Chapter PENTAGONAL NUMBERS

Chapter PENTAGONAL NUMBERS Part of this work contained in this chapter has resulted in the following publication: Gopalan, M.A. and Jayakumar, P. "On pentagonal numbers", International Journal Acta Ciencia Indica, 31M(3) (2005) 711-716. MR2235905

Chapter III PENTAGONAL NUMBERS The theory of numbers has been called the Queen of Mathematics because of its rich varieties of fascinating problems. Many numbers exhibit fascinating properties, they form sequences, they form patterns and so on. A pentagonal number is a figurate number that extends the concept of triangular and square numbers to the pentagon, but, unlike the first two, the patterns involved in the construction of pentagonal numbers are not rotationally symmetrical. The «" pentagonal number p^is the number of distinct dots in a pattern of dots consisting of the outlines of regular pentagons whose sides contain 1 to n dots, overlaid so that they share one vertex. For instance, the third one is formed from outlines comprising 1, 5 and 10 dots, but the 1, and 3 of the 5, coincide with 3 of the 10 - leaving 12 distinct dots, 10 in the form of a pentagon, and 2 inside... p is given by the formula: Pn = n(3n-v) for n>l. The first few pentagonal numbers are: 1,5, 12,22,35,...

42 The generating function for the pentagonal numbers is ^^^^^ (1-x)^ = x + 5x^ + l2x'+22xu Every pentagonal number is j^of a triangular number. There are conjectured to be exactly 210 positive integers that can not be represented using three pentagonal numbers, namely 4, 8, 9, 16, 19, 20, 21, 26, 30, 31, 33, 38, 42, 43, 50, 54,..., 20250, 33066. There six positive integers that can not be expressed using four pentagonal numbers: 9, 21, 31, 43, 55 and 89. All positive integers can be expressed using five pentagonal numbers. Letting x, be the set of numbers relatively prime to 6, the generalized pentagonal number is given by(x?-l)/24. Also, letting y, be the subset of thex, for which x, s 5 (mod 6) the usual pentagonal numbers are given by(^j; -l)/24. Generalized pentagonal numbers are obtained from the formula given above, but with n taking values in the sequence 0, 1,-1,2, -2,3,-3,4,-4... producing the sequence: 0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117,126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345,376, 392,425,442... Generalized pentagonal numbers are important to Euler's theory of partitions, as expressed in his pentagonal number theorem.

43 The number of dots inside the outermost pentagon of a pattern forming a pentagonal number is itself a generalized pentagonal number. Pentagonal numbers should not be confused with centered pentagonal numbers. Tests for pentagonal numbers One can test whether a positive integer x is a pentagonal number by computing «=-V24x + l +1. 6 If n is an integer, then x is the n* pentagonal number. If n is not an integer, then x is not pentagonal. In this chapter, explicit formulas for the ranks of pentagonal numbers which are simultaneously equal to hexagonal, heptagonal and octagonal numbers in turn presented, and determines pairs of pentagonal numbers whose ratios are non-square integers. A few interesting relations among pentagonal numbers are also presented [26]. First prove the follovnng theorems relating to explicit formula for pentagonal numbers. are

Equality of the pentagonal numbers with the hexagonal numbers Theorem 3.1 Explicit formula for the ranks of pentagonal numbers which are simultaneously equal to hexagonal numbers are given by 44 m = ^ 3 [(l + V3)2n+3 + (l_73)2n+3 + 2n+2]^ k= -L 2ri<l^{(l + V3)2n+3_. (l_73)2n+3+ ^3 2n+2]. v3 Proof It is well known that the pentagonal and hexagonal numbers are given by P,n= Om-\),...(3.1) Hk=k(2k-1),...(3.2) where m and k are the corresponding ranks. The identity P = Hi^, leads to the equation {?,m-\)^k{2k-\). Multiplying both sides by 24, we get 36w'-12w = 3(16A:'-8A:). By writing complete squares, we get (6AW-1)^-1 = 3[(4A:-1)'-1]. IfwetakeY=6m-l',...(3.3) X=4k-1,...(3.4) then we get Y2_3X2 = -2,...(3.5) which is the famous Pell's equeition. If y^ + V3xp is the fundamental solution of Y^ -3X^ = 1,

45 then the sequence of solutions can be obtained by y +Sx ^{y, +^x,r\n = 0, 1, 2... = (2 + V3r',n = 0, 1,2......(3.6) Further, we have yn -V3x =(2-/3)"^', n= 0, 1, 2......(3.7) Adding and subtracting the above equations (3.6-3.7), we get y =^[(2 + V3)"^' +(2-Sr'], -(3.8) x =^[(2 + V3)"^' -(2-V3r^]....(3.9) The general solutions of the equation (3.5) are given by ^.=^oyn+yox =y +x,...(3.10) Y =Y,+DX,x ^y +3x....(3.11) Using the equations (3.8), (3.9) in (3.10), (3.11), we get r.=i[(2.73r(vj.l)-(2-v3r(v3-l)] = ik^/3+1)'" -(73-1)"-],...(3.12) X =-^[(2 + V3r(V3+l) + (2-V3r(V3-1)] 2V3 1 (V3+l)"^^ ^ (^^-1)^"" ^\n+\ 2V3 2"^' 2 =!-^[(V3+ir*'+(V3-l)^'"']...(3.13) 2"^'V3 In view of (3.3) and (3.4), the ranks of pentagonal and hexagonal numbers are given by

46 m = -^3[Q + VS)'"^' + (1 --v^)'"^' + 2""'], A; =-^2"^'[(1 + V3)'""'-(1 - VS)'""'+ V32""']. The values of ranks m and k will be in integers only when n = 4(r-l), r=l, 2, 3... Some examples are presented in the table 3.1. s. No Values of n m k Table 3.1 Pentagonal number Hexagonal number Hk 1 0 1 1 1 1 2 4 165 143 40755 40755 3 8 31977 27693 1533776805 1533776805 4 12 6203341 5372251 5722156241751 5722156241751 Equality of the pentagonal numbers with the heptagonal numbers Theorem 3.2 Explicit formula for the ranks of pentagonal numbers which are simultaneously equal to heptagonal are given by m = ^[(4 + Vl5)"(21 + 5A/l5)-(4-Vr5)"(21-5Vr5) + 2Vl5], 12V15 k = W[(4 + Vl5)" (25 + 7Vl5) - (4 - Vl5)" (25-7Vl5) + 6715], I2V15 where r = 4(r-l), r =1,2,3...

Proof It is well known that the pentagonal and numbers are given by 47 heptagonal P =~(3m-\),...(3.14) H,=^(5k-3),...(3.15) where m and k are the corresponding ranks. The identity P = H^',leads to the equation in Jc (3w-l) = -(5w-3) 2 2?,m^ -m = 5k^ -3k. By writing complete squares, we get 3[{m--Y - ] = 5[(A:- )' - ]. ^ 6 36 10 100 [(6m-l)'-l] = --[(10A:-3)'-9]. 36 100 20[(6m-l)' -l] = 12[(10A:-3)' -9]. 5[(6w-l)^-]] = 3[(10yt-3)'-9]. Ifwe take X= 10k-1,...(3.16) Y = 6m-1,...(3.17) then we get 5(7'-l) = 3(X'-9). 57'-5 = 3Z'-27. 3X'-5F'=22....(3.18)

By introducing the linear transformations X=u+5t,...(3.19) Y=u+3t,...(3.20) the equation (3.18) leads to the equation 3(u + 5ty-5(u + 3ty=22. 48 3(u' +I0ut + 25t')-5(u' +6ut + 9t') = 22. 3M' + 30ut + ISe - 5u' - 30ut - 45/' = 22. Simplifying we get -2M'+30?'=22. Dividing by 2, we get -w'+lsr' =11. (or) u^ -I5t^ =-11,...(3.21) which is the famous Pell's equation. by The general solutions of the above equation (3.21) are given tn =toun+ UoTn = Un + 2Tn, Un =UoUn + DioTn = 2Un + l5tn, where MO +-4l5to =2 + vt5 is the fundamental solution of (3.21) and UQ + -JlSTg = 4 + ^ T5 is the fundamental solution of the Pellian u^-l5t^=l....(3.22) The infinite number of solutions of (3.22) can be obtained by i7 +vr57;=(f/o+vr5ror^'=(4+vi5)",...(3.23) Further, we have where n = 1, 2... U -VTST; =(U, -Vl57;r' =(4-Vl5)",...(3.24) where n = 1, 2...

49 Adding and subtracting the above two equations (3.23-3.24), we get U =i[(4 + VT5)'' +(4-Vi5)^],...(3.25) ^«=T^[(4+ 715)"-(4-^15)"]....(3.26) 2-vl5 Using (3.25), (3.26) in the general solutions of (3.21), we get u =±[(4 + Vl5)"(2 + ^/l5)-(4-vl5)"(2-^/i5)],...(3.27) t = ^=:^[(4 + VT5)"(2 + ^5)-(4-ViI)"(Vr5-2)]....(3.28) In view of (3.16), (3.17), (3.19) and (3.20), the ranks of pentagonal and heptagonal numbers are given by m = --^[(4 + Vl5)"(21 + 5Vl5)--(4-Vr5)"(21-5Vl5) + 2Vl5], 12-\/15 k = ^[(4 + Vl5)" (25 + 7V[5) - (4 - Vl5)" (25 - vvls) + 6^15]. I2V15 The values of ranks m and k will be in integers only when n = 4(r-l), where r = 1,2,3... s. Some examples are presented in the table 3.2. Table 3.2 No Values of n m k Pentagonal number Pm Hexagonal number Hk 1 0 1 1 1 1 2 2 54 42 4347 4347 3 4 3337 2585 16701685 16701685 4 6 206830 160210 64167869935 64167869935

Equality of the pentagonal numbers with the numbers 50 octagonal Theorem 3.3 Explicit formula for the ranks of pentagonal numbers which are simultaneously equal to octagonal numbers are given by m = {(3 + 2 V2)""' (1 + 2 V2') + (3-2 V2)""' (1-2 V2) + 2} k = ^^{(3 + 2V2)"*'(2V2 + l) + (3-2V2)""'(2V2-l)+4V2}. Proof It is well known that the pentagonal and octagonal numbers are given by Pm=-{3m-l),...(3.29) Ok = k(3k-2)...(3.30) where mi and k are the corresponding ranks. The identity Pm = Ok, leads to the equation (3m-l)=k(3k-2). Multiplying both sides by 24, we get 36m^ -I2m = 24{3k^-2k) ^2{36k'^-2Ak) By writing complete squares, we get {6m-\y -\ = 2[{6r-lY -A]. lfwetakey = 6m-1,...(3.31) X = 6r-2,...(3.32) then we get 7^ -1 = 2X^ - 8

51 Y^-2X^=-7,...(3.33) which is the famous Pell's equation. The general solutions of the equation (3.33) are given by Xn = -Xoyn + YoXn = 2y + x^, Y = Yoy + lxax =y + 4x, where Yo + is the fundamental solution of (3.33), >'o + V2xo = 3 + 2V2 is the fundamental solution of the Pellian 72 _ 2^2=1....(3.34) The infinite number of solutions of (3.34) can be obtained by y +^x =(3 + 2^y^\n=0,l,2,...(3.35) Further, we have yr, -V2x =(3-2.j2)"'-\n=0,l,2, (3.36) Adding and subtracting the above two equations (3.35) and (3.36), we get ;; =-[(3 + 2^2)"^' +(3-2/2r'],...(3.37) x = J_[(3 + 2V2r'-(3-2V2r*]....(3.38) 2V2 Using (3.37), (3.3.8) in the general solutions of (3.33), we get Y =-t(3 + 2V2r' +(3-2V2r']+ -^[(3 + 2V2)"^' -(3-2V2r'] = l[(3 + 2V2r\l + 2V2) + (3-2V2r'(l-2V2)],...(3.39) X =i[(3 + 2V2r +(3-2V2r']+-^[(3 + 2V2)"^' -(3-2V2r'] = J_[(3 + 2V2r'(2V2+l) + (3-2V2r'(2V2-l)]...(3.40} 2V2

52 In view of (3.31) and (3.32), the ranks of the pentagonal and octagonal numbers are given by 6w -1 = i[(3 + 2V2)""' (1 + 2V:>') + (3-2^2)"^' (1-2V2)] 6m =~[(3 + 2V2)"^' (1 + 2V2) + (3-2V2)"^' (1-2V2)]+1 Hence m = [(3 + 2V2)"*' (1 + 2V2) + (3-2V2)"^' (1-2V2) + 2] 6r-2 = -^[(3 + 2^y^\l42 +1)+<3-2^2)"^' (2^2-1)] 2V2 6r =-^[(3 + 2V2)""'(2V2 +1) + (3-2V2)""'(2V2-1)] + 2 2v2 Hence r = ^[(3 + 2V2)"^'(2V2+l) + (3-2V2)"^'(2V2-1) + 4V2] 12v2 s. No The values of ranks m and r will be in integers only when n = 4r-3, r =1, 2, 3,... Values of n Some examples are presented in the table 3.3. m r Table 3.3 Pentagonal number Pm Octagonal number Or 1 1 11 8 176 176 2 5 12507 8844 23463132 23463132 3 5 14432875 10205584 312461813932000 (15 digits) 4 13 16655525051 11777234708 416109772078405066376 (21 digits) 312461813932000 (15 digits) 416109772P78405066376 (21 digits)

Also obtain the following observations. Observation 53 Pairs of pentagonal numbers whose ratios are non-square integers Proof If P^ and Pf. are denoting pentagonal numbers, then the identity P =ap^, where a >1, a square-free number, leads to the equation -i3m-l) = a-(3k-l). 3m^ - m ^ a{3k^ - k). By writing complete squares, we get (6m-l)^-l = a[(6fe-l)^-l]. IfwetakeX = 6K-l,...(3.41) then we get Y = 6m-1,...(3.42) Y^ =ax^-(a-l)^ax^~n,wheren = a-\....(3.43) Y'-aX'=-N...(3.44) The general solutions of the equation (3.44) are given by Y^=Yoy + axox,where Yo+^faX^ is the fundamental solution of (3.44), yo+^faxo is the fundamental solution of the Pellian Y'-aX'=\....(3.45)

54 The infinite number of solutions of (3.45) are given by y +4^x =(>'o+^ov«r',«= 0,1,2...(3.46) Further we have, y -^x ^{y^-x^4^y\n = 0,1,2...(3.47) Adding and subtracting the above equations (3.46) and (3.47) we get yn =-[{yo+4alxqy^^+{yo-4axqy^'^],...(3.48) x«= j=[{yo +^[axqy^'^-{yo -Vaxo)"^']....(3.49) zvoc Using (3.48), (3.49) in the general solutions of (3.44), we get x =^[(yo +4^x,r' +{yo -V^^o)""'] +^[0o +4^x,r' -{y, ~4^x,r'] ^ 2^ a = -^[(yo+^x,y^\x ^ + Y,) + {y, -4^x,r\X,4^-Y,)] zvor Yn = vt(^o +4^xx'+{yo -^<r']...(3.50) +^[u+v^xor' -u -V^xor'] = -[{yo + 4ax,y'\Xo4a + Yo) + {y.-^xoy*\yo-xo4a)\...(3.51) In view of (3.41) and (3.42), the ranks of the pentagonal numbers are obtained as m= [iyo +Vaxb)'"'(ZoVa+ro) + 0;o -Vaxo)'"'(K -Xo4d)+2\,...(3.52) k = -^[{y^+4^xx\x,4^ + Y,^ + {y,-4^x,r\x,^-y,)+2^], 12Va where n = 0,1,2......(3.53)

Some particular cases 55 When a = 2, the equation (3.44) leads to the Pell's equation Y2-2X2 = -1. In view of (3.52) and (3.53), the ranks of the pentagonal numbers are given by m = j-[(3 + 2V2)"-'Hl + V2) f (3 -l^l)"^^(1 - V2) +1], ^ =-^[(3 + 272)""^'(1 + V2) + (3-2V2)"+VV2-1) + 2V2]. The values of ranks m and k will be in integers only when n = 4r+1, r = 0, 1, 2,3... s. No Some examples are presented in the table 3.4. Values of n m^ k Table 3.4 Pentagonal number Pentagonal number Pk 1 1 7 5 70 35 2 5 7887 5577 93303210 46651605 When a = 3, the equation (3.44) leads to the Pell's equation Y2-3X2 = -2. In view of (3.52) and (3.53), the ranks of the pentagonal numbers are given by - [(2 + V3 r+^s + 3 V3) + (2 - V3 r+1 (5-3 V3) + 2], m- 12 k = - ^ f(2 + Sf^ (5 + 3A/3) + (2 - V3)"^^ (3 V3-5) + 2^31. 12V3 ^

56 The values of ranks m and k will be in integers only when n = 6r+ l,r = 0, 1,2... Some examples are presented in the table 3.5. Table 3.5 s. No Values ofn m k Pentagonal number Pm Pentagonal number Pk 1 1 12 7 210 70 2 5 31977 18462 1533776805 511258935 When a = 5, the equation (3.44) leads to the Pell's equation Y2 _ 5X2 = -4. In view of (3.52) and (3.53), the ranks of the pentagonal numbers are obtained by! = [(9 + 4 V5)"'' (1 + S) + (9-4 V5 )"^' (1 - Vs) + 2], m- 12' k = y [(9 + 4A/5)"+^ (1 + VS) + (9-4VS )"+^ (VS -1) + 2^5 ]. The values of ranks m and k will be in integers only when n = (4r+ l),r = 0, 1,2... s. No. Some examples are presented in the table 3.6. Values ofn m k Table 3.6 Pentagonal number Pm Pentagonal number Pk 1 1 87 39 11310 2262 2 5 9003087 4026303 121583358792810 24316671758562

Observations A few interesting relations among pentagonal numbers 57 It presents below a few interesting relations among pentagonal numbers. Pm and Ts are denoting the pentagonal and the triangular numbers. First derives the recurrence relation of pentagonal numbers which satisfy the following recurrence relations. Recurrence relations 3.1 m+1 m~i m Proof By the definition of the pentagonal number, Z'. =f(3/«-l)....(3.54) Thus?,_,=-^^[3(m-l)-l] = i!!izl2[3w_4],...(3.55) = (!!L±l)[3; + 3-l] 2 From the equation (3.55), we get _ (w +1) [3m+ 2]...(3.56) 2 m{3m-\) (3m-V) 3 -(m-1) 2 2 2 37«I 3 3 " 2 2 2 2 = P -3W + 2....(3.57)

From the equation (3.56), we get 58 m(3m-l) 1 ^,,3 2 2 ^ 2 D 3 13 3 = P + m + m + 2 2 2 2 = P +3m + l...(3.58) Adding the equations (3.57), (3.58), we get P,r+P -,=P.-3m + 2 + P +3m + \ = 2P +3 thatis,p +?, _, -2P, =3....(3.59) This is the recurrence relation for the pentagonal numbers. Example 3.1 If we take m = 2, then T^ =l,the first pentagonal number, Tj =5, the second pentagonal number, T^ =12, the third pentagonal number. Thus P,,+P _,-2P =12+ 1-2(5) = 13-10 = 3. Hence the equation (3.59) is satisfied. Identity 3.1 The sum of the pentagonal numbers can be expressed as the polynomial in s. S 1 that is, y]p = s^(s + V) = st^,where T^ is the s"' triangular number. Proof The m"" pentagonal number is given by Hence X^.=i;[v(3«^-^)]

59 1 ' -^ m=l m=l - Ir3^(5 + 1)(25 + 1) ^(5 + 1) ~ 2 6 2 ^ = -5(5 + l)[25fl-l] 4 2 sis + D = 5[-^^ -^ = sts, where T^ is the s"" triangular number. Identity 3.2 Rencc±P =s'^^^st,. m=1 -^ The sum of squares of the pentagonal numbers can be expressed as the polynomial in s. That is, ^P '=^^^^^(27/+18^r^-135-2). m=l 60 triangular number. Proof = ^(27.y^+18^^-13^-2), where T, is the s"" 30 The m"" pentagonal number is given by Hence P '=[ (3^-1)]^ =-[m\3m-iy]. = -[w'(9m'-6w + l)]. 4

60 9 4 3 3 1 2 = m m + m. 4 2 4 m=\ ^ m=l ^ m=\ ^ m=l 4 ' ^' ^' 2^ 4 ^ 4 6 = -5(5 + l)[-(65' +95' +5-l)-3s(5 + l) + -(25 + l)] 8 5 3 = '^^'^^^^[54^^ + Sb' + 9^ -9-455' - 455 +10^ + 5]. 120 ^ ^ _ ^(^ + l)rc^ 3, o^ 2 [54;y'+365'-26^-4]. 120 = [27^^+185^-13^-2]. 30 -* Thus, ZP.'=^^^^^(27/+18/-135-2), 60 = (275^+185^-135-2). 30 Identity 3.3 The sum of cubes of the pentagonal numbers can be expressed as the polynomial in s. That is tpj = ^^^il^[1355'+1805"-1175'-1285'+585+ 12].»-=' 280 ^' [I355'+180/-1175'-1285'+585 + 12], where 7^ is the 140 5"' triangular number.

61 Proof The w"" pentagonal number is given by Hence pj=^(2m-iy. O = (27w' +9m-27m^ -1). 8 ^ = -(27m^ +9m^ -21m^ -m^) 8 Thus ±Pj^U21±m'-21±m^+9±m'-±m']. m=l O m=l m=i m=l m=l ^^ y(.y + l)(6^^+15/+6.y^-6.y^-^ + l) ^^(^ + 1)^(2^^+2^-1) 42 12 9.y(5 +1)(65^ +9s^+s-l) ^^ (^ +1)^ ^ 3 0 4 = ^^^^[90(65^ +I5s'+ 6s' -6s' -s + l)-3l5s(s + l)i2s' +2s-l) 1120 " +42(6s' +9/ +5-l)-355(5+l)]. _4s + l) [540^^ + 7205^ - 4685' - 512^' + 2325 + 48]. 1120 ^ ^4S + 1)L^^^5 ^jgq^4 _jj^^3 _i285^ +585 + 12]. 280 ^ [1355'+1805^-1175'-1285'+585 + 12] where Ts is the s" 140 triangular number.

62 Hence tpj = ^^^^^[1355^+ 1805^-117/-128^'+ 58^ + 121. -"= 280 ^ th Ts IS the s -^' [135/+1805^-1175'-1285'+58^+ 12], where 140 triangular number. Identity 3.4 Pm = a square + (m -1) ^ triangular number. Proof The m' pentagonal number is given by m Pm= {3>m-\). 2 3m m 2 2 2 m m = m +. 2 2 = m H (m-~l)= a square +(m- V)"" triangular number. It has seen that the explicit formulas for the ranks of pentagonal numbers ' which are simultaneously equal to hexagonal, heptagonal and octagonal numbers. One may search for such hexagonal, heptagonal and octagonal numbers. Further, it has observed that there exist pairs of pentagonal numbers whose ratios are non-square integers. One may search for such numbers whose ratios are non-square integers.