[k]>0 then if k=0 then d:=10*x[k];
else d:=10*x[k]+k; fi; y[3]:=y[3]*10^(ilog10(d)+1)+d; fi; od;
print(y[3]); od; end: P(10^2);
-
Nest[Append[#, FromDigits[Join @@ Map[Flatten@ Reverse@ # &, IntegerDigits@ Sort@ Tally[Join @@ IntegerDigits[#[[-3 ;; -1]]]]]]] &, {1, 2, 3}, 15] (* Michael De Vlieger, Mar 01 2018 *)
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