Krzysztof Ziemak - cp's OEIS frontend

User: Krzysztof Ziemak

Krzysztof Ziemak has authored 4 sequences.

A298365 Numbers k such that there exists at least one odd pseudoprime of order k.

10, 11, 14, 15, 16, 18, 20, 21, 22, 23, 24, 25, 26, 28, 29, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 97

Offset: 1

Krzysztof Ziemak, Jan 17 2018

nonn

A composite divisor d of M(m) := 2^m - 1 is called primitive if M(k) != 0 for any k < m.
A primitive composite divisor d of M(m) is said to have rank m, and we write rank(d)=m.
Let M(m)=2^m-1, and define D to be the set of all numbers d such that d|M(m), d==1 (mod m), and rank(d)=m. Then each element d from D is an odd pseudoprime, because if m|d-1, then M(m)|M(d-1) and thus d|M(d-1). The set D contains all composite and primitive divisors d|M(m) that have rank(d)=m and each odd pseudoprime d with rank(d)=m generates only one class [a(n)] with all pseudoprimes d, where a(n)=m, if a(n) is defined as below. See attached file with examples of pseudoprimes.

			10 is a term since 341 is an odd pseudoprime whose order is 10.

Krzysztof Ziemak, First 172 class [a(n)] of odd pseudoprime numbers
Krzysztof Ziemak, PARI code for generation sequence a(n)

Cf. A001567, A086249.

a(n) = min{k: k>a(n-1) and M(k) has a composite divisor d and rank(d)=k and d==1 (mod k)} for n = 1,2,3,... where M(k):=2^k-1.

A298299 a(n) is the smallest b > 0 such that b^(2n) + 1 has all prime divisors p == 1 (mod 2n).

Original entry on oeis.org

2, 2, 6, 2, 10, 6, 14, 2, 36, 20, 66, 18, 26, 28, 120, 2, 170, 6, 570, 140, 2184, 88, 184, 42, 110, 312, 1440, 42, 116, 9060, 124, 2, 7656, 34, 13650, 132, 74, 228, 7800, 40, 1066, 4158, 430, 132, 6283590, 46, 94, 12, 1246, 1960

Offset: 1

Krzysztof Ziemak and Thomas Ordowski, Jan 16 2018

All the terms are even.
The number a(n)^(2n) + 1 has all divisors d == 1 (mod 2n).
Conjecture: a(n) exists for every n. This is implied by the generalized Bunyakovsky conjecture (Schinzel's hypothesis H).
Theorem: a(n) = 2 if and only if n is a power of 2.
Note: rad(2n) divides rad(a(n)), where rad(m) = A007947(m).
Even numbers 2n such that a(n) = rad(2n) are powers of two and 6, 10, 12, 14, 26, 36, ... Are there infinitely many such numbers?
We have a(n) = 2n = 2, 6, 10, 14, 20, 26, 28, ...
Problem: are there infinitely many even numbers m <> 2^k such that the number m^m + 1 has all divisors d == 1 (mod m)?
From Kevin P. Thompson, Mar 13 2022: (Start)
Additional terms: a(46) = 46, a(47) = 94, a(48) = 12, a(49) = 1246, a(50) = 1960, a(52) = 208, a(53) = 636, a(55) = 17600, a(56) = 476.
a(45) > 1000000 (sequence A298398 likewise has a very large value for n=45).
a(51) >= 16524 (a C241 remains to be factored to verify b=16524).
a(54) >= 6864 (a C201 remains to be factored to verify b=6864).
(End)

			a(5) = 10, because 10^10 + 1 = 10000000001 = 101*3541*27961 and all the prime factors p == 1 (mod 2*5), so all divisors d == 1 (mod 10).

Kevin P. Thompson, Factorizations to support known terms of a(n) for n = 1..56

Cf. A007947, A298076 (see PARI subroutines used for a(48)), A298398.

find_a_ORDOWSKI2(n=2, a=1, B_START=2, LIM=10^11,DEBUG=1)={
  my(B,FF,LL);
  my(fn="_THOMAS_ORDOWSKI_b_a_n.txt");
  LL=R2('b,a,n);   \\ R(b,a,n)=(b^n+a)
  FF=factor(LL);
  if(DEBUG==1,
    print(FF);
    print(LL);
  );
  if(Mod(n,2)==0,  \\ n-EVEN
    B=FIND_BASE(n,BSTART=B_START,LIM,STEP=2,FF);
  );
  if(B>0,
     return([n,B,[subst(FF,'b,B)]]);
  );
  return(0);
}

a(n) = min{b > 1: for all prime p, if p | (b^(2n) + 1) then p == 1 (mod 2n)}.

a(31)-a(44) from Kevin P. Thompson, Mar 13 2022

a(45)-a(50) from Daniel Suteu, Jul 01 2022

A298310 Least k > 1 such that all divisors d of (k^(2n+1)+1)/(k+1) satisfy d == 1 (mod 2n+1).

Original entry on oeis.org

2, 3, 2, 2, 9, 2, 2, 15, 2, 2, 15, 2, 32, 81, 2, 2, 55, 21, 2, 39, 2, 2, 4141, 2, 18, 51, 2, 551, 39, 2, 2, 21267, 21, 2, 1012, 2, 2, 826, 330, 2, 729, 2, 136, 204, 2, 3, 280, 20, 2

Offset: 0

Thomas Ordowski and Krzysztof Ziemak, Jan 17 2018

a(n) is the smallest k > 1 such that Phi_m(-k) has all its divisors == 1 (mod n) for all m > 1 dividing 2n+1, where Phi_m(x) are the cyclotomic polynomials.
By Schinzel's hypothesis H, a(n) exists for every n (see A298076).
If 2n+1 is a prime > 3, then a(n) = 2.
We have a(n)^(2n+1) == a(n) (mod 2n+1), so every composite number 2n+1 is a weak Fermat pseudoprime to base a(n).
a(n) >= A239452(2n+1).
a(42) requires factorization of a 132 digit composite. - M. F. Hasler, Oct 16 2018
From Kevin P. Thompson, Mar 30 2022: (Start)
Additional nontrivial terms: a(55) = 111, a(61) = 165, a(64) = 216, a(66) = 49.
a(49) >= 656811 (a C322 remains to be factored to verify k=656811).
a(52) >= 3547020 (a C288 remains to be factored to verify k=3547020).
a(57) >= 4900 (a C258 remains to be factored to verify k=4900).
a(58) > 784720.
a(59) >= 714 (a C299 remains to be factored to verify k=714).
a(60) >= 233 (a C240 remains to be factored to verify k=233).
a(62) >= 126 (a C191 remains to be factored to verify k=126). (End)

			a(170) = 2 wherein 2*170 + 1 = 341 = 11*31 is the smallest psp(2).
From _M. F. Hasler_, Oct 15 2018: (Start)
a(0) = 2 is the least integer k > 1 for which (k+1)/(k+1) == 1 (mod 1). (Here we even have equality, but any integer is congruent to any other integer, modulo 1.)
a(1) = 3 is the least k > 1 for which (k^3+1)/(k+1) = k^2 - k + 1 = P3(-k) == 1 (mod 3). Indeed, P3(-3) = 7 == 1 (mod 3), while P3(-2) = 3 == 0 (mod 3). (End)

Kevin P. Thompson, Factorizations to support known terms of a(n) for n = 1..68

Cf. A239452, A298076 (see Ordowski's conjecture for b < -1 and odd n).

Table[SelectFirst[Range[2, 100], AllTrue[Divisors[(#^(2 n + 1) + 1)/(# + 1)], Mod[#, 2 n + 1] == 1 &] &], {n, 21}] (* Michael De Vlieger, Feb 01 2018 *)

isok(k, n) = {fordiv((k^(2*n+1)+1)/(k+1), d, if (Mod(d, (2*n+1)) != 1, return (0));); return(1);}
a(n) = {my(k = 2); while (!isok(k, n), k++); k;} \\ Michel Marcus, Jan 19 2018

A298310(n)={n=n*2+1;for(k=2,oo,fordiv(n,m,m>1&&vecmax(factor(polcyclo(m,-k))[,1]%n)!=1&& next(2));return(k))} \\ M. F. Hasler, Oct 14 2018

a(n) = min{k > 1: for all prime p, if p | (k^(2n+1)+1)/(k+1) then p == 1 (mod 2n+1)}. - Kevin P. Thompson, Mar 18 2022

a(22) corrected by Robert Israel, Jan 18 2018
a(1) corrected by Michel Marcus, Jan 19 2018
a(27)-a(30) from Robert Price, Feb 17 2018
a(31)-a(41) from M. F. Hasler, Oct 15 2018
a(42)-a(48) from Kevin P. Thompson, Mar 18 2022

A296104 Numbers k such that 2^k == 3 (mod k-1).

Original entry on oeis.org

2, 111482, 465794, 79036178, 1781269903308, 250369632905748, 708229497085910, 15673900819204068

Offset: 1

Krzysztof Ziemak and Max Alekseyev, Dec 04 2017

Also, numbers k such that 2^k - 2 is a Fermat pseudoprime, i.e., 2^k - 2 belongs to A015919 and A006935.
a(3) was found by McDaniel (1989).
Some larger terms (maybe not in order): 2338990834231272653582, 341569682872976768698011746141903924998969680638.
Discovered huge even PSP(2) numbers of the form 2*M(n), where n=p*q and M(n)=2^n-1, ensure that the following numbers are also even pseudoprimes of the form 2*M(p)*M(q): 2*M(37)*M(12589), 2*M(131)*M(17854891864360859951), 2*M(179)*M(1398713032993), 2*M(2111)*M(335494787819), 2*M(35267)*M(50508121). - Krzysztof Ziemak, Jan 01 2018

W. L. McDaniel, Some Pseudoprimes and Related Numbers Having Special Forms, Math. Comp. 53:187 (1989), 407-409.

Cf. A000079, A015919, A006935, A014741, A050259, A055685, A296370.

k = 2; lst = {2}; While[k < 1000000001, If[ PowerMod[2, k, k -1] == 3, AppendTo[lst, k]]; k += 10; If[ PowerMod[2, k, k -1] == 3, AppendTo[lst, k]]; k += 2]; lst (* Robert G. Wilson v, Jan 01 2018 *)

is_A296104(n) = Mod(2, n-1)^n == 3; \\ Iain Fox, Dec 07 2017

A296104_list = [n for n in range(2,10**6) if pow(2,n,n-1) == 3 % (n-1)] # Chai Wah Wu, Dec 06 2017

a(n) = A296370(n) + 1.

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User: Krzysztof Ziemak

A298365 Numbers k such that there exists at least one odd pseudoprime of order k.

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A298299 a(n) is the smallest b > 0 such that b^(2n) + 1 has all prime divisors p == 1 (mod 2n).

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A298310 Least k > 1 such that all divisors d of (k^(2n+1)+1)/(k+1) satisfy d == 1 (mod 2n+1).

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A296104 Numbers k such that 2^k == 3 (mod k-1).

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