cp's OEIS Frontend

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.

Showing 1-10 of 17 results. Next

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

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Keywords

Comments

This kind of counting sequence is always eventually periodic with period 1, 2 or 3. - Herve Lehning (lehning(AT)noos.fr), Oct 01 2003

Examples

			The third term is 1110 because the second term contains one 1 and one 0.
		

Crossrefs

Cf. A007890 (same as this, starting at 1), A001155 (same as this, but using method A047842: by increasing digit value), A005150 (as before, starting at 1), A036059 ("fibonacci" based on this), A036066.

Programs

  • PARI
    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
    
  • PARI
    a(n,a=0)={for(k=1,n,a==(a=A244112(a))&&break);a} \\ M. F. Hasler, Feb 25 2018

Formula

a(n+1) = A244112(a(n)), a(0) = 0. - M. F. Hasler, Feb 25 2018

A036066 The summarize Lucas sequence: summarize the previous two terms, start with 1, 3.

Original entry on oeis.org

1, 3, 1311, 2331, 331241, 14432231, 34433241, 54533231, 2544632221, 163534435221, 263544436231, 363554634231, 463554733221, 17364544733221, 37263554634231, 37363554734231, 37364544933221, 1937263554933221, 3927263544835231, 391827264534836231, 293827363544836231
Offset: 0

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Comments

After the 26th term the sequence goes into a cycle of 46 terms.
"Summarize" uses here method C = A244112: in order of decreasing digit value.

Crossrefs

Cf. A036059.
Cf. A244112 (summarizing as used here: by decreasing digit value), A047842 (alternative summarizing method: by increasing digit value), A047843 (another method: don't omit missing digits between smallest and largest one).

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 2*n+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
  • Mathematica
    a[0] = 1; a[1] = 3; 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, 17}] (* Jean-François Alcover, Dec 30 2017 *)
  • PARI
    {a=[1,3]; for(n=1,50,a=concat(a,A244112(eval(Str(a[n],a[n+1]))))); a} \\ M. F. Hasler, Feb 25 2018

Formula

a(n+1) = A244112(concat(a(n),a(n-1))). - M. F. Hasler, Feb 25 2018

A036106 A summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

1, 2, 1211, 2231, 133241, 14333231, 24632241, 1634534231, 261544434231, 262564533241, 363564435231, 464544634221, 463574533221, 17363574434221, 37263554634231, 37363554734231, 37364544933221, 1937263554933221
Offset: 0

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Comments

From the 26th term the sequence gets into a cycle of 46.

Examples

			a(24) = 293847463554538221;
a(25) = 294827365564537221 = first term of first period;
a(26) = 293837366554537221;
a(70) = 294837364554538221 = last term of first period != a(24);
a(71) = 294827365564537221 = a(25) = first term of second period;
a(72) = 293837366554537221 = a(26);
a(116) = 294837364554538221 = a(70) = last term of second period;
a(117) = 294827365564537221 = a(71) = first term of third period.
		

Crossrefs

Cf. A036059.

Programs

  • Haskell
    import Data.List (sort, group); import Data.Function (on)
    a036106 n = a036106_list !! n
    a036106_list = 1 : 2 : map (read . concatMap say . reverse . group . sort)
                   (zipWith ((++) `on` show) a036106_list $ tail a036106_list)
                   where say ws = (show $ length ws) ++ [head ws]
    -- Reinhard Zumkeller, Oct 05 2015
  • Mathematica
    a[0] = 1; a[1] = 2; a[n_] := a[n] = Reverse /@ (IntegerDigits /@ {a[n-2], a[n-1]} // Flatten // Tally // Sort // Reverse) // Flatten // FromDigits; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Nov 02 2016 *)

A036067 A summarize Fibonacci sequence starting with a(0)=a(1)=0: summarize the previous two terms!.

Original entry on oeis.org

0, 0, 20, 1220, 321120, 13423120, 1433424120, 4443424120, 8433423120, 187443422120, 28175443523120, 2827255433624120, 282716454443823120, 382726356443923120, 19382736256463823120, 29382756253483823120
Offset: 0

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From the 54th term the sequence goes into a cycle of 117 terms.
For n > 1, a(n) is of the form 2^(k+2) * 5 * r, where k is a nonnegative integer and r is a positive integer relatively prime to 10. Therefore, for n > 1, a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Mar 19 2020

Crossrefs

Cf. A036059.

A036103 A look-and-say sequence: each term summarizes the previous two terms.

Original entry on oeis.org

0, 1, 1110, 4110, 145120, 1524124120, 2534425120, 354413624120, 16355433523120, 26454463424120, 36456453523120, 46457443522120, 1736556443522120, 2736556433623120, 2756553463623120, 2766552473622120
Offset: 0

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Comments

A look-and-say sequence. a(3) says 4 ones, 1 zero, which is the count of ones and zeros in the two previous terms. a(4) says 1 four, 5 ones, 2 zeros.
From the 28th term the sequence gets into a cycle of 117.

Examples

			a(26) = 39283736554483723130;
a(27) = 39384726554493622120 = first term of first period;
a(28) = 39383736455493622120;
a(143) = 39283746553473823130 = last term of first period != a(26);
a(144) = 39384726554493622120 = first term of second period = a(27);
a(145) = 39383736455493622120 = a(28).
		

Crossrefs

Cf. A036059.
Cf. A036106.

Programs

  • Haskell
    import Data.List (sort, group); import Data.Function (on)
    a036103 n = a036103_list !! n
    a036103_list = 0 : 1 : map (read . concatMap say . group . reverse . sort)
                   (zipWith ((++) `on` show) a036103_list $ tail a036103_list)
                   where say w = (show $ length w) ++ [head w]
    -- Reinhard Zumkeller, Oct 05 2015
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := a[n] = Reverse /@ (Join[a[n-2] // IntegerDigits, a[n-1] // IntegerDigits] // Tally // SortBy[#, First]& // Reverse) // Flatten // FromDigits;
    Array[a, 16, 0] (* Jean-François Alcover, Jul 13 2016 *)

A036109 A summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

2, 1, 1211, 1241, 142251, 15243241, 2534134241, 2554334241, 3564434231, 163564533221, 363544633231, 463534833231, 183625441032221, 2826254473524110, 281726355433825120, 382726554443824120, 482726555453823120, 482726656443822120, 482746555433922120
Offset: 0

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Comments

From the 25th term the sequence gets into a cycle of 117.

Crossrefs

Cf. A036059.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 2-n, (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

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

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Author

Paolo P. Lava, Feb 28 2018

Keywords

Comments

From the 1458th term the sequence goes into a cycle of 850 terms.

Examples

			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.
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    Nest[Append[#, FromDigits[Join @@ Map[Flatten@ Reverse@ # &, IntegerDigits@ Sort@ Tally[Join @@ IntegerDigits[#[[-3 ;; -1]]]]]]] &, {1, 2, 3}, 15] (* Michael De Vlieger, Mar 01 2018 *)

A036104 A summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

0, 2, 1210, 222110, 424120, 24523120, 153413522120, 352433624120, 16353453624120, 36354463523120, 46454473423120, 1736357463422120, 3736257453523120, 4736453473623120, 4736454483522120, 183736357463522120
Offset: 0

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Comments

From the 50th term the sequence gets into a cycle of 117.

Crossrefs

Cf. A036059.

A036105 A summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

1, 0, 1110, 3120, 13124120, 1423324120, 3433525120, 253453523120, 552463522120, 16652443722120, 1736453423823120, 182736254453724120, 283726355463724120, 284736455453823120, 383736555463722120
Offset: 0

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Author

Keywords

Comments

From the 54th term the sequence gets into a cycle of 117.

Crossrefs

Cf. A036059.

A036107 A summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

1, 4, 1411, 2431, 24131241, 34233241, 44434241, 74433221, 1774333221, 3734534231, 371534733231, 372534932231, 19372524934231, 39272534736231, 3937162534736231, 2947362524935231, 3937363534836231
Offset: 0

Views

Author

Keywords

Comments

From the 50th term the sequence gets into a cycle of 3.

Crossrefs

Cf. A036059.
Showing 1-10 of 17 results. Next