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 14 results. Next

A102085 Remove the least number of commas from A093086 and concatenate digits so as to always have a(n) < a(n+1).

Original entry on oeis.org

0, 1, 12, 35, 81, 394, 1213, 5334, 8867, 12161, 41383, 377554, 1111610, 1412109, 5222771, 15533191, 47449148, 261086410, 1051111813, 1051210871, 8141051115, 62229944156, 331815899551, 562261184411, 1813856119649, 9613171814106, 61184872912852, 299411131172101
Offset: 0

Views

Author

Eric Angelini, Feb 13 2005

Keywords

Comments

The zeros at the first digits of a term are omitted. 35th term was such a number (01152299487468610171) so it is written without the zero in the first digit. - Hakan Icoz, Sep 03 2020

Examples

			a(4) = 81, the next term must be bigger than the previous term with least commas removed.
81 < 3 is not true; 81 < 39 is not true; 81 < 394 = a(5) is the next term since it satisfies a(n) < a(n+1).
		

Crossrefs

Cf. A093086.

Extensions

More terms from Hakan Icoz, Sep 03 2020

A093087 "Lucas in digits": start with a(1)=1, a(2)=3; repeatedly adjoin digits of sum of next two terms.

Original entry on oeis.org

1, 3, 4, 7, 1, 1, 8, 2, 9, 1, 0, 1, 1, 1, 0, 1, 1, 2, 2, 1, 1, 2, 3, 4, 3, 2, 3, 5, 7, 7, 5, 5, 8, 1, 2, 1, 4, 1, 2, 1, 0, 1, 3, 9, 3, 3, 5, 5, 3, 3, 1, 1, 4, 1, 2, 1, 2, 6, 8, 1, 0, 8, 6, 4, 2, 5, 5, 3, 3, 3, 8, 1, 4, 9, 1, 8, 1, 4, 1, 0, 6, 7, 1, 0, 8, 6, 6, 1, 1, 9, 5, 1, 3, 1, 0, 9, 9, 5, 5, 1, 6, 1, 3, 8, 1
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Comments

See A093086 (an analogous sequence) for a detailed description of how terms are calculated and links to similar sequences. - Paolo Xausa, Aug 19 2025

Crossrefs

Cf. A093086.

Programs

  • Mathematica
    Fold[Join[#, IntegerDigits[Total[#[[#2;; #2+1]]]]] &, {1, 3}, Range[100]] (* Paolo Xausa, Aug 19 2025 *)

Extensions

Name edited by Paolo Xausa, Aug 19 2025

A093094 "Products into digits": start with a(1)=2, a(2)=2; adjoin digits of product of a(k) and a(k+1) for k from 1 to infinity.

Original entry on oeis.org

2, 2, 4, 8, 3, 2, 2, 4, 6, 4, 8, 2, 4, 2, 4, 3, 2, 1, 6, 8, 8, 8, 1, 2, 6, 2, 6, 4, 8, 6, 4, 6, 4, 8, 2, 1, 2, 1, 2, 1, 2, 2, 4, 3, 2, 4, 8, 2, 4, 2, 4, 2, 4, 3, 2, 1, 6, 2, 2, 2, 2, 2, 2, 4, 8, 1, 2, 6, 8, 3, 2, 1, 6, 8, 8, 8, 8, 8, 1, 2, 6, 2, 6, 1, 2, 4, 4, 4, 4, 4, 8, 3, 2, 8, 2, 1, 2, 4, 8, 2, 4
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Comments

Only the digits 1,2,3,4,6,8 occur, infinitely often. The sequence is not periodic. Around a(800) there are many 8's.
From Giovanni Resta, Mar 16 2006: (Start)
Proof that sequence is not periodic:
Let us assume that somewhere in the sequence there is a subsequence of 3 adjacent 8': ...,8,8,8,....(which is true).
Then we know that in the following there will be the subsequence ...,6,4,6,4.. (i.e. 8x8, 8x8) again, there will be somewhere ...,2,4,2,4,2,4,... (i.e. 6x4, 4x6, 6x4) and finally ...,8,8,8,8,8,...
Analogously, starting from 8,8,8,8 we obtain 6,4,6,4,6,4 then 2,4,2,4,2,4,2,4,2,4 and finally 8,8,8,8,8,8,8,8,8.
Generalizing, if somewhere appears a run of k>2 8's, then in some future position will appear a run of at least 4*k-7 8's (where since k>2, 4*k-7>k).
So the sequence will contain arbitrary long runs of 8's, without being constantly equal to 8, thus it cannot be periodic. (End)
Essentially the same as A045777. [R. J. Mathar, Sep 08 2008]

Examples

			a(3)=a(1)*a(2), a(4)=a(2)*a(3), a(5)=first digit of (a(3)*a(4)), a(6)=2nd digit of (a(3)*a(4)), a(9)=a(6)*a(7)
		

Crossrefs

Programs

  • Haskell
    a093094 n = a093094_list !! (n-1)
    a093094_list = f [2,2] where
       f (u : vs@(v : _)) = u : f (vs ++
         if w < 10 then [w] else uncurry ((. return) . (:)) $ divMod w 10)
            where w = u * v
    -- Reinhard Zumkeller, Aug 08 2013
    
  • Mathematica
    Fold[Join[#, IntegerDigits[Times @@ #[[#2;; #2+1]]]] &, {2, 2}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)
  • Python
    from itertools import islice
    from collections import deque
    def agen(): # generator of terms
        a = deque([2, 2])
        while True:
            a.extend(list(map(int, str(a[0]*a[1]))))
            yield a.popleft()
    print(list(islice(agen(), 101))) # Michael S. Branicky, Feb 15 2024

Extensions

Definition revised by Franklin T. Adams-Watters, Mar 16 2006

A093088 "Fibonacci-like in digits": start with a(1)=1, a(2)=4; repeatedly adjoin digits of sum of previous two terms.

Original entry on oeis.org

1, 4, 5, 9, 1, 4, 1, 0, 5, 5, 1, 5, 1, 0, 6, 6, 6, 1, 6, 1, 2, 1, 2, 7, 7, 7, 3, 3, 3, 9, 1, 4, 1, 4, 1, 0, 6, 6, 1, 2, 1, 0, 5, 5, 5, 5, 1, 6, 1, 2, 7, 3, 3, 1, 5, 1, 0, 1, 0, 1, 0, 6, 7, 7, 3, 9, 1, 0, 6, 4, 6, 6, 1, 1, 1, 1, 1, 6, 1, 3, 1, 4, 1, 0, 1, 2, 1, 0, 1, 6, 1, 0, 1, 0, 1, 2, 7, 2, 2, 2, 2, 7, 7, 4, 4
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Fold[Join[#, IntegerDigits[Total[#[[#2;; #2+1]]]]] &, {1, 4}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

A093089 "Fibonacci in pairs": start with a(1)=1, a(2)=1; repeatedly adjoin sum of previous two terms but chopped from the right into pairs of 2 digits.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1, 44, 90, 45, 1, 34, 1, 35, 46, 35, 35, 36, 81, 81, 70, 71, 1, 17, 1, 62, 1, 51, 1, 41, 72, 18, 18, 63, 63, 52, 52, 42, 1, 13, 90, 36, 81, 1, 26, 1, 15, 1, 4, 94, 43, 14, 1, 3, 1, 26, 1, 17, 82, 27, 27, 16, 16, 5, 98, 1, 37, 57, 15, 4, 4, 27, 27, 18
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Comments

Do all pairs of digits appear infinitely often? The sequence is not periodic.

Examples

			... a(11)=a(9)+a(10), a(12)=left pair of (a(10)+a(11)=55+89=1 44), a(13)=right pair of (a(10)+a(11)=55+89=1 44), a(14)=a(11)+a(12) ...
		

Crossrefs

A093090 Start with a(1)=1, a(2)=3, apply rule of A093089.

Original entry on oeis.org

1, 3, 4, 7, 11, 18, 29, 47, 76, 1, 23, 77, 24, 1, 0, 1, 1, 25, 1, 1, 2, 26, 26, 2, 3, 28, 52, 28, 5, 31, 80, 80, 33, 36, 1, 11, 1, 60, 1, 13, 69, 37, 12, 12, 61, 61, 14, 82, 1, 6, 49, 24, 73, 1, 22, 75, 96, 83, 7, 55, 73, 97, 74, 23, 97, 1, 71, 1, 79, 90, 62, 1, 28, 1, 70, 1, 71, 97, 1, 20
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Crossrefs

Extensions

More terms from Bodo Zinser, Mar 21 2004

A214365 Digit-wise Fibonacci: Start with 0,1; then the next term is always the sum of the earliest two consecutive digits not yet summed so far.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 9, 4, 12, 13, 5, 3, 3, 4, 8, 8, 6, 7, 12, 16, 14, 13, 8, 3, 3, 7, 7, 5, 5, 4, 11, 11, 6, 10, 14, 12, 10, 9, 5, 2, 2, 2, 7, 7, 1, 1, 5, 5, 3, 3, 1, 9, 14, 7, 4, 4, 9, 14, 8, 2, 6, 10, 8, 6, 4, 10, 10, 5, 11, 11, 8, 13, 10, 5, 12, 10, 8, 7, 1, 8, 14, 10, 5
Offset: 0

Views

Author

Eric Angelini and M. F. Hasler, Feb 16 2013

Keywords

Comments

Offset chosen in analogy to the classical Fibonacci sequence. But the present sequence has no term larger than 9+9=18.
As observed by H. Havermann in reply to the original post (cf. link), a run of k consecutive 18's will yield a run of (at least) 2k-1 consecutive 9's somewhere later, which in turn will yield (at least) 2k-2 consecutive 18's. Since there are such runs of sufficient length (Z. Seidov pointed out that a(n)=9 for 78532 < n < 78598), the sequence cannot become periodic.
In what precedes, a value of k >= 3 is sufficient for infinite growth. But a run of only three 9's is also sufficient because the 2 consecutive 18's will be followed by a number >= 10, which then yields four 9's and subsequently infinitely long runs of 9's, cf Example.
From Michael S. Branicky, Dec 14 2020: (Start)
Likewise, a run of k consecutive 6's will yield (at least) k-1 consecutive 12's, then 2k-3 3's, then 2k-4 6's, leading to infinite growth for k > 4. Five consecutive 6's first occur at a(17072).
Similarly, a run of k 8's will yield k-1 16's, 2k-3 7's, 2k-4 14's, 4k-9 5's, 4k-10 10's, 8k-21 1's, 8k-22 2's, 8k-23 2's, then 8k-24 8's, leading to infinite growth for k > 3. Four consecutive 8's first occur at a(9606). (End)

Examples

			The sequence starts in the same way as the Fibonacci sequence A000045. But after 5+8=13 follows the digit-wise continuation, viz: 8+1=9, 1+3=4, 3+9=12, ... (Due to the presence of 2-digit terms, the summed digits lag more and more behind the correspondingly computed term.)
The first run of 3 consecutive 9's occurs at a(3862)=a(3863)=a(3864)=9, which then yield a(4975)=a(4976)=18, a(4977)=14 and the first run of four 9's at 6392 <= n <= 6395. [_M. F. Hasler_, Feb 17 2013]
		

Crossrefs

Cf. A093086.

Programs

  • Mathematica
    Module[{rd = {0, 1}, a}, Join[rd, Table[rd = Join[Rest[rd], IntegerDigits[a = Total[Take[rd, 2]]]]; a, 100]]] (* Paolo Xausa, Aug 22 2025 *)
  • PARI
    A214365(n,show=0,d=[0,1])={show & print1(d[1]","d[2]); for(i=2,n, n=d[1]+d[2]; show & print1(","n); d=concat(vecextract(d,"^1"),digits(n))); n}
    
  • Python
    def aupto(n):
      alst, remaining = [0, 1], [0, 1]
      for i in range(2, n+1):
        an = remaining.pop(0) + remaining[0]
        alst.append(an)
        remaining.extend(list(map(int, str(an))))
      return alst    # use alst[n] for a(n)
    print(aupto(89)) # Michael S. Branicky, Dec 14 2020

A093091 "Fibonacci in pairs from left": start with a(1)=1, a(2)=1; repeatedly adjoin sum of previous two terms but chopped from the left into pairs of 2 digits.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 14, 4, 10, 3, 18, 14, 13, 21, 32, 27, 34, 53, 59, 61, 87, 11, 2, 12, 0, 14, 8, 98, 13, 14, 12, 14, 22, 10, 6, 11, 1, 27, 26, 26, 36, 32, 16, 17, 12, 28, 53, 52, 62, 68, 48, 33, 29, 40, 81, 10, 5, 11, 4, 13, 0, 11, 6, 81, 62, 69, 12, 1, 91, 15, 16
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Comments

Do all pairs of digits appear infinitely often? The sequence is not periodic.

Examples

			... a(11)=a(9)+a(10), a(12)=left pair of (a(10)+a(11)=55+89=14 4),
a(13)=right pair of (a(10)+a(11)=55+89=14 4),
a(14)=left pair of (a(11)+a(12)=89+14=10 3),
a(15)=right pair of (a(11)+a(12)=89+14=10 3), a(16)=a(12)+a(13) ...
		

Crossrefs

A093092 "Fibonacci in digits - up and down": start with a(1)=1, a(2)=1; repeatedly adjoin either the sum of the two previous terms (if that sum happens to be even) or else adjoin digits of the sum of previous two terms (if that sum happens to be odd).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 1, 3, 9, 4, 12, 1, 3, 16, 1, 3, 4, 1, 9, 1, 7, 4, 7, 5, 10, 10, 8, 1, 1, 1, 1, 12, 1, 5, 20, 18, 9, 2, 2, 2, 1, 3, 1, 3, 6, 2, 5, 38, 2, 7, 1, 1, 4, 4, 3, 4, 4, 4, 9, 8, 7, 4, 3, 40, 9, 8, 2, 5, 8, 7, 7, 8, 8, 1, 3, 1, 7, 1, 5, 1, 1, 7, 4, 3, 4, 9, 1, 7, 10, 7, 1, 3, 1, 5, 14, 1, 5, 16, 9, 4
Offset: 1

Views

Author

Bodo Zinser, Mar 20 2004

Keywords

Examples

			... a(6)=a(4)+a(5), a(7)=left digit of (a(5)+a(6)=5+8=1 3) as 13 is odd, a(8)=right digit of (a(5)+a(6)=5+8=1 3) as 13 is odd, a(11)=a(8)+a(9) as even ...
		

Crossrefs

Programs

  • Mathematica
    a = {0, 1}; f[n_] := Block[{k = a[[n - 1]] + a[[n - 2]]}, If[ EvenQ[k], AppendTo[a, k], a = Join[a, IntegerDigits[k]] ]]; Do[ f[n], {n, 3, 100}]; a (* Robert G. Wilson v, Mar 27 2004 *)

A093099 "Fibonacci-digits": start with "11", append sum of first 2 digits to the preceding number, drop first digit.

Original entry on oeis.org

11, 12, 23, 35, 58, 813, 139, 394, 9412, 41213, 12135, 21353, 13533, 35334, 53348, 33488, 34886, 48867, 886712, 8671216, 67121614, 712161413, 121614138, 216141383, 161413833, 614138337, 141383377, 413833775, 138337755, 383377554, 8337755411, 33775541111
Offset: 1

Views

Author

Bodo Zinser, Mar 22 2004

Keywords

Examples

			a(6): sum of first 2 digits of a(5)=13, append 13 to 58 and get 5813, drop first digit and get 813.
		

Crossrefs

Programs

  • Maple
    f:= proc(x) local t,q;
      t:= floor(x/10^(ilog10(x)-1));
      t:= (t mod 10) + floor(t/10);
      t:= t+x*10^(ilog10(t)+1);
      q:= 10^ilog10(t);
      t - floor(t/q)*q
    end proc:x:= 11: R:= x:
    for n from 2 to 100 do x:= f(x); R:= R,x od:R; # Robert Israel, May 21 2020
    # second Maple program:
    a:= proc(n) option remember; `if`(n=1, 11, (s->parse(cat(
          s[2..-1], parse(s[1])+parse(s[2]))))(""||(a(n-1))))
        end:
    seq(a(n), n=1..35);  # Alois P. Heinz, May 21 2020
  • Mathematica
    fd[n_]:=Module[{idn=IntegerDigits[n]},FromDigits[Rest[Join[idn,IntegerDigits[Total[Take[idn,2]]]]]]]
    NestList[fd,11,30]  (* Harvey P. Dale, Mar 06 2011 *)
Showing 1-10 of 14 results. Next