A056031 -id:A056031 - cp's OEIS frontend

A056021 Numbers k such that k^4 == 1 (mod 5^2).

1, 7, 18, 24, 26, 32, 43, 49, 51, 57, 68, 74, 76, 82, 93, 99, 101, 107, 118, 124, 126, 132, 143, 149, 151, 157, 168, 174, 176, 182, 193, 199, 201, 207, 218, 224, 226, 232, 243, 249, 251, 257, 268, 274, 276, 282, 293, 299, 301, 307, 318, 324, 326, 332, 343, 349

Offset: 1

Robert G. Wilson v, Jun 08 2000

Numbers congruent to {1, 7, 18, 24} mod 25.

These terms (apart from 1) are tetration bases characterized by a constant convergence speed strictly greater than 1 (see A317905). - Marco Ripà, Jan 25 2024

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Colin Barker)
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A056022, A056024, A056025, A056026, A056027, A056028, A056031, A056034, A056035, A317905.

Cf. A381319 (general case mod n^2)

Select[ Range[ 400 ], PowerMod[ #, 4, 25 ]==1& ]
(* or *)
LinearRecurrence[{1, 0, 0, 1, -1}, {1, 7, 18, 24, 26}, 100] (* Mike Sheppard, Feb 18 2025 *)

a(n) = (-25 - (-1)^n + (9-9*I)*(-I)^n + (9+9*I)*I^n + 50*n) / 8 \\ Colin Barker, Oct 16 2015

Vec(x*(x^2+3*x+1)^2/((1+x)*(x^2+1)*(x-1)^2) + O(x^100)) \\ Colin Barker, Oct 16 2015

for(n=0, 1e3, if(n^4 % 5^2 == 1, print1(n", "))) \\ Altug Alkan, Oct 16 2015

isok(k) = Mod(k, 25)^4 == 1; \\ Michel Marcus, Jun 30 2021

G.f.: x*(x^2+3*x+1)^2 / ((1+x)*(x^2+1)*(x-1)^2). - R. J. Mathar, Oct 25 2011
a(n) = (-25 - (-1)^n + (9-9*i)*(-i)^n + (9+9*i)*i^n + 50*n) / 8, where i = sqrt(-1). - Colin Barker, Oct 16 2015
From Mike Sheppard, Feb 18 2025: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5).
a(n) ~ (5^2/4)*n. (End)

A056022 Numbers k such that k^6 == 1 (mod 7^2).

Original entry on oeis.org

1, 18, 19, 30, 31, 48, 50, 67, 68, 79, 80, 97, 99, 116, 117, 128, 129, 146, 148, 165, 166, 177, 178, 195, 197, 214, 215, 226, 227, 244, 246, 263, 264, 275, 276, 293, 295, 312, 313, 324, 325, 342, 344, 361, 362, 373, 374, 391, 393, 410, 411, 422, 423, 440, 442

Offset: 1

Robert G. Wilson v, Jun 08 2000

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).

Cf. A056021, A056024, A056025, A056026, A056027, A056028, A056031, A056034, A056035.

Cf. A381319 (general case mod n^2).

Select[ Range[ 500 ], PowerMod[ #, 6, 49 ]==1& ]
LinearRecurrence[{1, 0, 0, 0, 0, 1, -1}, {1, 18, 19, 30, 31, 48, 50}, 61] (* Mike Sheppard, Feb 18 2025 *)

isok(k) = Mod(k, 49)^6 == 1; \\ Michel Marcus, Jun 30 2021

From Mike Sheppard, Feb 18 2025 : (Start)
a(n) = a(n-1) + a(n-6) - a(n-7).
a(n) = a(n-6) + 7^2.
a(n) ~ (7^2/6)*n.
G.f.: (1 + x*(17 + x + 11*x^2 + x^3 + 17*x^4 + x^5))/(1 - x - x^6 + x^7). (End)

A056024 Numbers k such that k^10 == 1 (mod 11^2).

Original entry on oeis.org

1, 3, 9, 27, 40, 81, 94, 112, 118, 120, 122, 124, 130, 148, 161, 202, 215, 233, 239, 241, 243, 245, 251, 269, 282, 323, 336, 354, 360, 362, 364, 366, 372, 390, 403, 444, 457, 475, 481, 483, 485, 487, 493, 511, 524, 565, 578, 596, 602, 604, 606, 608, 614, 632

Offset: 1

Robert G. Wilson v, Jun 08 2000

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

Cf. A056021, A056022, A056025, A056026, A056027, A056028, A056031, A056034, A056035.

Cf. A381319 (general case mod n^2).

Select[ Range[ 800 ], PowerMod[ #, 10, 121 ]==1& ]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 3, 9, 27, 40, 81, 94, 112, 118, 120, 122}, 65] (* Mike Sheppard, Feb 19 2025 *)

From Mike Sheppard, Feb 19 2025: (Start)
a(n) = a(n-1) + a(n-10) - a(n-11).
a(n) = a(n-10) + 11^2.
a(n) ~ (11^2/10)*n. (End)

A056025 Numbers k such that k^12 == 1 (mod 13^2).

Original entry on oeis.org

1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168, 170, 188, 191, 192, 239, 249, 258, 268, 315, 316, 319, 337, 339, 357, 360, 361, 408, 418, 427, 437, 484, 485, 488, 506, 508, 526, 529, 530, 577, 587, 596, 606, 653, 654, 657, 675, 677, 695, 698, 699, 746

Offset: 1

Robert G. Wilson v, Jun 08 2000

From 19 to 168 inclusive, these are the numbers that 'fool' the strong pseudoprimality test described in Wilf (1986) in regard to determining whether 169 is composite. - Alonso del Arte, Feb 05 2012

Herbert S. Wilf, Algorithms and Complexity, Englewood Cliffs, New Jersey: Prentice-Hall, 1986, pp. 158-160.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A056021, A056022, A056024, A056026, A056027, A056028, A056031, A056034, A056035.

Cf. A381319 (general case mod n^2).

Select[ Range[ 800 ], PowerMod[ #, 12, 169 ]==1& ]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168, 170}, 56] (* Mike Sheppard, Feb 19 2025 *)

is(k)=Mod(k,169)^12==1 \\ Charles R Greathouse IV, Feb 07 2018

From Mike Sheppard, Feb 19 2025 : (Start)
a(n) = a(n-1) + a(n-12) - a(n-13).
a(n) = a(n-12) + 13^2.
a(n) ~ (13^2/12)*n. (End)

Definition corrected by T. D. Noe, Aug 23 2008

A056027 Numbers k such that k^16 == 1 (mod 17^2).

Original entry on oeis.org

1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288, 290, 327, 329, 354, 364, 399, 420, 423, 444, 447, 468, 503, 513, 538, 540, 577, 579, 616, 618, 643, 653, 688, 709, 712, 733, 736, 757, 792, 802, 827, 829, 866, 868, 905, 907, 932, 942

Offset: 1

Robert G. Wilson v, Jun 08 2000

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A056021, A056022, A056024, A056025, A056026, A056028, A056031, A056034, A056035.

x=17; Select[ Range[ 1000 ], PowerMod[ #, x-1, x^2 ]==1& ]
(* or *)
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288, 290}, 55] (* Mike Sheppard, Feb 17 2025 *)

From Mike Sheppard, Feb 17 2025: (Start)
a(n) = a(n-1) + a(n-16) - a(n-17).
a(n) ~ (17^2/16) * n.
G.f.: (x*(1 + 37*x + 2*x^2 + 25*x^3 + 10*x^4 + 35*x^5 + 21*x^6 + 3*x^7 + 21*x^8 + 3*x^9 + 21*x^10 + 35*x^11 + 10*x^12 + 25*x^13 + 2*x^14 + 37*x^15 + x^16))/((1-x)*(1-x^16)).
(End)

A056034 Numbers k such that k^28 == 1 (mod 29^2).

Original entry on oeis.org

1, 14, 41, 60, 63, 137, 190, 196, 221, 236, 267, 270, 374, 416, 425, 467, 571, 574, 605, 620, 645, 651, 704, 778, 781, 800, 827, 840, 842, 855, 882, 901, 904, 978, 1031, 1037, 1062, 1077, 1108, 1111, 1215, 1257, 1266, 1308, 1412, 1415, 1446, 1461, 1486, 1492

Offset: 1

Robert G. Wilson v, Jun 08 2000

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A056021, A056022, A056024, A056025, A056026, A056027, A056028, A056031, A056035.

Cf. A381319 (general case mod n^2).

x=29; Select[ Range[ 2000 ], PowerMod[ #, x-1, x^2 ]==1& ]

isok(k) = Mod(k, 29^2)^28 == 1; \\ Michel Marcus, Apr 10 2025

From Mike Sheppard, Feb 20 2025 : (Start)
a(n) = a(n-1) + a(n-28) - a(n-29).
a(n) = a(n-28) + 29^2.
a(n) ~ (29^2/28)*n. (End)

A056035 Numbers k such that k^30 == 1 (mod 31^2).

Original entry on oeis.org

1, 115, 117, 145, 229, 235, 333, 338, 374, 388, 414, 430, 439, 440, 448, 513, 521, 522, 531, 547, 573, 587, 623, 628, 726, 732, 816, 844, 846, 960, 962, 1076, 1078, 1106, 1190, 1196, 1294, 1299, 1335, 1349, 1375, 1391, 1400, 1401, 1409, 1474, 1482, 1483

Offset: 1

Robert G. Wilson v, Jun 08 2000

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A056021, A056022, A056024, A056025, A056026, A056027, A056028, A056031, A056034.

x=31; Select[ Range[ 2000 ], PowerMod[ #, x-1, x^2 ]==1& ]

From Mike Sheppard, Feb 20 2025 : (Start)
a(n) = a(n-1) + a(n-30) - a(n-31).
a(n) = a(n-30) + 31^2.
a(n) ~ (31^2/30)*n. (End)

A143548 Irregular triangle of numbers k < p^2 such that p^2 divides k^(p-1)-1, with p=prime(n).

Original entry on oeis.org

1, 1, 8, 1, 7, 18, 24, 1, 18, 19, 30, 31, 48, 1, 3, 9, 27, 40, 81, 94, 112, 118, 120, 1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168, 1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288, 1, 28, 54, 62, 68, 69, 99, 116, 127, 234, 245, 262, 292, 293, 299, 307, 333, 360

Offset: 1

T. D. Noe, Aug 24 2008

Row n begins with 1 and has prime(n)-1 terms. The first differences of each row are symmetric. For k > p^2, the solutions are just shifted by m*p^2 for m > 0. An open question is whether every integer appears in this sequence. For instance, 2 does not appear until the prime 1093 and 5 does not appear until the prime 20771.
For row n > 1, the sum of the terms in row n is (p-1)*p^2*(p+1)/2, which is A138416. - T. D. Noe, Aug 24 2008, corrected by Robert Israel, Sep 27 2016
Max Alekseyev points out that there is a much faster method of computing these numbers. Let p=prime(n) and let r be a primitive root of p (see A001918 and A060749). Then the terms in row n are r^(k*p) (mod p^2) for k=0..p-2. - T. D. Noe, Aug 26 2008
The numbers in this sequence are the bases to Euler pseudoprimes q, which are squares of prime numbers, such that n^((q-1)/2) == +-1 mod q. An exception is the first number 9 = 3*3, which is, following the strict definition in Crandall and Pomerance, no Fermat pseudoprime and hence no Euler pseudoprime. - Karsten Meyer, Jan 08 2011
For row n > 1, the sum is zero modulo p^2 (rows are antisymmetric due to Binomial Theorem). - Peter A. Lawrence, Sep 11 2016

			(2)   1,
(3)   1, 8,
(5)   1, 7, 18, 24,
(7)   1, 18, 19, 30, 31, 48,
(11)  1, 3, 9, 27, 40, 81, 94, 112, 118, 120,
(13)  1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168,
(17)  1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288,

R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2005

T. D. Noe, Rows n=1..50 of triangle, flattened

Cf. A039678, A056020, A056021, A056022, A056024, A056025, A056027, A056028, A056031, A056034, A056035, A096082, A138416.

f:= proc(n) local p,j,x;
  p:= ithprime(n);
  x:= numtheory:-primroot(p);
  op(sort([seq(x^(i*p) mod p^2, i=0..p-2)]))
end proc:
map(f, [$1..20]); # Robert Israel, Sep 27 2016

Flatten[Table[p=Prime[n]; Select[Range[p^2], PowerMod[ #,p-1,p^2]==1&], {n,50}]] (* T. D. Noe, Aug 24 2008 *)
Flatten[Table[p=Prime[n]; r=PrimitiveRoot[p]; b=PowerMod[r,p,p^2]; Sort[NestList[Mod[b*#,p^2]&,1,p-2]], {n,50}]] (* Faster version from T. D. Noe, Aug 26 2008 *)

A056026 Numbers k such that k^14 == 1 (mod 15^2).

Original entry on oeis.org

1, 26, 199, 224, 226, 251, 424, 449, 451, 476, 649, 674, 676, 701, 874, 899, 901, 926, 1099, 1124, 1126, 1151, 1324, 1349, 1351, 1376, 1549, 1574, 1576, 1601, 1774, 1799, 1801, 1826, 1999, 2024, 2026, 2051, 2224, 2249, 2251, 2276, 2449, 2474, 2476, 2501

Offset: 1

Robert G. Wilson v, Jun 08 2000

Numbers congruent to {1, 26, 129, 224} mod 225.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Colin Barker)
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A056021, A056022, A056024, A056025, A056027, A056028, A056031, A056034, A056035.

Select[ Range[ 3000 ], PowerMod[ #, 14, 225 ]==1& ]
LinearRecurrence[{1,0,0,1,-1},{1,26,199,224,226},50] (* Harvey P. Dale, Nov 11 2011 *)

a(n) = (-225 - 125*(-1)^n + (171-171*I)*(-I)^n + (171+171*I)*I^n + 450*n)/8 \\ Colin Barker, Oct 16 2015

Vec(x*(1+25*x+173*x^2+25*x^3+x^4)/((1+x)*(1+x^2)*(x-1)^2) + O(x^100)) \\ Colin Barker, Oct 16 2015

G.f.: x*(1+25*x+173*x^2+25*x^3+x^4) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 25 2011
a(1)=1, a(2)=26, a(3)=199, a(4)=224, a(5)=226, a(n) = a(n-1)+a(n-4)-a(n-5). - Harvey P. Dale, Nov 11 2011
a(n) = (-225 - 125*(-1)^n + (171-171*i)*(-i)^n + (171+171*i)*i^n + 450*n)/8 where i=sqrt(-1). - Colin Barker, Oct 16 2015

A381319 Order of linear recurrence with constant coefficients of solutions of k satisfying k^(n-1) == 1 (mod n^2) for a given n.

Original entry on oeis.org

2, 3, 2, 5, 2, 7, 2, 3, 2, 11, 2, 13, 2, 5, 2, 17, 2, 19, 2, 5, 2, 23, 2, 5, 2, 3, 4, 29, 2, 31, 2, 5, 2, 5, 2, 37, 2, 5, 2, 41, 2, 43, 2, 9, 2, 47, 2, 7, 2, 5, 4, 53, 2, 5, 2, 5, 2, 59, 2, 61, 2, 5, 2, 17, 6, 67, 2, 5, 4, 71, 2, 73, 2, 5, 4, 5, 2, 79, 2, 3, 2, 83, 2, 17, 2, 5, 2, 89

Offset: 2

Mike Sheppard, Feb 20 2025

nonn

For a given n, the solutions for k have the linear recurrence with constant coefficients k(m) = k(m-1) + k(m-(a(n)-1)) - k(m-a(n)), with order a(n). If a(n)=2 then the term k(m-1) appears twice and is k(m) = 2*k(m-1) - k(m-2).

Also, k(m) - k(m-(a(n)-1)) = n^2 = k(m-1) - k(m-a(n)), so all have nonhomogeneous linear recurrence of k(m) = k(m-(a(n)-1)) + n^2. Equivalently, k(m) = k(m-A063994(n)) + n^2, with order A063994(n). Thus, k(m) ~ (n^2 / A063994(n)) * m = (n^2 / (a(n)-1)) * m.

			For n=5 the congruence equation k^4 ==1 mod (5^2) has solutions of k (A056021) which satisfy k(m) = k(m-1) + k(m-4) - k(m-5), the order being 5, a(5)=5.
For n=9, k^8==1 mod (9^2) has solutions of k with recurrence k(m) = k(m-1) + k(m-2) - k(m-3), order 3, a(9)=3.

Paolo Xausa, Table of n, a(n) for n = 2..10000

Cf. A063994, A056020 (n=3), A056021 (n=5), A056022 (n=7), A056024 (n=11), A056025 (n=13), A056028 (n=19), A056031 (n=23), A056034 (n=29), A056035 (n=31).

A381319[n_] := Times @@ GCD[FactorInteger[n][[All, 1]] - 1, n - 1] + 1;
Array[A381319, 100, 2] (* Paolo Xausa, Mar 05 2025 *)

a(n) = 1 + A063994(n).

a(p) = p if p is prime.

cp's OEIS Frontend

A056021 Numbers k such that k^4 == 1 (mod 5^2).

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A056022 Numbers k such that k^6 == 1 (mod 7^2).

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A056024 Numbers k such that k^10 == 1 (mod 11^2).

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A056025 Numbers k such that k^12 == 1 (mod 13^2).

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A056027 Numbers k such that k^16 == 1 (mod 17^2).

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A056034 Numbers k such that k^28 == 1 (mod 29^2).

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A056035 Numbers k such that k^30 == 1 (mod 31^2).

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A143548 Irregular triangle of numbers k < p^2 such that p^2 divides k^(p-1)-1, with p=prime(n).

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