Paul Underwood (Another of the Prime Pages' resources) View this page in: GIMPS has discovered a new largest known prime number: 282589933-1 (24,862,048 digits) A titan, as defined by Samuel Yates, is anyone who has found a titanic prime. This page provides data on those that have found these primes. The data below only reflect on the primes currently on the list. (Many of the terms that are used here are explained on another page.)

 Proof-code(s): p6, x43, p98, p102, p62, L97, L158, L256, p308, p367, p372, c70, c73, c76, c81, c84, c89, p406 E-mail address: paulunderwood@mindless.com Web page: http://arxiv.org/abs/1706.01265 Username: Underwood (entry created on 01/18/2000) Database id: 181 (entry last modified on 11/26/2019) Active primes: on current list: 15 (unweighted total: 25), rank by number 62 Total primes: number ever on any list: 2579.95 (unweighted total: 6133) Production score: for current list 46 (normalized: 15), total 46.8135, rank by score 315 Largest prime: 3 · 23136255 - 1 ‏(‎944108 digits) via code L256 on 03/08/2007 Most recent: primV(67, - 1, 13081)/65419672274940815357 ‏(‎23451 digits) via code c84 on 10/03/2019 Entrance Rank: mean 73457.44 (minimum 12, maximum 103514)

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 Probable Prime Tests: Trinomial Test Define F(a,m,r) = am-ar-1 where a,m,r in N a > 1 m > 2 m>r>0 except F(2,m,m-1) Conjecture: If aF=a modulo F then F is prime If aF=a modulo F then amF-arF-1 = 0 modulo F. The roots satisfy: sum(rootsk) = sum(rootsFk) modulo F for all k. (symmetric pseudoprime) For the specific example F(a,3,1) = a3-a-1 the roots satisfy: sum(rootsF) = 0 modulo F (Perrin pseudoprime.) What is the probability of a Perrin pseudoprime? What is the probability that composite n satisfies an=a modulo n? (Click here for answer) How many tests are expected to refute the conjecture? If you find a counterexample, please let me know. 8 Aug 2001: f=a3-a-1 tested for all a less than 1.4*10^9 with 15 Miller Rabin rounds. 17 Nov 2001: f=a3-a-1 proved for all a less than 10^8. 14 Mar 2002: f=a3+a-1 proved for all a less than 10^8. 20 Sep 2002: f=a3-a-1 tested for all a less than 10^10+2 with 5 Miller Rabin rounds (527,345,506 PrPs.) 01 Jan 2003: f=a3-a-1 tested for all a less than 10^11 with 5 Miller Rabin rounds (4,772,369,646 PrPs.) 01 Jan 2003: f=a3-a-1 tested for all a from 10^11 to 223,490,000,000 with 5 Miller Rabin rounds by Michael Angel. Quadratic Test 05 Jun 2005: f=a2-2 tested with 5 Miller-Rabin rounds for a base-a PSP ; none found for all odd a from 3 to 10^11 (3,809,286,968 PRPs.) 26 Apr 2006: f=a2-2 further tested by Carlos Eduardo to a=344,360,000,003 (12,480,999,468 PRPs.) 29 Mar 2015: tested with (L+2)^(f+1)==5 (mod f, L^2+1) for odd a < 10^13 (320,120,182,301 PRPs.) Unifying Test For integers a>1, s>=0, all r>0, all t>0, odd and irreducible {a^s\times\prod{(a^r-1)^t}}-1 is a-PRP, except for the cases a^2-a-1 and a-2 and a-1 and -1. FLT-type Conjecture There are no non-zero integer solutions to A*x^n+B*y^n=C*z^n where |A|+|B|+|C|<=n and x,y,z are distinct. 5-Selfridge Q=-5 and Q=5 Lucas Test N>5 and any P such that gcd(P,n)==1 and KroneckerSymbol(P^2-4*5,N)==-1 and KroneckerSymbol(P^2+4*5,N)==-1 and L^(N+1)==5 (mod N, L^2-P*L+5) and L^(N+1)==-5 (mod N, L^2-P*L-5). 6-Selfridge A-2 and A+2 Fermat-Lucas Test N>5 coprime to 30, any A such that JacobiSymbol((A-2)^2-4,N)==-1 and JacobiSymbol((A+2)^2-4,N)==-1, test (A-2)^N==A-2 (mod N) and (A+2)^N==A+2 (mod N) and L^(N+1)==1 (mod N, L^2-(A-2)*L+1) and L^(N+1)==1 (mod N, L^2-(A+2)*L+1). Verified for N < 2*10^8 and for Carmichael numbers < 2^32. 6-Selfridge A-1 and A+1 Fermat-Lucas Test N>5 coprime tp 210*A, any A such that JacobiSymbol((A-1)^2-4,N)==-1 and JacobiSymbol((A+1)^2-4,N)==-1, test (A-1)^N==A-1 (mod N) and (A+1)^N==A+1 (mod N) and L^(N+1)==1 (mod N, L^2-(A-1)*L+1) and L^(N+1)==1 (mod N, L^2-(A+1)*L+1). Verified for N < 2*10^8 and for Carmichael numbers < 2^32. Derived 5-Selfridge A-2 and A+2 Fermat-Lucas Test N>5 coprime to 30, find A such that JacobiSymbol((A-2)^2-4,N)==-1 and JacobiSymbol((A+2)^2-4,N)==-1, test 4^N==4 (mod N) L^(N+1)==1 (mod N, L^2-(A-2)*L+1) and L^(N+1)==1 (mod N, L^2-(A+2)*L+1). Counterexample: N=105809903; A=15164718 Derived 5-Selfridge A-1 and A+1 Fermat-Lucas Test N>5 coprime to 210*A find x such that JacobiSymbol((A-1)^2-4,N)==-1 and JacobiSymbol((A+1)^2-4,N)==-1, test 2^N==2 (mod N) and L^(N+1)==1 (mod N, L^2-(A-1)*L+1) and L^(N+1)==1 (mod N, L^2-(A+1)*L+1). Counterexample: N=2499327041; A=20003797 1st 2.X-Selfridge Composite Test Algorithm For N>5 coprime to 30, find the minimal integer x>0 where JacobiSymbol(x^2-4,N)==-1 and perform the probable prime test (x*L-3)^(N+1)==9-2*x^2 (mod N, L^2-x*L+1). Verified for N < 2.481*10^12. 2nd 2.X-Selfridge Composite Test Algorithm For N find minimal integer x>=0 where KroneckerSymbol(x^2-4,N)==-1 and perform the probable prime test (L+2)^(N+1)==2*x+5 (mod N, L^2-x*L+1). Verified for odd N < 2^50. L-2 and L+2 Test N>1, for any integer x such that KoneckerSymbol(x^2-4,N)==-1, test (L-2)^(N+1)==5-2*x (mod N, L^2-x*L+1) and (L+2)^(N+1)==5+2*x (mod N, L^2-x*L+1) Verified for odd N < 10^8. 5-Selfridge Fermat-Euler-Lucas Test N>5 coprime to 30, for any integer x: gcd(x^3-x,N)==1 and JacobiSymbol(x^2-4,N)==-1, test (x-2)^((N-1)/2)==JacobiSymbol(x-2,N) (mod N) (Euler) and (x+2)^((N-1)/2)==JacobiSymbol(x+2,N) (mod N) (Euler) and x^(N-1)==1 (mod N) (Fermat) and L^(N+1)==1 (mod N, L^2-x*L+1) (Lucas) Verified for N< 2.6*10^7 and Carmichael numbers < 2^32. Links to David Broadhurst's counterexamples: 1, 2, 3, 4, 5 and 6 Quartic Test for L+x^2-2 N coprime to 210, any x indivisible by n and JacobiSymbol(x^2-4,N)==-1 and gcd((x^3-x)*(x^2-2)*(x^2-3),N)==1, test (L+x^2-2)^N==-L^3+(x^2-2)*L+x^2-2 (mod N, (L^2-x*L+1)*(L^2+x*L+1)). Verified for N < 10^8. Quartic Test for L+x+1 Odd N>5, for any x such that N does not divide x and JacobiSymbol(x^2-4,N)==-1 and gcd(x^2+x,N)==1, test (L+x+1)^N==-L^3+(x^2-2)*L+x+1 (mod N, (L^2-x*L+1)*(L^2+x*L+1)). Verified for N < 10^8. Quartic Test for L+x^2-1 Odd N>7, for any x such that N does not divide x and JacobiSymbol(x^2-4,N)==-1 and gcd(x^3-x,N)==1, test (L+x^2-1)^N==-L^3+(x^2-2)*L+x^2-1 (mod N, (L^2-x*L+1)*(L^2+x*L+1)). Verified for N < 2.6*10^7. Counterexample: N=62164241, x=2290208. Quartic Test for L+x^2 Odd N>7, for any x such that N does not divide x and JacobiSymbol(x^2-4,N)==-1 and gcd(x^2-1,N)==1, test (L+x^2)^N==-L^3+(x^2-2)*L+x^2 (mod N, (L^2-x*L+1)*(L^2+x*L+1)). Verified for N < 10^8. x and 2*x or x^2 or x+2 Double Quadratic Test N coprime to 30, for any x such that JacobiSymbol(x^2-4,N)==-1 and gcd(x^3-x,N)==1, test (L+x)^(N+1)==1+2*x^2 (mod N, L^2-x*L+1) and (L+A)^(N+1)==1+A^2+x*A (mod N, L^2-x*L+1) where A = 2*x or x^2 or x+2. Verified for N < 10^8. L-1 and L+1 Test N odd, for any a such that gcd(a,N)==1 and JacobiSymbol((a+1)^2-4),N)==-1 and JacobiSymbol((a-1)^2-4,N)==-1, test (L-1)^(N+1)==1-a (Mod N, L^2-(a+1)*L+1) and (L+1)^(N+1)==1+a (Mod N, L^2-(a-1)*L+1). Verified for N < 2*10^8 and for Carmichael numbers < 2^32. Quad Test N coprime to 6, for any x, a and b such that gcd(a*b*x,N)==1 and gcd(a^2-b^2,N)==1 and JacobiSymbol(x^2-4,N)==-1, test (L+a)^(N+1)==1+a^2+x*a and (L-a)^(N+1)==1+a^2-x*a and (L+b)^(N+1)==1+b^2+x*b and (L-b)^(N+1)==1+b^2-x*b all (mod N, L^2-x*L+1). Verified for N < 1.4*10^5. David's counterexamples: 1, 2 and 3 [N,P=x,a,b] Fermat+Lucas+Frobenius Test N coprime to 30, for any x such that gcd(x^3-x,N)==1 and JacobiSymbol(x^2-4,N)==-1 test (2*x)^(N-1)==1 (mod N) and L^(N+1)==1 (mod N, L^2-x*L+1) and (L+x)^(N+1)==2*x^2+1 (mod N, L^2-x*L+1). Verified for N < 10^8. Double Fermat+Frobenius Test N coprime to 30, for any x and y such that gcd(x^3-x,N)==1 and JacobiSymbol(x^2-4,N)==-1 and gcd(y^3-y,N)==1 and JacobiSymbol(y^2-4,N)==-1 and gcd(x^2-y^2,n)==1 test (2*x)^(N-1)==1 (mod N) and (L+x)^(N+1)==2*x^2+1 (mod N, L^2-x*L+1) and (2*y)^(N-1)==1 (mod N) and (L+y)^(N+1)==2*y^2+1 (mod N, L^2-y*L+1) and Verified for N < 10^8. David's 104 counterexamples a^2+k Test For n=a^2+k coprime to 6 and n>42 where k>0 and a>1 and gcd(k^3-k,n*a)==1 find x such that gcd(x,n)==1 and jacobiSymbol(x^2-4,n)==-1 and test: (L+a)^(n+1)==a^2+1+a*x (mod n, L^2-x*L+1) and (L-a)^(n+1)==a^2+1-a*x (mod n, L^2-x*L+1). Verified for n< 1.3*10^6. Counterexample: [n, a, k, x]=[1432999, 641, 1022118, 381129] Q only Test For n coprime to 6 and n>5 find Q : gcd(Q+1,n)==1 and jacobiSymbol(Q,n)==-1 and jacobiSymbol(Q-4,n)==1 and jacobiSymbol(Q+4,n)==1 and test: Q^((n-1)/2)==-1 (mod n) and x^((n+1)/2)==-1 (mod n, x^2-(Q-2)*x+1) and x^((n+1)/2)==-jacobiSymbol(-1,n) (mod n, x^2+(Q+2)*x+1). Verified for n < 2.3*10^7 David's counterexamples: {[[1581223211, 1093877615], [13078382623, 6926732822], [21577464971, 6280410804], [33816027671, 12420727435], [56004757043, 39375517400], [65963127163, 38771733269], [89249862731, 63319378549], [92225622499, 91249250716], [97457318983, 76930744545]]} 2*x+1 and 2*x-1 Test For odd n, find a such that jacobiSymbol(a^2-4,n)==-1 and test (2*x+1)^(n+1)==5+2*a (mod n, x^2-a*x+1) and (2*x-1)^(n+1)==5-2*a (mod n, x^2-a*x+1). Double Parameter Test For n coprime to 30 find a and b: jacobiSymbol(a^2-4,n)==-1 and gcd(b^2-4,n)==1 and gcd(a^2-b^2,n)==1 and gcd((a^3-a)*(b^3-b),n)==1 and gcd(a^3-a,b^3-b)==6 and test: (b*x+1)^(n+1)==1+b^2+a*b (mod n, x^2-a*x+1) and (b*x-1)^(n+1)==1+b^2-a*b (mod n, x^2-a*x+1) David's counterexamples: {[[107752185411149, 83669262602538, 19154350998482], [159132028928489, 38034692864164, 56837927663062], [2544844174968929, 2391864938130562, 575637140842943], [1484926074920009, 1014048410273000, 421042641324018], [2333564831617769, 191389787226297, 210732564557270], [11149098312734969, 3776761935674103, 1265100831985174], [8353109377649009, 2445829862238038, 3451336874803033], [4627936813857689, 2353076244593330, 2940176822193232], [5099517882159749, 3276100765569083, 2684765653980526], [7308466463154149, 6633660182802507, 5895989769974006]]} x-Q and x+Q Test n coprime to 6, find Q: gcd(Q^2-1,n)==1 and jacobiSymbol(1-4*Q,n)==-1 and test (x-Q)^(n+1)==Q^2 and (x+Q)^(n+1)==(Q+1)^2-1 both (mod n, x^2-x+Q). Verified for n < 1.8*10^7. 5 Selfridge Q Test n coprime to 6, find Q: gcd(Q^2-1,n)==1 and jacobiSymbol(1-4*Q,n)==-1 and jacobiSymbol(Q,n)==-1 and test Q^((n-1)/2)==-1 (mod n) and x^((n+1)/2)==-1 (mod n, x^2-(1-2*Q)/Q*x+1) and x^((n+1)/2)==jacobiSymbol(2+Q,n) (mod n, x^2-(5-2*Q)/(2+Q)*x+1). Verified for n < 5*10^6. Counterexample: n=5073799 and Q=2484110. David's counterexamples: {[[76537243, 62685438], [156524443, 1148764], [486819343, 352255858], [724361923, 38713830], [1001493799, 859390320], [7445895763, 916298043], [15728383999, 4890179599]]} Complex test For n, test (2+x*I)^n == 2+x*I^((n+1)/2) (mod n, x^2+I) Claim: pseudoprimes exist only for n == 1 mod 8. Further complex test To test n, with integer k such that gcd(k,n)==1, test (2+x)^n == 2+x^(n%(2*k)) (mod n, x^k+1). Claim: pseudoprimes exist only for n == 1 mod 2*k. Base x+1 test To test n, with integer k (>3) such that gcd(k,n)==1 and n%(2*k)!=1 and n%(2*k)!=2*k-1 test the following: (x+1)^n == x^(n%(2*k))+1 (mod n, x^k+1) 2.X-Selfridge Composite Test Algorithm Revisited For N find integer 0<=x<=ceil(N^0.25) where KroneckerSymbol(x^2-4,N)==-1 and perform the probable prime test (L+2)^(N+1)==2*x+5 (mod N, L^2-x*L+1). Verified for odd N < 10^12. 8 Selfridge test For odd N>6 find a such that gcd(a^3+a,N)==1 and JacobiSymbol((a*I)^2-4,N)==-1 and perform the probable prime test: (x+1)^(N+1)==2+a*I mod(N, x^2-a*I*x+1) if N==1 mod 4; (x+1)^(N+1)==2-a*I*x mod(N, x^2-a*I*x+1) if N==3 mod 4 Verified for N < 10^6. (Note: if a is small this test is 8 selfridges.) Counterexample: [n,a]=[1037623, 134383] 2 Selfridge Baillie-PSW test For non-square odd N find minimal a>2 such that JacobiSymbol(a^2-4,N)==-1 and perform the probable prime test: (2*x)^((N+1)/2)==JacobiSymbol(2*(a+2),N)*2 mod(N, x^2-a*x+1) Verified for N < 2^64 6^r test For non-square N with gcd(210,N)==1 and let b=6 and a=6^r (any r), where gcd(a^3-a,N)==1 and JacobiSymbol(a^2-4,N)==-1. Then test: (b*x)^((N+1)/2)==b*JacobiSymbol(b*(a+2),N) (mod N, x^2-a*x+1) Verified for N < 1.4*10^13

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