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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
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