The beam and the double-pan balance
Put the beam of a double-pan balance between any two successive terms of the hereunder sequence D. D = 1,2,3,3,4,5,6,3,7,5,8,7,9,9,1,8,10,11,12,3,13,11,14,7,… There are always k terms to the left of the beam with sum t and k terms to the right of the beam with sum 2t . Examples (the beam is the ampersand & ) t=1. D =1 & 2 (sum 2 on the right pan = 2t) t=3. D =1,2 & 3,3 (sum 6 on the right pan = 2t) t=6. D =1,2,3 & 3, 4,5 (sum 12 on the right pan = 2t) t=9. D =1,2,3,3 & 4,5, 6,3 (sum 18 on the right pan = 2t) t=13. D =1,2,3,3,4 & 5,6,3, 7,5 (sum 26 on the right pan = 2t) t=18. D =1,2,3,3,4,5 & 6,3,7,5, 8,7 (sum 36 on the right pan = 2t) t=24. D =1,2,3,3,4,5,6 & 3,7,5,8,7, 9,9 (sum 48 on the right pan = 2t) t=27. D =1,2,3,3,4,5,6,3 & 7,5,8,7,9...