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%I #20 Apr 11 2022 21:46:05
%S 1,3,8,20,43,98,212,465,1000,2144,4497,9504,19872,41455,85356,178630,
%T 363467,757085,1541998,3183600,6515066,13357593,27432649,55914902,
%U 114683858,233517515,478061719,972479046,1986013932
%N Form a triangle with n numbers in top row; all other numbers are the sum of their parents. E.g.: 4 1 2 7; 5 3 9; 8 12; 20. The numbers must be positive and distinct and the final number is to be minimized. Sequence gives final number.
%C Suggested by Problem 401 of the All-Soviet-Union Mathematical Competitions 1961-1986. Two different links are available for this collection.
%H Mauro Fiorentini, <a href="/A028307/a028307.txt">Further comments</a>
%H Vladimir A. Pertsel, <a href="https://olympiads.win.tue.nl/imo/soviet/RusMath.html">Problems of the All-Soviet-Union Mathematical Competitions 1961-1986</a>
%F From _A.H.M. Smeets_, Feb 25 2022: (Start)
%F a(n) > 2*a(n-1). Proof: Let x, y be the numbers in the second last row, then x >= a(n-1), y >= a(n-1) and x != y, so a(n) = x + y > 2*a(n-1).
%F It seems that a(n) > (4/3)*(2*a(n-1)-a(n-2)). (End)
%e Solutions for n = 1, 2, ... are:
%e 1;
%e 1, 2;
%e 2, 1, 4;
%e 4, 1, 2, 7;
%e 7, 2, 1, 4, 6;
%e 8, 6, 1, 3, 2, 10;
%e ...
%Y Cf. A028308, A062896, A340389.
%K nonn,nice
%O 1,2
%A _Mauro Fiorentini_
%E More terms from the author, Jul 03 2001
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