Chapter II TRIANGULAR NUMBERS

Chapter II TRIANGULAR NUMBERS Part of this work contained in this chapter has resulted in the following publications: Gopalan, M.A. and Jayakuar, P. "Note on triangular nubers in arithetic progression", International Journal Acta Ciencia Indica, 30(4) (2004) 717-718. MR2145051 Raaraj, T. and Jayakuar, P. "On triangular nubers", InternationalJournal of Acta Ciencia Indica, 32M(3) (2007) 867-873.

Chapter- II TRIANGULAR NUMBERS A triangular nuber is a nuber obtained by adding all positive integers is equal to a given positive integer n, i.e., Tn = Eft = «(«+1)= = *=' 2 2 12 As shown in the rightost ter of this forula, every triangular nuber is a binoial coefficient: the triangular is the nuber of distinct pairs to be selected fro n +1 objects. In this foi- it solves the ^handshake proble' of counting the nuber of handshakes if each person in a roo shakes hands once uu'ith eclcvi 0 person. The sequence of triangular nubers for n=l, 2, 3...is: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55... Relations to other figurate nubers Triangular nubers have a wide variety of relations to other figurate nubers. Most siply, the su of two consecutive triangular nubers is a square nuber. Algebraically, " "-' ^2 2^ 2 2 ^ 2 2 2 2 ^ There are infinitely any triangular nubers that are also square nubers; e.g., 1, 36. Soe of the can be generated by a siple recursive forula:

20 S,,=4S (SS +l)withs,=^l. All square triangular nubers are found fro the recursion S = 345 _, -S _2 + 2 with So =0 and Si =1. Also, the square of the triangular nuber is the sae as the su of the cubes of the integers 1 to n. nuber, The su of the n first triangular nubers is the tetrahedral in)(n + l)(n + 2) 6 More generally, the difference between the n"" -gonal nuber and the n"" (+l)-gonal nuber is the (n-l)"" triangular nuber. For exaple, the sixth heptagonal nuber (81) inus the sixth hexagonal nuber (66) equals the fifth triangular nuber (15). Every other triangular nuber is a hexagonal nuber. Knowing the triangular nubers, one can reckon any centered polygonal nuber: the centered k-gonal nuber is obtained by the forula Ck =kt _,+\ where T is a triangular nuber. Other properties Every even perfect nuber is triangular (given by forula, M 2"'' =M {M +\)I2 = TM, wherem is a Mersenne prie), and no odd perfect nubers are known, hence all known perfect nubers are triangular. In base 10, the digit root of a triangular nuber is always 1, 3, 6, or 9.

21 Hence every triangular nuber is either divisible by three or has a reainder of 1 when divided by nine: 6=3x2, 10=9x 1 + 1, 15 = 3x5, 21 =3x7, The inverse of the stateent above is, however, not always true. For exaple, the digital root of 12, which is not a triangular, is 3 and divisible by three. The su of the reciprocals all the triangular nubers is: OO 1 00 "1 223^ = 2. 2 This can be shown by using the basic su of a telescoping series: CO 1 tf«(«+!) Two other interesting forulas regarding triangular nubers are: and both of which can be established either by looking at dot patterns or with soe siple algebra. In 1796, Geran atheatician and scientist Carl Friedrich Gauss discovered that every positive integer is representable as a su of at ost three triangular nubers, writing in his diary his faous words,

22 "Heureka! nu = A + A + A." Note that this theore does not iply that the triangular nubers are different (as in the case of 20 =10 +10), nor that a solution with three nonzero triangular nubers ust exist. This is a special case of Ferat's polynoial nuber Theore. The largest triangulai- nuber of the for 2* -1 is 4095. Tests for triangular nubers One can efficiently test whether a positive integer x is a triangular nuber by coputing VSx + l-l n =. 2 If n is an integer, then x is the triangular nuber. If n is not an integer, then x is not triangular. In this chapter, two paraetric representations of the ranks of three triangular nubers T^,T^,T^ satisfying the equation r +r^ =27^ are found. Also explicit forulas for the ranks of triangular nuber which are siultaneously equal to pentagonal, hexagonal, heptagonal, and octagonal nubers in turn are presented, and deterine pairs of triangular nubers whose ratios are non-square integers. Also a few interesting relations aong triangular nubers are presented [25] and [42].

23 Theore 2.1 The two paraetric representations of the ranks of three triangular nubers T^,T^,T^ satisfying the equation T^+T^=2T^ are a = ~{^ -n^ -ln-\),h = -{^ +n^ -1), c = -(^ -n^ +2n-\), where, n are integers such that >n,-nsl (-n)2 > 2n2+ 1. {od2) and Proof By the standard definition of triangular nuber, r. = ^....(2.1) Using (2.1), the equation Ta+T,-2T,...(2.2) is written as 2 2 2 ' Multiplying both sides by 4, we get 4a (a+1) + 4c (c+1) = 2 x 4 b (b+1). By writing coplete squares, we get (2a +!)'-! + (2c +1)' -1 = 2[(2b +1)' -1]. (2a + l)^+(2c + l)' =2(26 + 1)^ If we take 2a+l = X, 2c+l = Z and 2b+l = Y,...(2.3) then we get X^+Z^=2Y\...(2.4) which is a Diophantine equation of second degree.

24 This shows thaxx^,y^,z^ are in Arithetic progression. To find three squares X^,Y^,Z^ are in Arithetic progression, we ust have Y^ -X^ =Z^ -Y^ (or) 2Y^=X^+Z\ Introducing the linear transforations X = p-q, Z = p + q,...(2.5) the equation (2.4) becoes X'+Z'=ip-qy+(p + qy=2{p'+q') = 2Y\ i.e., p'+q'=y\ which is the well-known Pythagorean equation and whose solution is given by p = 2 - n2, q = 2n, Y = ^ + n^, where and n are distinct positive integers. In view of (2.5), we have X = 2 - n2-2n,'z = 2 - n2 + 2n. Thus, fro the equation (2.3), we obtain a = (^ -n^ -2n-\), 2 cz=-(^ -rp- +2n-l). 2 For a, b, c to be integers, the values of and n should be chosen such that > n, - n SE 1 (od2) and ( - n) 2 > 2n2 + 1.

A few exaples are presented in the table 2.1. Table 2.1 n a b c Ta Tb Tc 4 1 3 8 11 6 36 66 7 2 8 26 36 36 351 666 8 3 3 36 51 6 666 1326 9 2 20 42 56 210 903 1596 10 3 15 54 75 120 1485 2850 25 Next prove the following theores. Theore 2.2 Explicit forula for the ranks of triangular nuber which are siultaneously equal to pentagonal nubers are given by r = J-p2 + V3)""(V3+l)+(2-V3)""(l-V3)+2],«= 0,l,2... Proof It is well- known that the triangular and pentagonal nubers are given by T = { +1)...(2.6) P.=f(3-1),...(2.7) where and r are the corresponding ranks. The identity r = P,, leads to

{ +1) r(3r -1) 2 2. 1-S-, ^ + -3r^ -r. By writing coplete squares, we get (».4)-i = 3(.'- ). = 3[(,-i)--±, 26 That is 3(2w + l)'-3 = (6r-l)'-l. If we take X=6r-1,...(2.8) Y=2+1,...(2.9) then we get X'-37'=-2,...(2.10) which is the faous Pell's equation. The general solutions of the equation (2.10) are obtained by ^.=i[(2 + Vl)"*'(V3+l)+(2-V3)"'(l-V5)],...(2.11) i; = A-[(2 + V3)"'(V3+l)+(2-,/3)"*'(.73-l)],...(2.12) where n=0,l, 2,... In view of (2.8) and (2,9), the ranks of triangular and pentagonal nuber are given by = _l^p2 + V3)""(v^+l)+(2-V3)""(V3-l)-2V^" r ^[(2 + V3) "^' (V3 +1)+ (2 - S) "^'(1 - V3)+ 2], where n = 0,1,2.. 12 The values of ranks and r will be in integers only when n = 2r, r = 1, 2, 3...

Soe exaples are presented in the table 2.2. Table 2.2 27 s.. Values Rank Rank Triangular Pentagonal No. of n r nuber T nuber Pr 1 2 20 12 210 210 2 4 285 165 40755 40755 3 6 3976 2296 7906276 7906276 4 8 55385 31977 1533776805 1533776805 5 10 771420 445380 297544793910 297544793910 Theore 2.3 Explicit forula for the ranks of triangular nuber which aire siultaneously equal to heptagonal nuber are given by. 1 ^[(9 + 4V5)""(3-fV5)+(9-4V5)""(V5-3)-2V5" r = ^ ^(9 + 4V^)"" (3 + V5)+ (9-4V5 )"'' (3 - V5)+ 6, where «= 0,1,2... 20 L Proof It is well- known that the triangular and heptagonal nubers are given by 7;= (. + l),//.=^(5r-3), where and r are the corresponding ranks. The identity T = //,., leads to ( + l) = -(5r-3). 2 2 i.e.jw^ + = 5r^ -3r.

By writing coplete squares, we get 28 i-e., (+ -)'-- = 5[(r- Y - ]. 2 4 10 100 (2 + l)^-l^[(10r-3)^-9] 4 20 If wetakex= lor-3,...{2.13) Y = 2+1,...{2.14) then we get 5(F^-l) = X'-.9. X^-5Y'=4,...{2.15) which is the faous Pell's equation. The general solutions of the equation (2.15) are obtained by " 2 (9 + 4VJ)"'(3 + V5)+(9-4V?)"*'(3-^)],...(2.16)!'.=^P5+4V5)"'(3 + Vl).(9-4,/f)"'(V?-3)]....(2.171 where n = 0, 1, 2... In view of {2.13) and {2.14), the ranks of triangular and heptagonal nuber are given by = J U9 + 4^/5)"" (3 + V5)+ (9-4V5)"'' (V5-3)- 2V5 r L[(9 + 4~5)"" (3 + V?)+ (9-4V^)"" (3-4~5)+6],n = 0,1,2... 20 The values of ranks and r will be in integers only when n = 2r, r =1, 2, 3...

s. No. Soe exaples are presented in the table 2.3. Values of n Rank ni Rank r Table 2.3 Triangular nuber T Heptagonal nuber Hr 1 0 10 5 55 55 2 2 3382 1513 5720653 5720653 3 4 1089154 487085 593128762435 59312876235 4 6 350704366 156839761 61496776341083161 61496776341083161 29 Theore 2.4 Explicit forula for the ranks of triangular nuber which are siultaneously equal to hexagonal nuber are given by = 2k + l, r = k + l,k = 0,1,2,... Proof It is well- known that the triangular and hexagonal nubers are given by " 2 H^=r(2r-\), v^here and r are the corresponding ranks....(2.18)...(2.19) The identity T =H^, leads to the equation X^=Y^orX = Y,...(2.20) where X = 2+1,....(2.21) Y=4r-1....(2.22) The general solutions of the equation (2.20) are obtained by X =7 =^ where t = 0, 1,2, 3...

30 In view of (2.21) and (2.22), the ranks of triangular and hexagonal nuber are given by =, r =,t = 0,1,2,3... The values of ranks and r will be in integers only when t = 4r-l, r =1, 2, 3... s. No Soe exaples are presented in the table 2.4 Values of n Rank Rank r Table 2.4 Triangular nuber T Hexagonal nuber Hr 1 3 1 1 1 1 2 7 3 2 6 6 3 11 5 3 15 15 4 15 7 4 28 28 5 19 9 5 45 45 Theore 2.5 Explicit forula for the ranks of triangular nuber which are siultaneously equal to octagonal nubers are given by ^ ''(5 + V24) "*' (s + 724)+ (5 - V24) "*' (V24 - s)- 2 V24 = - 4V24 1 r = (5 + V24) "^' (3 + V24)+ (5 - V24) "^' (V24-3)+ 2V24, n = 0,1,2... 6V24 Proof It is well- known that the triangular and octagonal nubers are given by T^= { + l),0^= r{3r-2), where and r are the corresponding ranks.

The identity T^=0^, leads to the equation 31 (w + l) = r(3r-2).- i.e., ^ \- = 2r{3r-2) = e{r^ ^). By writing coplete squares, we get { + -y-- 2 4 = 6[{r-~f--l 3 9 (2+ 1)^ 1^ (3r-l)^ 1 4 4 9 9 3[(2/w + l)'-l] = 8[(3r-l)'-l]. 3(2/n + l)'-3 = 8(3r-l)^-8. If wetakex=2+l,...(2.23) Y=3r-1,...(2.24) then we get 2>X^ -87^ = -5,...(2.25) which is the faous Pell equation. By introducing the linear transforations X = u + '^t,...(2.26) Y^u + 3t,...(2.27) the equation (2.25) leads to 3(u+8t)'-8(u+3t)'=-5. 3(u^+l6ut + 64t^)-8{u^+6ut + 9t^) = -5 3u^ +48ut + l92t^ -Su^ -48ut-72t^ =-5-5u^+\20t^ =-5 -u^+24t^ =-l (or) M'-24/'=1,...(2.28) which is the faous Pell equation. by The general solutions of the above equation (2.28) are given

32 u = (5 + V24)""+(5-V2^f)""l,...(2.29) 1 2V24 where n= 0, 1, 2, 3... (5 + V24)""-(5-V^)"^'],...(2.30) In view of (2.23-2.24) and (2.29-2.30), the ranks of triangular and octagonal nubers are given by ^ ""(s+v24) "^' (g+v:24)+ (5 - V24) "^' (V24 - s)- 2V24 4V24L r = -^[(5 + V24) "^' (3 + V24)+ (5 - V24) "^' (V24-3)+ 2V24 The values of ranks and r will be in integers only when n = 2r, r=l, 2, 3... s. No 1 2 3 4 Soe exaples are presented in the table 2.5. Values of n 0 2 4 6 Rank 6 638 62566 6130878 Table 2.5 Rank r 3, 261 25566 2502921 Triangular nuber T ' 21 203841 1957283461 18793835590881 Octagonal nuber Or 21 203841 1957283461 18793835590881 Also obtain the following observations. Observation Pair of triangular nubers whose ratios are non-square integers Proof If T and Tk are denoting triangular nubers, then the identityr =at,^, where a>\ is a square free nuber, leads to (OT + 1) = (k + V). 2 2 ^ + = a(k^ +k).

By writing coplete squares, we get 33 2 4 2 4 {2 + \f 1 M + lf 1 4 4 4 4 {2 + \y -\ = a[{2k + \f-\]. {2 + \y -a{2k + \f =l-a. Ifwe takex=2k+r,...(2.31) then we get Y=2+1,...(2.32) 7'-«^'=-JV, where N=a-1....(2.33) The general solutions of the equation (2.33) are given by...(2.35) where J^ + Va^o is the fundaental solution of (2.33) and yq+4axf^ is the fundaental solution of the Pellian Y^-aX^^\...(2.36) In view of (2.31) and (2.32) the ranks of the triangular nubers are obtained as

1 4Va [[Y, + 4^X,) "^' (yo + Vax'o)+ {Y^ - V^Xo) "^' (xo Va - ;^o)- 2Va}, where n = 0, 1, 2, 3 Soe Particular cases...(2.38) Whena = 2, the equation (2.33) leads to the Pell's equation r'-2x'=-l. In view of (2.37) and (2.38), the ranks of the triangular nubers are given by! = - {(3 + 2V2)"" (1 + 42)+ (3-2 V2 ) "^' (1 - V2)- 2} w A: = ^ {(3 + 2V2) "^'(1 + 72)+ (3-2V2) "^' (V2-1)- 2V2}, where n = 0,1,2....s. No Soe exaples are presented in the table2. 6. Values of n Rank Table 2.6 Rank k Triangular nuber T Triangular nuber Tk 1 0 3 2 6 3 2 1 20 14 210 105 3 2 107 72 7140 3570 4 3 696 492 242556 121278 5 4 4059 2870 8239770 4119885 Y^-3X^=-2. When a-?>, the equation (2.33) leads to the Pell's equation In view of (2.37) and (2.38), the ranks of the triangular nubers are given by A=4. {(2+V3)"" (1+v^K (2 - v^)"" (v^ -1)-2V3}, 4V3 where n = 0,1,2,3 34

Soe exaples are presented in the table 2.7. Table 2.7 35 s. Values Rank Rank Triangular Triangular No of n K nuber T nuber Tk 1 0 2 1 3 1 2 1 9 5 45 15 3 2 35 20 630 210 4 3 132 76 8778 2926 5 4 494 285 122265 40755 Whena = 5, the equation (2.33) leads to the Pell's equation 7^-5X'=-4. In view of (2.37) and (2.38), the ranks of the triangular nubers are given by! = - (9+4V5)"''(l+V5)+(9-4V5)""(l-V5)-2L - A:=^{(9+V5)"'(l+V5)+(9-4V5)""(V5-l)-2V5J, where n = 0, 1,2... s. No 1 2 3 4 5 Soe exaples are presented in the following table 2.8. Values of n 0 1 2 3 4 Rank 14 260 4674 83880 1505174 Rank k 6 116 2090 37512 673134 Table 2.8 Triangular Nuber T 105 33930 10925475 3517969140 1132751377^^ Triangular Nuber Tk 21 6786 2185095 703593828 2655027545 Observation A few interesting relations aong triangular nubers It presents below a few interesting relations aong triangular nubers. T and Ts are denoting the triangular

36 nubers. First derives the recurrence relation aong triangular nubers. The triangular nubers satisfy the following recurrence relation. Recurrence relation 2.1 Proof By the standard definition of triangular nubers, the "" triangular nuber is given by 7;= (,«+ 1)....(2.39) Replacing by -1, +] in (2.39), we get T x=- -{-\ + \) = (-\),^ ^ "-' 2 2,...(2.40) T,,=^-Y^( + 2) = -( + l) + ( + l) = T^+ + l....(2.41) The equation (2.39) can be written as T =-(-l + 2) = -(-l) + -T,_,+....(2.42) Fro the equation (2.41), we get T..^-T,=+1...(2.43) Fro the equation (2.42), we get T.-T.r=...(2.44) Fro the equations (2.43) and (2.44), we get ^+1 "~2r^ +r _, = 1, which is the required relation.

37 Exaple 2.1 If we take =2, then 73= 6, third triangular nuber,7^2 =3, the second triangular nuber and T^ =l,the first triangular nuber. Thus r,,-27;+r., =6-6 + 1 = 1.. Identity 2.1 The su of triangular nubers can be expressed as the polynoial in s. i.e., Yj'^=-s{s + \){s + 2) = ^^^^Ts, where T^is the s'^ triangular nuber. Proof The "' triangular nuber is given by Thus tt, = t~{ + \) = -t[rn +\ 1 ^ -y ^ =\ =l = U\s{s + \)Cls + \) + ]-s{s + X)\ 2 6 2 = ^s{s + \){^{2s + \) + \\ = 5(5 + l)(25 + 4) 12 = -s{s + \){s + 2). 6

38 Hence :'--^<-'><-^'-^fr(-'>l = ^^» Identity 2.2 The su of squares of triangular expressed as a polynoial in s. nubers can be i-e.,y,tj = s(s + l)[3s' +12s^ +I3s + 2]^^[3s' +12s^ +13S + 2], =l 60 30 where T^ is the s"' triangular nuber. Proof The "" triangular nuber is given by s Thus I^.'=Etf('"-^l)l'42'"'('"-^')' 1. = -IY'+2 + l) ^ =l =l tn=\ I s{s + l)i6s'+9s'''+s-\) 2^'(5 + 1)' 5(;y +1X2^ + 1) ~4 30 "^ 4 "^ 6 s(s + r)j6s'+9s^+s-l),,, (2^ + 1)^ 8 15 3 ^s(s^ 3^ 24s 2 +26s+ 4] 120 = [65'+245'+265 + 41. 60 = [35^+125^+135 + 2]. 30

39 Identity 2.3 The su of cubes of triangular nubers can be expressed as a polynoial in s. i.e., -^7;'= r,(30^'+180/+366^'+26%'+165-16), where T^ is the s"' triangular nuber. Proof The "' triangular nuber is given by r =-(«j+i). Thus ±Tj =Y\-'^{ + \)f =\±\ + \) =1 =1 '^ O n,=1 1 * = -J^\^ +3^ +3 + l) o =\ =l =l =l _ 1 s(s + l)(6s' +I5s' + 6/ - 6s' -s + l) 3s\s + l)(2s'+2s-l) ~8 42 ^ 12 + 3s(s + l)(6s' +9s'+s-r)+ ^ ^ ]. 1, ^^,,6s'+l5s'+6s'-6s'-s + l) s(s + l)(2s' + 2s) = ~s(s +1)[( + - 16 ^ 21 2 5 2 = -^[60^' +3605" +7325' +528^' +325-32] ^ -[305^ +1805" +3665^+2645^ +I65-I6]. 840

Thus ttj = ^ Z (3O5' +1805" + 3665' + 2645' +165-16). ".=1 R40 ^ ^ 40 It is interesting to note that there are soe perfect nubers like 5, 28, and 490 aong triangular nubers. There are soe repeated digits of nubers like 66, 666 aong triangular nubers. It has been observed that there exist triplets of the triangular nubers in Arithetic progression. But there are no triplets of the triangular nubers in geoetric progression except 1,6,36. One ay search for the other triangular nubers which are in arithetic progression. It has seen that the presentation of explicit forulas for the ranks of triangular nuber which are siultaneously equal to pentagonal, hexagonal, heptagonal, and octagonal nubers. The researching for the other pentagonal, hexagonal, heptagonal and octagonal nubers could be relatively seen. The further observation of pairs of triangular nubers whose ratios are non-square integers. One ay search for the other triangular nubers whose ratios are non-square integers.