A151972 -id:A151972 - cp's OEIS frontend

A151971 Numbers n such that n^2 - n is divisible by 21.

0, 1, 7, 15, 21, 22, 28, 36, 42, 43, 49, 57, 63, 64, 70, 78, 84, 85, 91, 99, 105, 106, 112, 120, 126, 127, 133, 141, 147, 148, 154, 162, 168, 169, 175, 183, 189, 190, 196, 204, 210, 211, 217, 225, 231, 232, 238, 246, 252, 253, 259, 267, 273, 274, 280, 288, 294, 295, 301, 309

Offset: 1

N. J. A. Sloane, Aug 23 2009

Equivalently, numbers that are congruent to {0, 1, 7, 15} mod 21. - Bruno Berselli, Aug 06 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

For m^2 == m (mod n), see: n=2: A001477; n=3: A032766; n=4: A042948; n=5: A008851; n=6: A032766; n=7: A047274; n=8: A047393; n=9: A090570; n=10: A008851; n=11: A112651; n=12: A112652; n=13:A112653; n=14: A047274; n=15: A151972; n=16: A151977; n=17: A151978; n=18: A090570; n=19: A151979; n=20: A151980; n=21: A151971; n=22: A112651; n=24: A151973; n=26: A112653; n=30: A151972; n=32: A151983; n=34: A151978; n=38: A151979; n=42: A151971; n=48: A151981; n=64: A151984.

Cf. A215202.

[n: n in [0..309] | IsZero((n^2-n) mod 21)]; // Bruno Berselli, Aug 06 2012

A151971:=n->(42*n+14*I^((n-1)*n)-3*I^(2*n)-3)/8-7: seq(A151971(n), n=1..100); # Wesley Ivan Hurt, Jun 07 2016

Select[Range[0,400], Divisible[#^2-#,21]&] (* Harvey P. Dale, Jun 04 2012 *)

makelist((42*n+14*%i^((n-1)*n)-3*(-1)^n-3)/8-7, n, 1, 60); /* Bruno Berselli, Aug 06 2012 */

From Bruno Berselli, Aug 06 2012: (Start)
G.f.: x^2*(1+6*x+8*x^2+6*x^3)/((1+x)*(1-x)^2*(1+x^2)).
a(n) = (42*n +14*i^((n-1)*n) -3*(-1)^n -3)/8 -7, where i=sqrt(-1). (End)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - Wesley Ivan Hurt, Jun 07 2016
E.g.f.: (24 + (21*x - 31)*cosh(x) + 7*(sin(x) + cos(x) + (3*x - 4)*sinh(x)))/4. - Ilya Gutkovskiy, Jun 07 2016

A215202 Irregular triangle in which n-th row gives m in 1, ..., n-1 such that m^2 == m (mod n).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 4, 1, 1, 1, 1, 5, 6, 1, 1, 4, 9, 1, 1, 7, 8, 1, 6, 10, 1, 1, 1, 9, 10, 1, 1, 5, 16, 1, 7, 15, 1, 11, 12, 1, 1, 9, 16, 1, 1, 13, 14, 1, 1, 8, 21, 1, 1, 6, 10, 15, 16, 21, 25, 1, 1, 1, 12, 22, 1, 17, 18, 1, 15, 21, 1, 9, 28, 1, 1, 19, 20, 1, 13

Offset: 2

Eric M. Schmidt, Aug 05 2012

The n-th row has length A034444(n) - 1.
If m appears in row n, then gcd(n,m) appears in the n-th row of A077610. Moreover, if m', distinct from m, also appears in row n, then gcd(n, m) does not equal gcd(n, m').
For odd n and any integer m, m^2 == m (mod n) iff m^2 == m (mod 2n).
Let P(1)={1} and for integers x > 1, let P(x) be the set of distinct prime divisors of x. We can define an equivalence relation ~ on the set of elements in the ring (Z_n, +mod n,*mod n): for all a,b in Z_n (where a,b are the least nonnegative residues modulo n) a ~ b iff P(gcd(a,n)) intersect P(n) is equal to P(gcd(b,n)) intersect P(n). If we include 0 in each row then these elements can represent the equivalence classes. They form a commutative monoid. - Geoffrey Critzer, Feb 13 2016

			Triangle begins:
1;
1;
1;
1;
1, 3, 4;
1;
1;
1;
1, 5, 6;
1;
1, 4, 9;
1;
1, 7, 8;
1, 6, 10;
1;
1;
1, 9, 10; etc.  - _Bruno Berselli_, Aug 06 2012

Eric M. Schmidt, Rows 2..2000, flattened

For m^2 == m (mod n), see: n=2: A001477; n=3: A032766; n=4: A042948; n=5: A008851; n=6: A032766; n=7: A047274; n=8: A047393; n=9: A090570; n=10: A008851; n=11: A112651; n=12: A112652; n=13: A112653; n=14: A047274; n=15: A151972; n=16: A151977; n=17: A151978; n=18: A090570; n=19: A151979; n=20: A151980; n=21: A151971; n=22: A112651; n=24: A151973; n=26: A112653; n=30: A151972; n=32: A151983; n=34: A151978; n=38: A151979; n=42: A151971; n=48: A151981; n=64: A151984; n=100: A008852; n=1000: A008853.

[m: m in [1..n-1], n in [2..40] | m^2 mod n eq m]; // Bruno Berselli, Aug 06 2012

Table[Select[Range[n], Mod[#^2, n] == # &], {n, 2, 30}] // Grid (* Geoffrey Critzer, May 26 2015 *)

def A215202(n) : return [m for m in range(1, n) if m^2 % n == m];

A379211 List of positive integers that are congruent to {2, 7, 8, 13} mod 15.

Original entry on oeis.org

2, 7, 8, 13, 17, 22, 23, 28, 32, 37, 38, 43, 47, 52, 53, 58, 62, 67, 68, 73, 77, 82, 83, 88, 92, 97, 98, 103, 107, 112, 113, 118, 122, 127, 128, 133, 137, 142, 143, 148, 152, 157, 158, 163, 167, 172, 173, 178, 182, 187, 188, 193, 197, 202, 203, 208, 212, 217, 218, 223, 227, 232, 233, 238, 242, 247, 248, 253, 257, 262

Offset: 1

Peter Bala, Dec 18 2024

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001622, A072703, A151972, A204542, A315211, A379210.

a := proc(n) option remember;
      `if`(n < 5, [0, 2, 7, 8, 13][n+1], 15 + a(n-4))
     end:
seq(a(n), n = 1..70);

LinearRecurrence[{1, 0, 0, 1, -1}, {2, 7, 8, 13, 17}, 70] (* Amiram Eldar, Dec 24 2024 *)

a(n) = 15 + a(n-4); a(n) = - a(1-n).
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 6.
G.f.: x*(x^2 + 3*x + 1)*(2*x^2 - x + 2)/((1 + x)*(1 - x)^2*(1 + x^2)).
a(n)^2 = 15 * A379210(n) + 4.
For n >= 2, a(n-1) + a(n+1) = A072703(n).
It appears that a(n) + a(n+1) = (3/2) * A315211(n).
E.g.f.: (8 - 3*cos(x) + 5*(3*x - 1)*cosh(x) + 3*sin(x) + 5*(3*x - 2)*sinh(x))/4. - Stefano Spezia, Dec 23 2024
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(5*sqrt(3)*phi), where phi is the golden ratio (A001622). - Amiram Eldar, Dec 24 2024

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A151971 Numbers n such that n^2 - n is divisible by 21.

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A215202 Irregular triangle in which n-th row gives m in 1, ..., n-1 such that m^2 == m (mod n).

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A379211 List of positive integers that are congruent to {2, 7, 8, 13} mod 15.

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Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula