A136175 Tribonacci array, T(n,k).

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

Author

Clark Kimberling, Dec 18 2007

Keywords

• nonn
• tabl

Comments

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]

Examples

			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
		

Crossrefs

Cf. A035513, A353083, A353084.

Programs

Maple

# 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

Formula

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.)

Extensions

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