Miscellaneous Papers. 



251 



From these series are constructed the following perfect 

 squares : 



(1) 



0=7. d = l. n=2, ff = 3. 

 G=2.5. 



(2) 



a=7, d=2, n = l. s/ = 1.5. 



G = 1.5. 



(3 and 4) 



a=7, d=1.5, M=1.5, ff=2.5, 



G=1.5. 



77 77 

 No. 33. 



77 



77 77 

 No. 34. 



77 77 



77 77 77 77 77 

 Nos. 35 and 36. 



Two arrangements are shown for the third series. One (3) 

 is laid according to model square No. 15, ante, and the other 

 (4) is laid according to model square No. 19. Both are pre- 

 cisely alike; the only difference is in the order that the terms 

 are taken. Some of the others may also be taken in different 

 orders. 



As shown in the preceding pages every sum of a row in any 

 square of 4 is equal to twice the continued sum of double the 

 first term of the series plus eight times the difference between 

 the first and second terms, plus four times the difference be- 

 tween the second and third terms, plus twice the difference 

 between the fourth and fifth terms, plus the difference between 

 the eighth and ninth terms. Using the same letters as before 

 to represent these several differences, the formula becomes, as 

 already shown, 



S = 2(2a + 8d + 4n + 25f + G). 



To obtain 111 as the sum in a perfect square, as for any 

 other number, we take half the amount, in this case 55.5, 

 divide it up to suit fancy or convenience by making the several 



111 

 111 

 111 

 111 



111 111 111 111 111 



No. 37. 



Differences. 



2a= 2 

 8d=16 

 4w=18 

 2ff = 14 



a = l 

 d=2 

 71=4.5 

 ff=7 



Series. 



1 3 7.5 9.5 



16.5 18.5 23 25 



30.5 32.5 37 39 



46 48 52.5 54.5 



G= 5.5 G=5.5 



55.5 



