A278038 -id:A278038 - cp's OEIS frontend

A003146 Positions of letter c in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

4, 11, 17, 24, 28, 35, 41, 48, 55, 61, 68, 72, 79, 85, 92, 98, 105, 109, 116, 122, 129, 136, 142, 149, 153, 160, 166, 173, 177, 184, 190, 197, 204, 210, 217, 221, 228, 234, 241, 247, 254, 258, 265, 271, 278, 285, 291, 298, 302, 309, 315, 322, 329, 335, 342, 346, 353, 359

Offset: 1

N. J. A. Sloane

nonn

Comment from Philippe Deléham, Feb 27 2009: A003144, A003145, A003146 may be defined as follows. Consider the map psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite word a, b, a, c, a, b, a, a, b, a, c, a, b, a, b, a, c, ... (setting a = 1, b = 2, c = 3 gives A092782). The indices of a, b, c give respectively A003144, A003145, A003146.
The infinite word may also be defined as the limit S_oo where S_1 = a, S_n = psi(S_{n-1}). Or, by S_1 = a, S_2 = ab, S_3 = abac, and thereafter S_n = S_{n-1} S_{n-2} S_{n-3}. It is the unique word such that S_oo = psi(S_oo).
Also, indices of c in the sequence closed under a -> abac, b -> aba, c -> ab; starting with a(1) = a. - Philippe Deléham, Apr 16 2004
Theorem: A number m is in this sequence iff the tribonacci representation of m-1 ends with 11. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, Mar 02 2019

Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 1..10609
Elena Barcucci, Luc Belanger and Srecko Brlek, On tribonacci sequences, Fib. Q., 42 (2004), 314-320.
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Fibonacci representations of higher order, Fib. Quart., 10 (1972), 43-69. The present sequence is called c.
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Eric Duchêne and Michel Rigo, A morphic approach to combinatorial games: the Tribonacci case. RAIRO - Theoretical Informatics and Applications, 42, 2008, pp 375-393. doi:10.1051/ita:2007039. [Also available from Numdam]
A. J. Hildebrand, Junxian Li, Xiaomin Li, Yun Xie, Almost Beatty Partitions, arXiv:1809.08690 [math.NT], 2018.
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.

Cf. A003145, A003144, A080843, A092782, A058265, A276791, A276798, A276801, A277721.
First differences are A276792. A278041 (subtract 1 from each term, and use offset 0).
For tribonacci representations of numbers see A278038.

M:=17; S[1]:=`a`; S[2]:=`ab`; S[3]:=`abac`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: l:=length(t0); t1:=[];
for i from 1 to l do if substring(t0,i..i) = `c` then t1:=[op(t1),i]; fi; od:
# N. J. A. Sloane, Nov 01 2006

StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "c", {#}][[1]], "c"][[All, 1]] &@ 11 (* Michael De Vlieger, Mar 30 2017, Version 10.2, after JungHwan Min at A003144 *)

It appears that a(n) = floor(n*t^3) + eps for all n, where t is the tribonacci constant A058265 and eps is 0, 1, 2, or 3. See A277721. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

More terms from Philippe Deléham, Apr 16 2004

Entry revised by N. J. A. Sloane, Oct 13 2016

A003726 Numbers with no 3 adjacent 1's in binary expansion.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 48, 49, 50, 51, 52, 53, 54, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 80, 81, 82

Offset: 1

N. J. A. Sloane

Positions of zeros in A014082. Could be called "tribbinary numbers" by analogy with A003714. - John Keith, Mar 07 2022

The sequence of Tribbinary numbers can be constructed by writing out the Tribonacci representations of nonnegative integers and then evaluating the result in binary. These numbers are similar to Fibbinary numbers A003714, Fibternary numbers A003726, and Tribternary numbers A356823. The number of Tribbinary numbers less than any power of two is a Tribonacci number. We can generate Tribbinary numbers recursively: Start by adding 0 and 1 to the sequence. Then, if x is a number in the sequence add 2x, 4x+1, and 8x+3 to the sequence. The n-th Tribbinary number is even if the n-th term of the Tribonacci word is a. Respectively, the n-th Tribbinary number is of the form 4x+1 if the n-th term of the Tribonacci word is b, and the n-th Tribbinary number is of the form 8x+3 if the n-th term of the Tribonacci word is c. Every nonnegative integer can be written as the sum of two Tribbinary numbers. Every number has a Tribbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Index entries for 2-automatic sequences.

Cf. A278038 (binary), A063037, A000073, A014082 (number of 111).
Cf. A004781 (complement).
Cf. A007088; A003796 (no 000), A004745 (no 001), A004746 (no 010), A004744 (no 011), A003754 (no 100), A004742 (no 101), A004743 (no 110).

a003726 n = a003726_list !! (n - 1)
a003726_list = filter f [0..] where
   f x = x < 7 || (x `mod` 8) < 7 && f (x `div` 2)
-- Reinhard Zumkeller, Jun 03 2012

Select[Range[0, 82], SequenceCount[IntegerDigits[#, 2], {1, 1, 1}] == 0 &] (* Michael De Vlieger, Dec 23 2019 *)

is(n)=!bitand(bitand(n, n<<1), n<<2) \\ Charles R Greathouse IV, Feb 11 2017

There are A000073(n+3) terms of this sequence with at most n bits. In particular, a(A000073(n+3)+1) = 2^n. - Charles R Greathouse IV, Oct 22 2021

Sum_{n>=2} 1/a(n) = 9.516857810319139410424631558212354346868048230248717360943194590798113163384... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 13 2022

A136175 Tribonacci array, T(n,k).

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 11, 9, 8, 13, 20, 17, 15, 10, 24, 37, 31, 28, 19, 12, 44, 68, 57, 51, 35, 22, 14, 81, 125, 105, 94, 64, 41, 26, 16, 149, 230, 193, 173, 118, 75, 48, 30, 18, 274, 423, 355, 318, 217, 138, 88, 55, 33, 21, 504, 778, 653, 585, 399, 254, 162, 101, 61, 39, 23

Offset: 1

Clark Kimberling, Dec 18 2007

As an interspersion (and dispersion), the array is, as a sequence, a permutation of the positive integers. Column k consists of the numbers m such that the least summand in the tribonacci representation of m is T(1,k). For example, column 1 consists of numbers with least summand 1. This array arises from tribonacci representations in much the same way that the Wythoff array, A035513, arises from Fibonacci (or Zeckendorf) representations.
From Abel Amene, Jul 29 2012: (Start)
(Row 1) = A000073 (offset=4) a(0)=0, a(1)=0, a(2)=1
(Row 2) = A001590 (offset=5) a(0)=0, a(1)=1, a(2)=0
(Row 3) = A000213 (offset=4) a(0)=1, a(1)=1, a(2)=1
(Row 4) = A214899 (offset=5) a(0)=2, a(1)=1, a(2)=2
(Row 5) = A020992 (offset=6) a(0)=0, a(1)=2, a(2)=1
(Row 6) = A100683 (offset=6) a(0)=-1,a(1)=2, a(2)=2
(Row 7) = A135491 (offset=4) a(0)=2, a(1)=4, a(2)=8
(Row 8) = A214727 (offset=6) a(0)=1, a(1)=1, a(2)=2
(Row 9) = A081172 (offset=8) a(0)=1, a(1)=1, a(2)=0
(column 1) = A003265
(column 2) = A353083
(End) [Corrected and extended by John Keith, May 09 2022]

			Northwest corner:
1  2   4   7   13  24   44   81  149 274 504
3  6   11  20  37  68   125  230 423 778
5  9   17  31  57  105  193  355 653
8  15  28  51  94  173  318  585
10 19  35  64  118 217  399
12 22  41  75  138 254
14 26  48  88  162
16 30  55 101
18 33  61
21 39
23

Cf. A035513, A353083, A353084.

# maximum index in A73 such that A73 <= n.
A73floorIdx := proc(n)
    local k ;
    for k from 3 do
        if A000073(k) = n then
            return k ;
        elif A000073(k) > n then
            return k -1 ;
        end if ;
    end do:
end proc:
# tribonacci expansion coeffs of n
A278038 := proc(n)
    local k,L,nres ;
    k := A73floorIdx(n) ;
    L := [1] ;
    nres := n-A000073(k) ;
    while k >= 4 do
        k := k-1 ;
        if nres >= A000073(k) then
            L := [1,op(L)] ;
            nres := nres-A000073(k) ;
        else
            L := [0,op(L)] ;
        end if ;
    end do:
    return L ;
end proc:
A278038inv := proc(L)
    add( A000073(i+2)*op(i,L),i=1..nops(L)) ;
end proc:
A135175 := proc(n,k)
    option remember ;
    local a,known,prev,nprev,kprev,freb ;
    if n =1 then
        A000073(k+2) ;
    elif k>3 then
        procname(n,k-1)+procname(n,k-2)+procname(n,k-3) ;
    else
        if k = 1 then
            for a from 1 do
                known := false ;
                for nprev from 1 to n-1 do
                    for kprev from 1 do
                        if procname(nprev,kprev) > a then
                            break ;
                        elif procname(nprev,kprev) = a then
                            known := true ;
                        end if;
                    end do:
                end do:
                if not known then
                    return a ;
                end if;
            end do:
        else
            prev := procname(n,k-1) ;
            freb := A278038(prev) ;
            return A278038inv([0,op(freb)]) ;
        end if;
    end if;
end proc:
seq(seq(A135175(n,d-n),n=1..d-1),d=2..12) ; # R. J. Mathar, Jun 07 2022

T(1,1)=1, T(1,2)=2, T(1,3)=4, T(1,k)=T(1,k-1)+T(1,k-2)+T(1,k-3) for k>3. Row 1 is the tribonacci basis; write B(k)=T(1,k). Each row satisfies the recurrence T(n,k)=T(n,k-1)+T(n,k-2)+T(n,k-3). T(n,1) is least number not in an earlier row. If T(n,1) has tribonacci representation B(k(1))+B(k(2))+...+B(k(m)), then T(n,2) = B(k(2))+B(k(3))+...+B(k(m+1)) and T(n,3) = B(k(3))+B(k(4))+...+B(k(m+2)). (Continued shifting of indices gives the other terms in row n, also.)

T(3, 4) corrected and more terms by John Keith, May 09 2022

A352103 a(n) is the maximal (or lazy) tribonacci representation of n using a binary system of vectors not containing three consecutive 0's.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 111, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10010, 10011, 10100, 10101, 10110, 10111, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100100, 100101, 100110, 100111, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110010

Offset: 0

Amiram Eldar, Mar 05 2022

Each nonnegative integer has 2 unique representations as sums of distinct positive tribonacci numbers (A000073): 1, 2, 4, 7, 13, 24, ...: the minimal (or greedy, A278038) representation in which there are no 3 consecutive 1's (i.e., no 3 consecutive tribonacci numbers appear in the sum), and the maximal (or lazy) representation of n in which no 3 consecutive 0's appear.

			a(5) = 101 = 4 + 1.
a(6) = 110 = 4 + 2.
a(7) = 111 = 4 + 2 + 1.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000073, A007088, A003796, A278038.

Similar sequences: A104326 (Fibonacci), A130311 (Lucas).

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; trib[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; IntegerDigits[Total[2^(s - 1)], 2]]; a[n_] := Module[{v = trib[n]}, nv = Length[v]; i = 1; While[i <= nv - 3, If[v[[i ;; i + 3]] == {1, 0, 0, 0}, v[[i ;; i + 3]] = {0, 1, 1, 1}; If[i > 3, i -= 4]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, 0, FromDigits[v[[i[[1, 1]] ;; -1]]]]]; Array[a, 100, 0]

a(n) = A007088(A003796(n+1)).

A136808 Numbers k such that k and k^2 use only the digits 0, 1 and 2.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 1000, 1001, 1010, 1011, 1100, 1101, 10000, 10001, 10010, 10011, 10100, 10110, 11000, 11001, 11010, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 101000, 101001, 101100, 110000, 110001, 110010, 110100, 1000000, 1000001, 1000010, 1000011, 1000100

Offset: 1

Jonathan Wellons (wellons(AT)gmail.com), Jan 22 2008

Generated with DrScheme.
Subsequence of A136809, A136816, ..., A136836. - M. F. Hasler, Jan 24 2008
A278038(18) = 10101, A136827(294) = 10110001101, A136831(1276) = 101100010001101 resp. A136836(1262) = 101090009991101 are the first terms from where on these four sequences differ from the present one. - M. F. Hasler, Nov 15 2017
From Jovan Radenkovicc, Nov 15 2024: (Start)
A nonnegative integer n is in this sequence iff 10*n is also in this sequence.
Not a subsequence of A278038 (binary numbers without '111'). A counterexample is 10^2884 + 10^2880 + 10^2872 + 10^2857 + 10^2497 + 10^2426 + 10^2285 + 10^2004 + 10^1443 + 10^1442 + 10^1441 + 10^881 + 10^600 + 10^459 + 10^388 + 10^27 + 10^12 + 10^4 + 1. There are infinitely many counterexamples not divisible by 10. This counterexample follows from the fact that 111^2+2000*4+200*4=12321+8000+800=21121. In fact, every binary substring will eventually occur in this sequence. Also, if n is a term containing only the digits 0 and 1, then 10^k*n+1 and n+10^k are also in this sequence for any sufficiently large integer k. (End)

			101000100100001^2 = 10201020220210222010200200001.

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 1359 terms from Jonathan Wellons)
Jonathan Wellons, Tables of Shared Digits [archived]
Index to sequences related to squares.

A subsequence of the binary numbers A007088.
Cf. A278038.
Cf. A136809, A136810, ..., A137147 for other digit combinations.
See also A058412 = A058411^2: squares having only digits {0,1,2}, A277946 = A277959^2 = squares whose largest digit is 2.

isA136808 := proc(n) local ndgs,n2dgs ; ndgs := convert(convert(n,base,10),set) ; n2dgs := convert(convert(n^2,base,10),set) ; if ( (ndgs union n2dgs) minus {0,1,2} ) = {} then true ; else false ; fi ; end: LtonRev := proc(L) local i ; add(op(i,L)*10^(i-1),i=1..nops(L)) ; end: A007089 := proc(n) convert(n,base,3) ; LtonRev(%) ; end: n := 1: for i from 0 do n3 := A007089(i) ; if isA136808(n3) then printf("%d %d ",n,n3) ; n := n+1 ; fi ; od: # R. J. Mathar, Jan 24 2008

Select[FromDigits/@Tuples[{0,1},7],Union[Take[DigitCount[#^2],{3,9}]]=={0}&] (* Harvey P. Dale, May 29 2013 *)

for(n=1,999,vecmax(digits((N=fromdigits(binary(n),10))^2))<3 && print1(N",")) \\ M. F. Hasler, Nov 15 2017

A317204 Expansion of n in the p-system based on convergents to sqrt(2).

Original entry on oeis.org

0, 1, 10, 11, 20, 100, 101, 110, 111, 120, 200, 201, 1000, 1001, 1010, 1011, 1020, 1100, 1101, 1110, 1111, 1120, 1200, 1201, 2000, 2001, 2010, 2011, 2020, 10000, 10001, 10010, 10011, 10020, 10100, 10101, 10110, 10111, 10120, 10200, 10201, 11000, 11001, 11010, 11011

Offset: 0

N. J. A. Sloane, Aug 07 2018

This is the minimal (or greedy) representation of nonnegative numbers in terms of the positive Pell numbers (A000129). - Amiram Eldar, Mar 12 2022

A. F. Horadam, Zeckendorf representations of positive and negative integers by Pell numbers, Applications of Fibonacci Numbers, Springer, Dordrecht, 1993, pp. 305-316.

Amiram Eldar, Table of n, a(n) for n = 0..10000
L. Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Pellian Representations, The Fibonacci Quarterly, Vol. 10, No. 5 (1972), pp. 449-488.
Aviezri S. Fraenkel, How to beat your Wythoff games' opponent on three fronts, Amer. Math. Monthly, Vol. 89, No. 6 (1982), pp. 353-361. See Table 2.

Cf. A000129, A169683.
Similar to, but different from, A014418.
Similar sequences: A014417, A130310, A278038.

pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; pellp[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; FromDigits @ IntegerDigits[Total[3^(s - 1)], 3]]; Array[pellp, 50, 0] (* Amiram Eldar, Mar 12 2022 *)

a(n) = { my (p=[1,2]); for (k=2, oo, if (n<=p[k], my (v=0, d); while (n, v+=10^k*d=n\p[k]; n-=d*p[k]; k--); return (v/10), p = concat(p, 2*p[k]+p[k-1]))) } \\ Rémy Sigrist, Mar 12 2022

More terms from Amiram Eldar, Mar 12 2022

A352090 Numbers k such that k and k+1 are both tribonacci-Niven numbers (A352089).

Original entry on oeis.org

1, 6, 7, 12, 13, 20, 26, 27, 39, 68, 75, 80, 81, 87, 115, 128, 135, 149, 176, 184, 185, 195, 204, 215, 224, 230, 236, 243, 264, 278, 284, 291, 344, 364, 399, 447, 506, 507, 519, 548, 555, 560, 575, 595, 615, 635, 656, 664, 665, 684, 704, 725, 744, 777, 804, 824

Offset: 1

Amiram Eldar, Mar 04 2022

Numbers k such that A278043(k) | k and A278043(k+1) | k+1.

The odd tribonacci numbers, A000073(A042964(m)), are all terms.

			6 is a term since 6 and 7 are both tribonacci-Niven numbers: the minimal tribonacci representation of 6, A278038(6) = 110, has 2 1's and 6 is divisible by 2, and the minimal tribonacci representation of 7, A278038(7) = 1000, has one 1 and 7 is divisible by 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000073, A042964, A278038, A278043.
Subsequence of A352089.
Subsequences: A352091, A352092.
Similar sequences: A330927, A328205, A328209, A328213, A330931, A331086, A333427, A334309, A331820, A342427, A344342, A351715, A351720.

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; Divisible[n, DigitCount[Total[2^(s - 1)], 2, 1]]]; Select[Range[1000], q[#] && q[# + 1] &]

A352091 Starts of runs of 3 consecutive tribonacci-Niven numbers (A352089).

Original entry on oeis.org

6, 12, 26, 80, 184, 506, 664, 1602, 1603, 1704, 3409, 6034, 9830, 15723, 16744, 19088, 21230, 21664, 22834, 33544, 39424, 40662, 40730, 51190, 55744, 56224, 60710, 61264, 63734, 66014, 66055, 67144, 67248, 73024, 78064, 81150, 84790, 94086, 95094, 109087, 111880

Offset: 1

Amiram Eldar, Mar 04 2022

			6 is a term since 6, 7 and 8 are all tribonacci-Niven numbers: the minimal tribonacci representation of 6, A278038(6) = 110, has 2 1's and 6 is divisible by 2, the minimal tribonacci representation of 7, A278038(7) = 1000, has one 1 and 7 is divisible by 1, and the minimal tribonacci representation of 8, A278038(8) = 1001, has 2 1's and 8 is divisible by 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A278038.
Subsequence of A352089 and A352090.
A352092 is a subsequence.
Similar sequences: A154701, A328206, A328210, A328214, A330932, A331087, A333428, A334310, A331822, A342428, A344343, A351716, A351721.

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; triboNivenQ[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; Divisible[n, DigitCount[Total[2^(s - 1)], 2, 1]]]; seq[count_, nConsec_] := Module[{tri = triboNivenQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ tri, c++; AppendTo[s, k - nConsec]]; tri = Join[Rest[tri], {triboNivenQ[k]}]; k++]; s]; seq[30, 3]

A352092 Starts of runs of 4 consecutive tribonacci-Niven numbers (A352089).

Original entry on oeis.org

1602, 218349, 296469, 1213749, 1291869, 1896630, 1952070, 2153709, 2399550, 3149109, 3753870, 3809310, 3983229, 4226208, 4256790, 4449288, 4711482, 5707897, 5727708, 6141750, 6589230, 6969429, 7205757, 7229208, 7276143, 7292943, 7454710, 7752588, 7937109, 8877069

Offset: 1

Amiram Eldar, Mar 04 2022

Conjecture: There are no runs of 5 consecutive tribonacci-Niven numbers (checked up to 10^10).

			1602 is a term since 1602, 1603, 1604 and 1605 are all divisible by the number of terms in their minimal tribonacci representation:
     k    A278038(k)  A278043(k)  k/A278043(k)
  --------------------------------------------
  1602  110100011010           6           267
  1603  110100011011           7           229
  1604  110100100000           4           401
  1605  110100100001           5           321

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A278038, A278043.
Subsequence of A352089, A352090 and A352091.
Similar sequences: A141769, A328211, A328207, A328215, A330933, A331824, A334311, A342429, A344344.

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; triboNivenQ[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; Divisible[n, DigitCount[Total[2^(s - 1)], 2, 1]]]; seq[count_, nConsec_] := Module[{tri = triboNivenQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ tri, c++; AppendTo[s, k - nConsec]]; tri = Join[Rest[tri], {triboNivenQ[k]}]; k++]; s]; seq[6, 4]

A058412 Squares composed of digits {0,1,2}, not ending with zero.

Original entry on oeis.org

1, 121, 10201, 22201, 1002001, 1022121, 1212201, 100020001, 100220121, 121022001, 210221001, 10000200001, 10002200121, 10020210201, 10201202001, 12100220001, 100021020121, 1000002000001, 1000022000121, 1000202010201

Offset: 1

Patrick De Geest, Nov 15 2000

All terms but the first one have their largest digit equal to 2, cf. A277946 = A277959^2. - M. F. Hasler, Nov 15 2017

P. De Geest, Index to related sequences
Hisanori Mishima, Sporadic tridigital solutions
OEIS Index to sequences related to squares.

Cf. A058411.
Cf. A063009, A066139. - Zak Seidov, Jul 01 2013
Cf. A136808, A136809 and A136810, ..., A137147 for other digit combinations.
See also A277946 = A277959^2 = squares whose largest digit is 2.
The first 1261 terms are also a subsequence of A278038 (binary numbers without '111'), in turn a subsequence of the binary numbers A007088.

apply( t->t^2, vector(100,i,N=if(i>1,next_A058411(N),1))) \\ M. F. Hasler, Nov 15 2017

a(n) = A058411(n)^2. - Zak Seidov, Jul 01 2013

cp's OEIS Frontend

A003146 Positions of letter c in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

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A003726 Numbers with no 3 adjacent 1's in binary expansion.

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A136175 Tribonacci array, T(n,k).

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A352103 a(n) is the maximal (or lazy) tribonacci representation of n using a binary system of vectors not containing three consecutive 0's.

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A136808 Numbers k such that k and k^2 use only the digits 0, 1 and 2.

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A317204 Expansion of n in the p-system based on convergents to sqrt(2).

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A352090 Numbers k such that k and k+1 are both tribonacci-Niven numbers (A352089).

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