>Let Mp = 2^p - 1 be a Mersenne prime, where p > 2. >Denote S[1] = 4 and S[k+1] = S[k]^2 - 2 for k >= 1. >Then S[p-2] == +- 2^((p+1)/2) mod Mp. >Predict which congruence occurs. Dear Peter and All, This is as far as I can go in Ubasic: p Result 3 + 5 + 7 - 13 + 17 - 19 - 31 + 61 + 89 - 107 - 127 + 521 - 607 - 1279 - 2203 + 2281 - 3217 - 4253 + The algebra suggests two values to consider 1) Consider q=((p+1)/2) mod n Taking the p pairwise where signs differ eliminates the following possible n: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27, 28,29,30,32,33,34,35,36,37,38,39,40,42,43,45,46,47,48,49,51,52,54,55,57, 58,59,60,61,63,64,66,67,72,73,74,75,77,78,80,81,84,86,87,89,91,96,99,103, 104,111,114,115,120,122,125,126,127,129,131,133,144,146,151,154,156,162, 169,177,178,182,183,185,189,192,193,197,203,208,211,222,225,230,231,240, 245,254,258,259,262,263,266,267,273,288,297,301,302,309,311,312,319,347, 353,359,364,366,370,375,378,399,462,493,507,515,518,524,526,531,534,546, 549,555,567,569,576,609,622,624,633,637,638,691,694,706,789,798,801,803, 841,933,986,1041,1048,1057,1059,1077,1092,1093,1098,1110,1125,1134,1138, 1139,1487,1545,1578,1593,1602,1606,1607,1823,1866,2073,2082,2117,2118, 2123 That first gap at 31 is interesting... Conjecture: take ((p+1)/2) mod 31 if in (0,2,3,7,16,17,19) then sign(S[p-2]) = + if in (4,9,10,13,14,20,23,25,28) then sign(S[p-2]) = - if in (1,5,6,8,11,12,15,18,21,22,24,26,27,29,30,31) then no data 2) Consider q=(p-2) mod n Taking the p pairwise where signs differ eliminates the following possible n: 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27, 28,29,30,32,33,34,35,36,37,38,39,40,42,43,44,45,46,47,48,49,50,51,52,54, 55,56,57,58,59,60,61,63,64,66,67,68,70,72,73,74,75,76,77,78,80,81,84,86, 87,89,90,91,92,94,96,98,99,102,103,104,108,110,111,114,115,116,118,120, 122,125,126,127,128,129,131,132,133,134,144,146,148,150,151,154,156,160, 162,168,169,172,174,177,178,182,183,185,189,192,193,197,198,203,206,208, 211,222,225,228,230,231,240,244,245,250,252,254,258,259,262,263,266,267, 273,288,292,297,301,302,308,309,311,312,319,324,338,347,353,354,356,359, 364,366,370,375,378,384,386,394,399,406,416,422,444,450,460,462,480,490, 493,507,508,515,516,518,524,526,531,532,534,546,549,555,567,569,576,594, 602,604,609,618,622,624,633,637,638,691,694,706,718,728,732,740,750,756, 789,798,801,803,841,924,933,986,1014,1030,1036,1041,1048,1052,1057,1059, 1062,1068,1077,1092,1093,1098,1110,1125,1134,1138,1139,1152,1218,1244, 1248,1266,1274,1276,1382,1388,1412,1487,1545,1578,1593,1596,1602,1606, 1607,1682,1823,1866,1972,2073,2082,2096,2114,2117,2118,2123,2154,2184, 2186,2196,2220,2250,2268,2276,2278,2974,3090,3156,3186,3204,3212,3214, 3646,3732,4146,4164,4234,4236,4246 Again a gap at n=31 Conjecture: take (p-2) mod 31 if in (0,1,3,4,11,28,29) then sign(S[p-2]) = + if in (5,6,12,15,16,17,22,23,25) then sign(S[p-2]) = - if in (2,7,8,9,10,13,14,18,19,20,21,24,26,27,30,31) then no data It's all a bit thin and arm-waving, but I would be interested to see if a continuation of the series confirms or denies either of these conjectures. Regards, Andy Steward Factorisations of generalised repunits at: <http://www.users.globalnet.co.uk/~aads/index.html> ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
