A001132 -id:A001132 - cp's OEIS frontend

A319040 Numbers k > 1 such that Pell(k) == 1 (mod k).

7, 17, 23, 31, 35, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127, 137, 151, 167, 169, 191, 193, 199, 223, 233, 239, 241, 257, 263, 271, 281, 311, 313, 337, 353, 359, 367, 383, 385, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487, 503, 521, 569, 577, 593, 599

Offset: 1

Jon E. Schoenfield, Sep 08 2018

nonn

It appears that most of the terms of this sequence are primes. The composite terms are 35, 169, 385, 899, 961, 1121, ... (A319042).

The primes in the sequence give A001132 (primes == +-1 (mod 8)), since for primes p we have Pell(p) == (2/p) (mod p) where (2/p) is the Legendre symbol. - Jianing Song, Sep 10 2018

			k = 7 is in the sequence since Pell(7) = 169 = 7 * 24 + 1 == 1 (mod 7).
k = 11 is not in the sequence: Pell(11) = 5741 = 11 * 522 - 1 !== 1 (mod 11).
k = 35 is in the sequence: Pell(35) = 8822750406821 = 35 * 252078583052 + 1 == 1 (mod 35).

Cf. A000129 (Pell numbers), A001132, A023173, A319041, A319042, A319043.

isA319040 := k -> simplify(2^(k-1)*hypergeom([1-k/2,(1-k)/2],[1-k],-1)) mod k = 1: A319040List := b -> select(isA319040, [$1..b]):
A319040List(600); # Peter Luschny, Sep 09 2018

Select[Range[500], Mod[Fibonacci[#, 2], #] == 1 &] (* Alonso del Arte, Sep 08 2018 *)

A366526 Prime powers (A246655) q such that 2 is a nonzero square in the finite field F_q.

Original entry on oeis.org

7, 9, 17, 23, 25, 31, 41, 47, 49, 71, 73, 79, 81, 89, 97, 103, 113, 121, 127, 137, 151, 167, 169, 191, 193, 199, 223, 233, 239, 241, 257, 263, 271, 281, 289, 311, 313, 337, 343, 353, 359, 361, 367, 383, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487, 503, 521, 529

Offset: 1

Jianing Song, Oct 12 2023

Prime powers q that are congruent to 1 or 7 modulo 8.
Odd prime powers q such that 2^((q-1)/2) = 1 in F_q.
Prime powers q such that x^2 - 2 splits into different linear factors in F_q[x].
Contains the powers of primes congruent to 1 or 7 modulo 8 and the even powers of primes congruent to 3 or 5 modulo 8.
Proposition 1: Suppose that q is not a power of 2, gcd(a,q) = 1, then a is a square in F_q if and only if the Jacobi symbol Jacobi(a,q) = 1.
Proof: a is a square if and only if a^((q-1)/2) == 1 (mod p). We have a^((q-1)/2) = (a^((p-1)/2))^((q-1)/(p-1)) == Jacobi(a,p)^((q-1)/(p-1)) (mod p). Write q = p^e, then by definition, we have Jacobi(a,q) = Jacobi(a,p)^e, so it remains to prove that (q-1)/(p-1) - e = Sum^{e-1}_{i=0} (p^i - 1) is always even, which is obvious.
A trivial corollary would be that if q is a square, then every integer a coprime to q is always a square in F_q (since Jacobi(a,q) = 1 in this case). Indeed, since F_q is the unique quadratic extension of F_{sqrt(q)}, every quadratic polynomial with coefficients in F_{sqrt(q)} splits in F_q.
Proposition 2: Suppose that a == 1 (mod 4), gcd(a,q) = 1, then x^2 - x - (a-1)/4 splits into different linear factors in F_q[x] if and only if Jacobi(q,a) = 1 (or Kronecker(a,q) = 1).
Proof: Proposition 1 deals with the case where q is odd. For even q, we have x^2 - x - (a-1)/4 = x^2 + x + 1, which is reducible over F_q[x] if and only if q is an even power of 2.

			9 is a term since 2 = -1 = (+-i)^2 in F_9 = F_3(i).

Jianing Song, Table of n, a(n) for n = 1..10000

Supersequence of A001132.

Prime powers q such that a is a nonzero square in F_q: A365082 (q=-2), A085759 (q=-1), this sequence (q=2), A365313 (q=3).

isA366526(n) = isprimepower(n) && (n%8==1 || n%8==7)

A035200 Coefficients in expansion of Dirichlet series Product_p (1-(Kronecker(m,p)+1)p^(-s)+Kronecker(m,p)p^(-2s))^(-1) for m = 18.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 2, 1, 1, 0, 0, 1, 0, 2, 0, 1, 2, 1, 0, 0, 2, 0, 2, 1, 1, 0, 1, 2, 0, 0, 2, 1, 0, 2, 0, 1, 0, 0, 0, 0, 2, 2, 0, 0, 0, 2, 2, 1, 3, 1, 2, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 2, 2, 1, 0, 0, 0, 2, 2, 0, 2, 1, 2, 0, 1, 0, 0, 0, 2, 0, 1

Offset: 1

N. J. A. Sloane

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001132, A003629.

a[n_] := DivisorSum[n, KroneckerSymbol[18, #] &]; Array[a, 100] (* Amiram Eldar, Nov 19 2023 *)

my(m = 18); direuler(p=2,101,1/(1-(kronecker(m,p)*(X-X^2))-X))

a(n) = sumdiv(n, d, kronecker(18, d)); \\ Amiram Eldar, Nov 19 2023

From Amiram Eldar, Nov 19 2023: (Start)
a(n) = Sum_{d|n} Kronecker(18, d).
Multiplicative with a(p^e) = 1 if Kronecker(18, p) = 0 (p = 2 or 3), a(p^e) = (1+(-1)^e)/2 if Kronecker(18, p) = -1 (p is in A003629 \ {3}), and a(p^e) = e+1 if Kronecker(18, p) = 1 (p is in A001132).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 4*log(sqrt(2)+1)/(3*sqrt(2)) = 0.830966986853... . (End)

A097958 Primes p such that p divides 6^((p-1)/2) - 3^((p-1)/2).

Original entry on oeis.org

3, 7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127, 137, 151, 167, 191, 193, 199, 223, 233, 239, 241, 257, 263, 271, 281, 311, 313, 337, 353, 359, 367, 383, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487, 503, 521, 569, 577, 593, 599, 601, 607, 617

Offset: 1

Cino Hilliard, Sep 06 2004

Apart from the first term, the same as A001132 or A038873. - Jianing Song, Apr 21 2022

Jianing Song, Table of n, a(n) for n = 1..10000 (terms 1..998 from Harvey P. Dale)

Cf. A001132, A038873.

Select[Prime[Range[150]],Divisible[6^((#-1)/2)-3^((#-1)/2),#]&] (* Harvey P. Dale, Dec 25 2021 *)

\s = +-1,d=diff ptopm1d2(n,x,d,s) = { forprime(p=3,n,p2=(p-1)/2; y=x^p2 + s*(x-d)^p2; if(y%p==0,print1(p","))) }

isA097958(p) = (p==3) || (isprime(p) && kronecker(p,2)==1) \\ Jianing Song, Apr 21 2022

Equals {3} union A001132. - Jianing Song, Apr 21 2022

Definition corrected by Cino Hilliard, Nov 10 2008
Definition clarified by Harvey P. Dale, Dec 25 2021
Offset corrected by Jianing Song, Apr 21 2022

A201916 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+2737)^2 = y^2.

Original entry on oeis.org

0, 75, 203, 323, 552, 708, 1020, 1127, 1311, 1428, 1608, 1820, 1955, 2336, 2675, 3128, 3311, 3627, 3927, 4140, 4508, 4743, 5535, 6003, 6800, 7280, 7848, 8211, 8588, 9240, 9860, 11063, 11895, 13583, 14168, 15180, 15827, 16827, 18011, 18768, 20915, 22836

Offset: 1

T. D. Noe, Feb 09 2012

Note that 2737 = 7 * 17 * 23, the product of the first three distinct primes in A058529 (and A001132) and hence the smallest such number. This sequence satisfies a linear difference equation of order 55 whose 55 initial terms can be found by running the Mathematica program.
There are many sequences like this one. What determines the order of the linear difference equation? All primes p have order 7. For those p, it appears that p^2 has order 11, p^3 order 15, and p^i order 3+4*i. It appears that for semiprimes p*q (with p > q), the order is 19. What is the next term of the sequence beginning 3, 7, 19, 55, 163? This could be sequence A052919, which is 1 + 2*3^f, where f is the number of primes.
The crossref list is thought to be complete up to Feb 14 2012.

Colin Barker, Table of n, a(n) for n = 1..1000

Cf. A001652 (1), A076296 (7), A118120 (17), A118337 (23), A118674 (31).
Cf. A129288 (41), A118675 (47), A118554 (49), A118673 (71), A129289 (73).
Cf. A118676 (79), A129298 (89), A129836 (97), A157119 (103), A161478 (113).
Cf. A129837 (119), A129992 (127), A129544 (137), A161482 (151).
Cf. A206426 (161), A130608 (167), A161486 (191), A185394 (193).
Cf. A129993 (199), A198294 (217), A130609 (223), A129625 (233).
Cf. A204765 (239), A129991 (241), A207058 (263), A129626 (281).
Cf. A205644 (287), A207059 (289), A129640 (313), A205672 (329).
Cf. A129999 (337), A118611 (343), A130610 (359), A207060 (401).
Cf. A129641 (409), A207061 (433), A130645 (439), A130004 (449).
Cf. A129642 (457), A129725 (521), A101152 (569), A130005 (577).
Cf. A207075 (479), A207076 (487), A207077 (497), A207078 (511).
Cf. A111258 (601), A115135 (617), A130013 (647), A130646 (727).
Cf. A122694 (761), A123654 (809), A129010 (833), A130647 (839).
Cf. A129857 (857), A130014 (881), A129974 (937), A129975 (953).
Cf. A130017 (967), A118630 (2401), A118576 (16807).

d = 2737; terms = 100; t = Select[Range[0, 55000], IntegerQ[Sqrt[#^2 + (#+d)^2]] &]; Do[AppendTo[t, t[[-1]] + 6*t[[-27]] - 6*t[[-28]] - t[[-54]] + t[[-55]]], {terms-55}]; t

a(n) = a(n-1) + 6*a(n-27) - 6*a(n-28) - a(n-54) + a(n-55), where the 55 initial terms can be computed using the Mathematica program.

G.f.: x^2*(73*x^53 +116*x^52 +100*x^51 +171*x^50 +104*x^49 +184*x^48 +57*x^47 +92*x^46 +55*x^45 +80*x^44 +88*x^43 +53*x^42 +139*x^41 +113*x^40 +139*x^39 +53*x^38 +88*x^37 +80*x^36 +55*x^35 +92*x^34 +57*x^33 +184*x^32 +104*x^31 +171*x^30 +100*x^29 +116*x^28 +73*x^27 -363*x^26 -568*x^25 -480*x^24 -797*x^23 -468*x^22 -792*x^21 -235*x^20 -368*x^19 -213*x^18 -300*x^17 -316*x^16 -183*x^15 -453*x^14 -339*x^13 -381*x^12 -135*x^11 -212*x^10 -180*x^9 -117*x^8 -184*x^7 -107*x^6 -312*x^5 -156*x^4 -229*x^3 -120*x^2 -128*x -75) / ((x -1)*(x^54 -6*x^27 +1)). - Colin Barker, May 18 2015

A350964 a(n) is the largest prime factor of 2^p - p^2 where p is the n-th prime.

Original entry on oeis.org

7, 79, 47, 113, 130783, 523927, 1198297, 240641, 641, 575058377, 1519711993, 65929327, 20105355479017, 9007199254738183, 7633399, 33189241, 21081993227096629777, 951850902549409, 4978773308244222679, 501615233613780359, 9671406556917033397642519, 8251206137, 3818597055399121, 13314319257913, 521211122055087383048446607

Offset: 3

N. J. A. Sloane, Mar 02 2022

nonn

All prime factors of 2^p - p^2 are congruent to 1 or 7 (mod 8). (See A001132.) - Robert G. Wilson v, Mar 14 2022

E.-B. Escott, Note #1642, L'Intermédiaire des Mathématiciens, 8 (1901), page 12.

Amiram Eldar, Table of n, a(n) for n = 3..95
Robert G. Wilson v, Factorization of 2^p - p^2 for n = 3..120

Cf. A001132, A006530, A072180, A098105, A117587.

a:= n-> max(numtheory[factorset]((p-> 2^p-p^2)(ithprime(n)))):
seq(a(n), n=3..27);  # Alois P. Heinz, Mar 03 2022

a[n_] := FactorInteger[2^(p = Prime[n]) - p^2][[-1, 1]]; Array[a, 25, 3] (* Amiram Eldar, Mar 03 2022 *)

a(n) = my(p=prime(n)); vecmax(factor(2^p - p^2)[,1]); \\ Michel Marcus, Mar 03 2022

a(n) = A006530(A098105(n)). - Amiram Eldar, Mar 03 2022

A066507 Numbers k such that there is a solution to x^2 == 2^k (mod k).

Original entry on oeis.org

1, 2, 4, 6, 7, 8, 10, 12, 14, 16, 17, 18, 20, 22, 23, 24, 26, 28, 30, 31, 32, 34, 36, 38, 40, 41, 42, 44, 46, 47, 48, 49, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 71, 72, 73, 74, 76, 78, 79, 80, 82, 84, 86, 88, 89, 90, 92, 94, 96, 97, 98, 100, 102, 103, 104, 106, 108, 110

Offset: 1

Benoit Cloitre, Jan 04 2002

nonn

All even numbers are in this sequence.

Odd terms in the sequence are numbers whose prime factors are +-1 (mod 8) (A058529), i.e., odd k such that x^2 == 2 (mod k) has a solution. - Jason Earls, Jan 22 2002

Cf. A058529, A001132, A057126.

isok(n) = {for (x=0, n-1, if (Mod(x, n)^2 == Mod(2, n)^n, return (1));); return (0);} \\ Michel Marcus, Nov 20 2013

Corrected by Vladeta Jovovic, Jan 22 2002

More terms from Jason Earls, Jan 22 2002

A352347 Least odd prime p such that q divides 2^p - p^2, where q is n-th prime of the form 8*k +- 1, or -1 if no such prime exists.

Original entry on oeis.org

5, 31, 29, 89, 11, 11, 13, 89, 7, 283, 29, 211, 13, 643, 2711, 491, 1627, 1699, 283, 727, 1493, 1663, 37, 89, 907, 1039, 73, 571, 2707, 149, 179, 197, 443, 463, 1187, 4133, 383, 359, 251, 1567, 4603, 3469, 2069, 313, 677, 1319, 2441, 647, 3733, 3623, 31, 1447

Offset: 1

Robert G. Wilson v, Mar 14 2022

nonn

Inspired by A350964.
In the first 100000 terms, greatest prime encountered was 204114067.
Records: 5, 31, 89, 283, 643, 2711, 4133, 4603, 8317, 23561, 25819, 45083, ...

			a(1) = 5 since A001132(1) | 2^5 - 5^2 = 32 - 25 = 7.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Robert G. Wilson v, Table of n, a(n) for n = 1..100000

Cf. A001132, A350964.

f(q) = forprime(p=5, oo, if(Mod(2, q)^p == Mod(p, q)^2, return(p)));
lista(nn) = forprime(q=7, nn, if((q+2)%8<4, print1(f(q), ", "))); \\ Jinyuan Wang, Jul 14 2022

Name edited by Jinyuan Wang, Jul 14 2022

A385449 Irregular triangle, read by rows: row n gives the pair of proper positive fundamental solutions (x, y) of the form x^2 - 2*y^2 representing -A057126(n).

Original entry on oeis.org

1, 1, 4, 3, 1, 2, 5, 4, 2, 3, 6, 5, 1, 3, 9, 7, 3, 4, 7, 6, 1, 4, 13, 10, 4, 5, 8, 7, 3, 5, 11, 9, 2, 5, 14, 11, 5, 6, 9, 8, 1, 5, 17, 13, 6, 7, 10, 9, 1, 6, 21, 16, 5, 7, 13, 11, 7, 8, 11, 10, 4, 7, 16, 13, 3, 7, 19, 15, 2, 7, 22, 17, 1, 7, 25, 19, 8, 9, 12, 11, 5, 8, 17, 14, 7, 9, 15, 13, 3, 8, 23, 18, 9, 10, 13, 12

Offset: 1

Wolfdieter Lang, Jul 11 2025

The number of (x, y) pairs in row n is 1 for n = 1 and 2, and 2^P, with P = P1 + P7, where P1 and P7 are the number of prime factors 1 modulo 8 and 7 modulo 8, respectively, of A057126(n), for n >= 3.
See A057126 for comments concerning its representation by x^2 - 2*y^2.
The numbers A057126 are given by 2^e_2 * Product_{i=1..P1} p_{1,i}^e_{1,i} * Product_{j=1..P7} p_{7,j}^e_{7,j}, with the odd primes p_{1,i} and p_{7,j} congruent to 1 and 7 modulo 8, respectively. See A007519 and A007522 for these odd primes. Together with 2 these primes are given in A038873, and without 2 in A001132. The exponents are e_2 = 0 or 1, and e_{1,i} and e_{7,j} are nonnegative. The a(1) = 1 is obtained if all exponents vanish. For the proof see Lemma 18 of the linked W. Lang paper, pp. 22 - 23.
The general solutions are obtained from each fundamental solution by application of integer powers of the matrix Auto' = Mat([3,4], [2,3]). See the linked paper eq (28), p. 14, and eq. (40), p. 17 for D = 2, and k = A057126(n). For the explicit form of the powers of Auto' in terms of Chebyshev polynomials S(n, 6) = A001109(n+1) see there eq. (38), and Lemma 10, eq. (43), p. 17.
The conversion to the pair of proper solutions (X, Y) of X^2 - 2*Y^2 = A057126(n) is given by (X, Y) = (2*y - x, x - y). This may result in solutions with negative Y values. They are then transformed to the fundamental positive proper solutions via the mentioned matrix Auto'. See the right part of the example below. For this conversion see also the Nov 09 2009 comment in A035251 by Franklin T. Adams-Watters.

			n, A057126(n) /k  1  2   3  4 ...   2^P | (X, Y) = (2*y - x, x - y)
-------------------------------------------------------------------
1,  1           | 1  1               1  |  1   0 (3   2)
2,  2           | 4  3               1  |  2   1
3,  7           | 1  2,  5  4        2  |  3  -1 (5   3),  3  1
4, 14 = 2*7     | 2  3,  6  5        2  |  4  -1 (8   5),  4  1
5, 17           | 1  3,  9  7        2  |  5  -2 (7   4),  5  2
6, 23           | 3  4,  7  6        2  |  5  -1 (11  7),  5, 1
7, 31           | 1  4, 13 10        2  |  7  -3 (9   5),  7  3
8, 34 = 2*17    | 4  5,  8  7        2  |  6  -1 (14  9),  6  1
9, 41           | 3  5, 11  9        2  |  7  -2 (13  8),  7  2
10, 46 = 2*23   | 2  5, 14 11        2  |  8  -3 (12  7),  8  3
11, 47          | 5  6,  9  8        2  |  7  -1 (17 11),  7  1
12, 49 = 7^2    | 1  5, 17 13        2  |  9  -4 (11  6),  9  4
13, 62 = 2*31   | 6  7, 10  9        2  |  8  -1 (20 13),  8  1
14, 71          | 1  6, 21 16        2  | 11  -5 (13  7), 11  5
15, 73          | 5  7, 13 11        2  |  9  -2 (19 12),  9  2
16, 79          | 7  8, 11 10        2  |  9  -1 (23 15),  9  1
17, 82 = 2*41   | 4  7, 16 13        2  | 10  -3 (18 11), 10  3
18, 89          | 3  7, 19 15        2  | 11  -4 (17 10), 11  4
19, 94 = 2*47   | 2  7, 22 17        2  | 12  -5 (16  9), 12  5
20, 97          | 1  7, 25 19        2  | 13  -6 (15  8), 13  6
21, 98 = 2*7^2  | 8  9, 12 11        2  | 10  -1 (26 17), 10  1
...
The corresponding fundamental positive proper solutions of X^2 - 2*Y^2 = +119 are: [13 -5 (19 11), 13, 5] and [11 -1 (29 19), 11 1].

Wolfdieter Lang, On Positive Integer Descartes-Steiner Curvature Quintuplets, arXiv:2503.08631 [nucl-th], 2025.

Cf. A001109, A007519, A007522, A035251, A057126, A038873.

A373560 a(n) is the smallest multiple of prime(n)^2 that starts a run of 5 consecutive integers with 6 divisors, or -1 if no such multiple exists.

Original entry on oeis.org

-1, -1, -1, 10093613546512321, -1, -1, 7700031346933907521, -1, 5344962129269790721, -1, 20453982425165652721, -1, 8163195338222675521, -1, 2467958104789157112721, -1, -1, -1, -1, 14666767069023896053921, 212170739123852995921, 287954235303137500060321, -1, 84769922583214545304321

Offset: 1

Jon E. Schoenfield, Jun 09 2024

sign

Terms were obtained using the b-file at A141621.
a(n) = -1 if prime(n) is not in A001132.
Conjecture: the converse is also true.

			a(1) = a(2) = a(3) = -1 because the first of five consecutive integers having six divisors is never a multiple of 2^2, 3^2, or 5^2.
a(4) = 10093613546512321 because it is the smallest term in A141621 that is a multiple of prime(4)^2 = 49.
a(9) = 5344962129269790721 because it is the smallest term in A141621 that is a multiple of prime(9)^2 = 23^2.

Cf. A001132, A030515, A042999, A054753, A141621.

cp's OEIS Frontend

A319040 Numbers k > 1 such that Pell(k) == 1 (mod k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

A366526 Prime powers (A246655) q such that 2 is a nonzero square in the finite field F_q.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A035200 Coefficients in expansion of Dirichlet series Product_p (1-(Kronecker(m,p)+1)*p^(-s)+Kronecker(m,p)*p^(-2s))^(-1) for m = 18.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A097958 Primes p such that p divides 6^((p-1)/2) - 3^((p-1)/2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions

A201916 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+2737)^2 = y^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A350964 a(n) is the largest prime factor of 2^p - p^2 where p is the n-th prime.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A066507 Numbers k such that there is a solution to x^2 == 2^k (mod k).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

Extensions

A352347 Least odd prime p such that q divides 2^p - p^2, where q is n-th prime of the form 8*k +- 1, or -1 if no such prime exists.

Views

Author

A035200 Coefficients in expansion of Dirichlet series Product_p (1-(Kronecker(m,p)+1)p^(-s)+Kronecker(m,p)p^(-2s))^(-1) for m = 18.