Alzhekeyev Ascar M - cp's OEIS frontend

A273618 Numbers m = 2*k+1 where k is odd with the property that 3^k mod m = 1 and k^k mod m = 1.

11, 59, 83, 107, 131, 179, 227, 251, 347, 419, 443, 467, 491, 563, 587, 659, 683, 827, 947, 971, 1019, 1091, 1163, 1187, 1259, 1283, 1307, 1427, 1451, 1499, 1523, 1571, 1619, 1667, 1787, 1811, 1907, 1931, 1979, 2003, 2027, 2099, 2243, 2267

Offset: 1

Alzhekeyev Ascar M, May 26 2016

nonn

All composites in this sequence are 2-pseudoprimes, see A001567, and strong pseudoprimes to base 2, A001262.
The subsequence of these composites begins: 143193768587, 440097066011, 1188059560451, 1392770336147, 1640446291859, 2526966350771, 3639120872171, 3989703695867, 4202422108523, ....
Perhaps this sequence contains all the terms of the sequence A107007 (except 3) or A168539.

			m=131; 131=2*65+1; 3^65 mod 131 = 1 and 65^65 mod 131 = 1.

Robert Israel, Table of n, a(n) for n = 1..10000

Subsequence of A176997.

Cf. A001262, A001567, A107007, A168539, A214151.

filter:= proc(n) local k;
  k:= (n-1)/2;
  3 &^ k mod n = 1 and k &^ k mod n = 1
end proc:
select(filter, [seq(i,i=3..3000, 4)]); # Robert Israel, Nov 28 2019

2#+1&/@Select[Range[1,1200,2],PowerMod[3,#,2#+1]==PowerMod[ #,#,2#+1] == 1&] (* Harvey P. Dale, May 05 2022 *)

A236373 Pseudoprimes to base 2 of the form 6p+1 such that 2^(p-1) == 1 (mod p).

Original entry on oeis.org

2047, 8388607, 140737488355327, 576460752303423487

Offset: 1

Alzhekeyev Ascar M, Jan 24 2014

The first four terms are A065341(1), A065341(2), A065341(7), A065341(9) and have the form 2^m-1. Are there terms not of this form?
Composite 2^n-1 belong to this sequence when n is in A005385 (e.g., 2^83-1, 2^167-1, etc.)
No other terms below 2^64. - Max Alekseyev, May 28 2014

			2047=6*341+1; 2^2046 == 1 (mod 2047); 2^340 == 1 (mod 341).

Subsequence of A001567.

a(4) from Max Alekseyev, May 28 2014

A214151 Numbers k from the set == 5 (mod 6) with the property that 3^((3*k-1)/2) == 3 (mod k) and 2^((k-1)/2) == (k-1) (mod k).

Original entry on oeis.org

11, 59, 83, 107, 131, 179, 227, 251, 347, 419, 443, 467, 491, 563, 587, 659, 683, 827, 947, 971, 1019, 1091, 1163, 1187, 1259, 1283, 1307, 1427, 1451, 1499, 1523, 1571, 1619, 1667, 1787, 1811, 1907, 1931, 1979, 2003, 2027, 2099, 2243, 2267

Offset: 1

Alzhekeyev Ascar M, Jul 05 2012

nonn

All composites in this sequence are 2-pseudoprimes, see A001567, and strong pseudoprimes to base 2, A001262.
The subsequence of these composites begins: 1441091, 3587553971, 4528686251, 23260036451, 47535120323, 61070250323, 90474845819, 143193768587, 162016315907, 173868807611, 180998962187, 238364070323, 285370693931, 298577370323, ...
Perhaps this sequence contains all the terms of the sequence A107007 or A168539.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Subsequence of A176997.

Cf. A003629, A006970, A175865.

isA214151 := proc(n)
    if (n mod 6 = 5) and modp(2 &^ ((n-1)/2),n)  = n-1 and modp(3 &^ ((3*n-1)/2),n)  = 3 then
        true;
    else
        false;
    end if;
end proc:
for n from 5 by 6 do
    if isA214151(n) then
        print(n) ;
    end if;
end do: # R. J. Mathar, Jul 20 2012

Select[Range[5,2500,6],PowerMod[3,(3#-1)/2,#]==3&&PowerMod[2,(#-1)/2,#] == #-1&] (* Harvey P. Dale, Mar 14 2022 *)

for(n=0, 200, b=6*n+5; if(Mod(3, b)^((3*b-1)/2)==3, if(Mod(2, b)^((b-1)/2)==b-1 , print1(b, ", "))));

A212876 Numbers of the form n^2+1 such that 3^(m+3)==9 (mod m) where m=n^4-1.

Original entry on oeis.org

5, 17, 37, 101, 257, 1297, 4357, 14401, 44101, 65537, 828101, 933157, 8122501, 8386817, 12362257, 41990401, 121220101, 157402117, 223502501, 318622501, 378146917, 506700101, 684345601, 702038017

Offset: 1

Alzhekeyev Ascar M, May 29 2012

nonn

All the known values of the sequence are prime (for a(n)<10^20). - Charles R Greathouse IV, Jun 07 2012
The values of n+1 are prime or composite 121, 11011, 108781, 170431...
All composite n+1 == 1 (mod 3) ???

			Let n = 10. Then m = n^4-1 = 9999. 3^10002 == 9 (mod 9999), so n^2+1 = 101 is a member of the sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..3556

v=List();for(n=2,1e6,m=n^4-1;if(Mod(3,m)^(m+3)==9, listput(v, n^2+1))); Vec(v) \\ Charles R Greathouse IV, May 29 2012

A212912 Numbers k such that 3^(m+3) == 9 (mod m) where m = (k-1)^2 - 1.

Original entry on oeis.org

3, 5, 7, 11, 17, 37, 47, 53, 67, 97, 101, 121, 211, 257, 367, 379, 457, 617, 911, 1091, 1237, 1297, 1361, 1549, 2003, 2557, 2851, 2897, 3517, 3733, 4201, 4357, 5209, 6481, 7621, 8461, 8647, 8689, 10253, 10457, 10631, 11953, 13729, 14401, 14951, 17431, 17837

Offset: 1

Alzhekeyev Ascar M, May 30 2012

nonn

Composites begin: 121, 108781, 155365, 267547, 2774521, 3166087, 3225601, 4907701, 8341201, 10712857, 11035921, 13216141, 17559829, 21708961, 29641921, 31116241, 31150351, ... are all composite terms congruent to 1 (mod 3)?

Join[{3},Select[Range[4,20000],PowerMod[3,(#-1)^2+2,(#-1)^2-1]==9&]] (* Harvey P. Dale, Dec 07 2019 *)

for(n=2, 1000, m=n^2-1; if(Mod(3, m)^(m+3)==9, print(n+1)));

More terms from Harvey P. Dale, Dec 07 2019

A182533 A symmetrical triangle. Read by rows: T(n,k) = 2*C(n-2,k-1) - C(n-2,k) - C(n-2,k-2), n >= 2, 0 <= k <= n, with T(0,0) = 0, T(1,0) = T(1,1) = 1.

Original entry on oeis.org

0, 1, 1, -1, 2, -1, -1, 1, 1, -1, -1, 0, 2, 0, -1, -1, -1, 2, 2, -1, -1, -1, -2, 1, 4, 1, -2, -1, -1, -3, -1, 5, 5, -1, -3, -1, -1, -4, -4, 4, 10, 4, -4, -4, -1, -1, -5, -8, 0, 14, 14, 0, -8, -5, -1, -1, -6, -13, -8, 14, 28, 14, -8, -13, -6, -1, -1, -7, -19, -21, 6, 42, 42, 6, -21, -19, -7, -1

Offset: 1

Alzhekeyev Ascar M, May 04 2012

			Triangle begins
   0;
   1,  1;
  -1,  2, -1;
  -1,  1,  1, -1;
  -1,  0,  2,  0, -1;
  -1, -1,  2,  2, -1, -1;
  -1, -2,  1,  4,  1, -2, -1;
  -1, -3, -1,  5,  5, -1, -3, -1;
  -1, -4, -4,  4, 10,  4, -4, -4, -1;

Muniru A Asiru, Table of n, a(n) for n = 1..5151 (rows n=1..100)
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Graphical representation

Cf. A000108, A120009, A137742.

T:=Concatenation([0,1,1],Flat(List([2..11],n->List([0..n],k->2*Binomial(n-2,k-1)-Binomial(n-2,k)-Binomial(n-2,k-2))))); # Muniru A Asiru, Nov 29 2018

Flatten[Join[{{0}, {1, 1}}, Table[2*Binomial[n - 2, k - 1] - Binomial[n - 2, k] - Binomial[n - 2, k - 2], {n, 2, 15}, {k, 0, n}]]] (* T. D. Noe, May 08 2012 *)

T(n,k) = 2*C(n-2,k-1) - C(n-2,k) - C(n-2,k-2).
Row sums = 0, for n >= 2.
T(2*n,n) = 2*A000108(n-1).
T(2*n+3,n) = -A120009(n).
T(n+10,3) = -A137742(n+2).
Sum_{k=0..n} T(n,k)*2^k = -(3^(n-2)) where n >= 2.
Sum_{k=0..n} T(n,k)*3^k = -(4^(n-1)) where n >= 2.
Sum_{k=0..n} T(n,k)*p^k = -((p+1)^(n-2))*((p-1)^2) where n >= 2.

A193447 a(n) = ((p - 2)! + p - 1)/(p*(p - 1)) where p is the n-th prime.

Original entry on oeis.org

3, 3299, 255877, 4807626353, 1040021719579, 100970241446066087, 13409937746820630739862069, 9507270961010432209186683871, 7757618593382991688938927430572972973, 12437732976339904486975781548721278876097561, 18522993694996570934756402022946152638511627907

Offset: 4

Alzhekeyev Ascar M, Jul 26 2011

nonn

Conjecture: for k >= 7, ((k - 2)! + k - 1)/(k*(k - 1)) is an integer iff k is prime.
Proof follows from Wilson's theorem. - Alois P. Heinz, Aug 07 2011
Note that a(1) = 1 is also an integer. - Jianing Song, Sep 17 2018

			a(4) = (5! + 6)/(7*6) = 126/42 = 3.
a(5) = (9! + 10)/(11*10) = 362890/110 = 3299.

Wikipedia, Wilson's theorem

Cf. A000040, A007619, A066161.

a(n) = my(p=prime(n)); ((p-2)!+p-1)/(p*(p-1)) \\ Jianing Song, Sep 17 2018

Name clarified by Jianing Song, Sep 17 2018

A190638 Numbers n such that with b=n(2n-1) two remainders x and y are defined via x = 2^(b-1) -1 mod b and y = (2n-1)^(b-1) - 1 mod b which satisfy x==y==0 (mod n) and y-x=n.

Original entry on oeis.org

5, 41, 257, 2309, 14621, 48821, 125429, 177269, 1595417, 5278001, 10596137, 15146069, 21523361, 63993929, 83629517, 111321257, 363526421, 375805589, 427518041, 446072909, 552010829, 752665649, 1980098177

Offset: 1

Alzhekeyev Ascar M, May 15 2011

nonn

The first 23 terms (at least) are primes.
Conjecture: All terms have the form 12*k+5.
The first composite numbers in the sequence are 2^80+1 and 2^512+1.
If we modify one of the conditions to y - x = 2*n, the sequence changes to 3, 7, 19, 31, 79, 139, 199, 211, 271, 283, 307, 331, 367, 379, 439, 499, 511, ...
or if we modify it to y - x = 64*n, the sequence becomes 89, 101, 197, 269, 317, 341, 461, 521, 569, 701, 821, 857, 881, 929, 1109, 1181, 1217, ...
There seem to be no solutions n if the condition is modified to any y - x <= 0.

			For n=41, b = 41*(2*41-1)=3321. So 2^3320 == 3199 (mod 3321) leads to x = 3199 - 1 = 3198 which satisfies x == 0 (mod 41), and 81^3320 == 3240 (mod 3321) leads to y = 3240 - 1 = 3239 which satisfies y == 0 (mod 41) and y - x = 41. Therefore n=41 is in the sequence.

isA190638 := proc(n) local b,x,y; b := n*(2*n-1) ; x := modp( 2 &^ (b-1),b) -1; y := modp( (2*n-1) &^ (b-1),b) -1; if y-x =n and modp(x,n) = 0 and modp(y,n) = 0 then true; else false; end if; end proc:
for n from 2 do if isA190638(n) then print(n); end if; end do: # R. J. Mathar, Jun 04 2011

A191485 Numbers n=k^2-k+1 such that 2^k == 1 (mod n).

Original entry on oeis.org

1, 3, 7, 73, 601, 8191, 262657, 8640661

Offset: 1

Alzhekeyev Ascar M, Jun 03 2011

nonn

The elements of this sequence are elements of the sequence A002061 (Central polygonal numbers).
The first composite number is 8640661 = 31 * 211 * 1321 (31 and 211 are elements of the sequence A002061).
No more terms up to 3773299855577673.

			k = 9;
n = k^2 - k + 1 = 81 - 9 + 1 = 73;
2^9 == 1 (mod 73).

for(k=1,10^9,n=k^2-k+1;if( lift(Mod(2,n)^k)==1,print1(n,", "))); /* Joerg Arndt, Jun 03 2011 */

A190213 Integers m such that m divides (2^m-2)^2 and (m-2)^((k-1)(1+k(m-1))) == 1 (mod k), where k = 2^m - 1.

Original entry on oeis.org

1, 3, 4, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839

Offset: 1

Alzhekeyev Ascar M, May 19 2011

Original definition: let k=2^n-1 and m=1+(k-1)*(n-1), x=m*k and define remainders a and b via 2^(x-1) == (a+1) (mod x) and m^(x-1) == (b+1) (mod x). If a == 0 (mod k) and b == 0 (mod k), n is in the sequence.
Conjecture: All odd entries are also Mersenne exponents (A000043): primes n such that 2^n-1 is prime.
Any exceptions to the conjecture are larger than 10^5. - Charles R Greathouse IV, Oct 03 2022

			For n=3, k=2^3-1=7, m=1+6*2=13, x=m*k=13*7=91, 2^(x-1)==(a+1) (mod x) with 2^90 == (63+1)(mod 91), fixes a=63. m^(x-1) == (b+1) (mod x) with 13^90 == (77+1) (mod 91) fixes b=77. The two conditions are satisfied: 63 == 0 (mod 7) and 77 == 0 (mod 7). Therefore n=3 is in the sequence.

A174265 is a subsequence.

Cf. A144671, A000043.

isA190213 := proc(n) local k,m,x,a,b ; k := 2^n-1 ; m := (k-1)*(n-1)+1 ; x := k*m ; a := modp( 2 &^ (x-1),x) -1 ; b := modp( m &^ (x-1),x) -1 ; return ( modp(a,k) = 0 and modp(b,k)=0 ) ; end proc:
for n from 2 do if isA190213(n) then printf("%d,\n",n); end if; end do; # avoids n=1 and undefined 0^0, R. J. Mathar, Jun 11 2011

okQ[n_] := Module[{k, m, x, a, b}, k = 2^n - 1; m = 1 + (k - 1)(n - 1); x = m k; a = PowerMod[2, x - 1, x] - 1; b = PowerMod[m, x - 1, x] - 1; Mod[a, k] == 0 && Mod[b, k] == 0];
Reap[For[n = 1, n < 10^4, n++, If[okQ[n], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 30 2019 *)

is(n)=my(k=2^n-1,m=(k-1)*(n-1)+1,e=m*k-1); Mod(2,k)^e==1 && Mod(m,k)^e==1 \\ Charles R Greathouse IV, Sep 16 2022

a(20)-a(23) from Jean-François Alcover, Oct 30 2019
a(24)-a(28) from Charles R Greathouse IV, Sep 16 2022
a(29) from Charles R Greathouse IV, Sep 29 2022
a(30)-a(33) from Bill McEachen, Jul 30 2024
Definition simplified by Max Alekseyev, Dec 04 2024

cp's OEIS Frontend

User: Alzhekeyev Ascar M

A273618 Numbers m = 2*k+1 where k is odd with the property that 3^k mod m = 1 and k^k mod m = 1.

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A236373 Pseudoprimes to base 2 of the form 6p+1 such that 2^(p-1) == 1 (mod p).

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A214151 Numbers k from the set == 5 (mod 6) with the property that 3^((3*k-1)/2) == 3 (mod k) and 2^((k-1)/2) == (k-1) (mod k).

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A212876 Numbers of the form n^2+1 such that 3^(m+3)==9 (mod m) where m=n^4-1.

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PARI

A212912 Numbers k such that 3^(m+3) == 9 (mod m) where m = (k-1)^2 - 1.

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Mathematica

PARI

Extensions

A182533 A symmetrical triangle. Read by rows: T(n,k) = 2*C(n-2,k-1) - C(n-2,k) - C(n-2,k-2), n >= 2, 0 <= k <= n, with T(0,0) = 0, T(1,0) = T(1,1) = 1.

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A193447 a(n) = ((p - 2)! + p - 1)/(p*(p - 1)) where p is the n-th prime.

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Extensions

A190638 Numbers n such that with b=n*(2n-1) two remainders x and y are defined via x = 2^(b-1) -1 mod b and y = (2*n-1)^(b-1) - 1 mod b which satisfy x==y==0 (mod n) and y-x=n.

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A190638 Numbers n such that with b=n(2n-1) two remainders x and y are defined via x = 2^(b-1) -1 mod b and y = (2n-1)^(b-1) - 1 mod b which satisfy x==y==0 (mod n) and y-x=n.

A190213 Integers m such that m divides (2^m-2)^2 and (m-2)^((k-1)(1+k(m-1))) == 1 (mod k), where k = 2^m - 1.