A151978 -id:A151978 - cp's OEIS frontend

A195047 Concentric 17-gonal numbers.

0, 1, 17, 35, 68, 103, 153, 205, 272, 341, 425, 511, 612, 715, 833, 953, 1088, 1225, 1377, 1531, 1700, 1871, 2057, 2245, 2448, 2653, 2873, 3095, 3332, 3571, 3825, 4081, 4352, 4625, 4913, 5203, 5508, 5815, 6137, 6461, 6800, 7141, 7497, 7855, 8228, 8603, 8993

Offset: 0

Omar E. Pol, Sep 27 2011

Also concentric heptadecagonal numbers or concentric heptakaidecagonal numbers.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Column 17 of A195040.

Cf. A032527, A032528, A061925, A069130, A151978, A195046, A195048, A195146, A195147.

LinearRecurrence[{2,0,-2,1},{0,1,17,35},50] (* Harvey P. Dale, Dec 23 2017 *)

a(n)=17*n^2/4+13*((-1)^n-1)/8 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 17*n^2/4+13*((-1)^n-1)/8. [Typo fixed by Ivan Panchenko, Nov 08 2013]
From R. J. Mathar, Sep 28 2011: (Start)
G.f.: -x*(1+15*x+x^2) / ( (1+x)*(x-1)^3 ).
a(n)+a(n+1) = A069130(n+1). (End)
From Bruno Berselli, Sep 29 2011: (Start)
a(n) = a(-n) = (34*n^2+13*(-1)^n-13)/8.
a(n) = A151978(A061925(n)). (End)
Sum_{n>=1} 1/a(n) = Pi^2/102 + tan(sqrt(13/17)*Pi/2)*Pi/sqrt(221). - Amiram Eldar, Jan 16 2023

A151972 Numbers that are congruent to {0, 1, 6, 10} mod 15.

Original entry on oeis.org

0, 1, 6, 10, 15, 16, 21, 25, 30, 31, 36, 40, 45, 46, 51, 55, 60, 61, 66, 70, 75, 76, 81, 85, 90, 91, 96, 100, 105, 106, 111, 115, 120, 121, 126, 130, 135, 136, 141, 145, 150, 151, 156, 160, 165, 166, 171, 175, 180, 181, 186, 190, 195, 196, 201, 205, 210, 211, 216, 220, 225

Offset: 1

N. J. A. Sloane, Aug 23 2009

Also, numbers n such that n^2 - n is divisible by 15.

Also, numbers n such that n^2 - n is divisible by 30.

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..1000] | n mod 15 in [0, 1, 6, 10]]; // Vincenzo Librandi, Apr 02 2011, simplified by Eric M. Schmidt, Aug 05 2012

[ n: n in [0..1000] | (n^2-n) mod (15) eq 0 ]; // Vincenzo Librandi, Apr 03 2011, altered by Eric M. Schmidt, Aug 05 2012

A151972:=n->(30*n-41-5*I^(2*n)+(3+3*I)*I^(-n)+(3-3*I)*I^n)/8: seq(A151972(n), n=1..100); # Wesley Ivan Hurt, Jun 07 2016

Select[Range[0, 300], Divisible[#^2-#, 15]&]  (* Harvey P. Dale, Apr 01 2011, altered by Eric M. Schmidt, Aug 05 2012 *)

G.f.: x^2*(1+5*x+4*x^2+5*x^3) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 25 2011
From Wesley Ivan Hurt, Jun 07 2016: (Start)
a(n) = (30*n-41-5*i^(2*n)+(3+3*i)*i^(-n)+(3-3*i)*i^n)/8 where i=sqrt(-1).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. (End)
E.g.f.: (20 + (15*x - 23)*cosh(x) + 3*(sin(x) + cos(x) + (5*x - 6)*sinh(x)))/4. - Ilya Gutkovskiy, Jun 07 2016

This is a merge of two identical sequences, A151972 and A151975.

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

Original entry on oeis.org

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];

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A195047 Concentric 17-gonal numbers.

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A151972 Numbers that are congruent to {0, 1, 6, 10} mod 15.

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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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