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A fractal tree, read by rows: for n > 2, T(n,1) = T(n-1,1)+2, T(n,n) = T(n-1,1)+3, and for k=2..n-1, T(n,k) = T(n-2,k-1).
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%I #39 Jan 02 2023 12:30:51

%S 1,2,3,4,1,5,6,2,3,7,8,4,1,5,9,10,6,2,3,7,11,12,8,4,1,5,9,13,14,10,6,

%T 2,3,7,11,15,16,12,8,4,1,5,9,13,17,18,14,10,6,2,3,7,11,15,19,20,16,12,

%U 8,4,1,5,9,13,17,21,22,18,14,10,6,2,3,7,11,15,19,23

%N A fractal tree, read by rows: for n > 2, T(n,1) = T(n-1,1)+2, T(n,n) = T(n-1,1)+3, and for k=2..n-1, T(n,k) = T(n-2,k-1).

%C It appears that:

%C 1) partial sums of terms, situated on the outer leftmost leftwise triangle diagonal are equal to A002061(k), k>=1;

%C 2) partial sums of terms, situated on the second (from the left) leftwise triangle diagonal represent recurrence a(k+1) = ((k-1)*a(k))/(k-3)-(2*(k+3))/(k-3), k>=3

%C 3) partial sums of terms, situated on the outer rightmost rightwise triangle diagonal are equal to A000290(k)=k^2, k>=1. - _Alexander R. Povolotsky_, Dec 28 2014

%H Reinhard Zumkeller, <a href="/A253146/b253146.txt">Rows n = 1..125 of triangle, flattened</a>

%H Éric Angelini, <a href="http://list.seqfan.eu/oldermail/seqfan/2014-December/014202.html">A fractal tree</a>, SeqFan list, Dec 27 2014.

%e . 1: 1

%e . 2: 2 3

%e . 3: 4 1 5

%e . 4: 6 2 3 7

%e . 5: 8 4 1 5 9

%e . 6: 10 6 2 3 7 11

%e . 7: 12 8 4 1 5 9 13

%e . 8: 14 10 6 2 3 7 11 15

%e . 9: 16 12 8 4 1 5 9 13 17

%e . 10: 18 14 10 6 2 3 7 11 15 19

%e . 11: 20 16 12 8 4 1 5 9 13 17 21

%e . 12: 22 18 14 10 6 2 3 7 11 15 19 23 .

%e Removing the first and last entries from each row gives the same tree back again.

%e From _N. J. A. Sloane_, Jan 04 2015: (Start)

%e Eric Angelini's original posting to the Sequence Fans mailing list gave a different sequence, as follows:

%e ..................................1,

%e .................................2,3,

%e ................................4,1,5,

%e ...............................6,2,3,7,

%e ..................................8,

%e ..............................9,4,1,5,10,

%e ............................11,6,2,3,7,12,

%e ................................13,14,

%e ...............................15,8,16,

%e ...........................17,9,4,1,5,10,18,

%e .........................19,11,6,2,3,7,12,20,

%e ........................21,13,8,4,1,5,9,14,22,

%e .............................23,15,16,24,

%e .................................25,

%e ...........................26,17,10,18,27,

%e ......................28,19,11,6,2,3,7,12,20,29,

%e .....................30,21,13,8,4,1,5,9,14,22,31,

%e ..........................32,23,15,16,24,33,

%e ................................34,35,

%e ..............................36,25,37,

%e ........................38,26,17,10,18,27,39,

%e ...................40,28,19,11,6,2,3,7,12,20,29,41,

%e ..................42,30,21,13,8,4,1,5,9,14,22,31,43,

%e ................44,32,23,15,10,6,2,3,7,11,16,24,33,45,

%e ...............46,34,25,17,12,8,4,1,5,9,13,18,26,35,47,

%e .......................48,36,27,19,20,28,37,49,

%e .............50,38,29,21,14,10,6,2,3,7,11,15,22,30,39,51,

%e ............52,40,31,23,16,12,8,4,1,5,9,13,17,24,32,41,53,

%e .....................54,42,33,25,18,26,34,43,55,

%e .............................56,44,45,57,

%e .................................58,

%e .....................................

%e The idea is that the n-th term is equal to the number of terms in the n-th row of the tree. This lovely sequence (whose precise definition is not clear to me) is not yet in the OEIS. (End)

%e The sequence referred to is A253028. - _Felix Fröhlich_, May 23 2016

%t T[n_, 1] := 2n - 2;

%t T[n_, n_] := 2n - 1;

%t T[n_, k_] := T[n, k] = T[n-2, k-1];

%t Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Sep 20 2021 *)

%o (Haskell)

%o a253146 n k = a253146_tabl !! (n-1) !! (k-1)

%o a253146_row n = a253146_tabl !! (n-1)

%o a253146_tabl = [1] : [2,3] : f [1] [2,3] where

%o f us vs@(v:_) = ws : f vs ws where

%o ws = [v + 2] ++ us ++ [v + 3]

%Y Cf. A253028. Row sums appear to be A035608.

%Y Cf. A000290, A002061.

%K nonn,tabl

%O 1,2

%A _Eric Angelini_ and _Reinhard Zumkeller_, Dec 27 2014