cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A165718 Integers of the form k*(k+7)/6.

Original entry on oeis.org

3, 5, 10, 13, 20, 24, 33, 38, 49, 55, 68, 75, 90, 98, 115, 124, 143, 153, 174, 185, 208, 220, 245, 258, 285, 299, 328, 343, 374, 390, 423, 440, 475, 493, 530, 549, 588, 608, 649, 670, 713, 735, 780, 803, 850, 874, 923, 948, 999, 1025, 1078, 1105, 1160, 1188
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Keywords

Comments

Integers of the form k + k*(k+1)/6 = k + A000217(k)/3; for k see A007494, for A000217(k)/3 see A001318. - R. J. Mathar, Sep 25 2009
Only 3 terms are prime numbers (3,5,13). Are all the rest composite?
The only prime terms in this sequence are 3, 5, and 13. If k=6j+1 or k=6j+4, k*(k+7) is congruent to 2 mod 6 and will never be an integer. If k=6j, k*(k+7)/6 = j*(6j+7) which is prime only for j=1 (i.e., 13 is in the sequence). If k=6j+2, k*(k+7)/6 = (3j+1)*(2j+3) which is prime only for j=0 (i.e., 3 is in the sequence). If k=6j+3, k*(k+7)/6 = (2j+1)*(3j+5) which is prime only for j=0 (i.e., 5 is in the sequence). If k=6j+5, k*(k+7)/6 = (6j+5)*(j+2) which is never prime. Thus {3,5,13} are the only primes in this sequence. - Derek Orr, Feb 26 2017
Conjecturally, the sequence terms are the exponents in the expansion of x/(1 + x) + Sum_{n >= 1} (-1)^n * x^(2*n-1) / Product_{k = 1..n+1} (1 + x^(2*k-1)) = x^3 - x^5 + x^10 - x^13 + x^20 - x^24 + - .... - Peter Bala, Nov 20 2024

Examples

			For k=1, 2, 3, ..., k*(k+7)/6 is 4/3, 3, 5, 22/3, 10, 13, 49/3, 20, 24, 85/3, 33, ..., and the integer values out of these become the sequence.

Links

Crossrefs

Cf. A000217, A001318, A007494, A165717.

Programs

  • Mathematica
    q=3;s=0;lst={};Do[s+=((n+q)/q);If[IntegerQ[s],AppendTo[lst,s]],{n,6!}];lst
  • PARI
    Vec(x*(-3-2*x+x^2+x^3) / ((1+x)^2*(x-1)^3) + O(x^60)) \\ Colin Barker, Feb 26 2017
  • PARI
    a(n)=if(n%2, 3*n^2 + 16*n + 5, 3*n^2 + 14*n)/8 \\ Charles R Greathouse IV, Feb 27 2017

Formula

From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(-3-2*x+x^2+x^3)/((1+x)^2 * (x-1)^3). (End)
a(n) = Sum_{i=1..n} numerator(i/2) + denominator(i/2). - Wesley Ivan Hurt, Feb 26 2017
From Colin Barker, Feb 26 2017: (Start)
a(n) = (3*n^2 + 14*n) / 8 for n even.
a(n) = (3*n^2 + 16*n + 5) / 8 for n odd. (End)
From Peter Bala, Dec 15 2020: (Start)
a(n) = A001318(n+2) - 2.
Exponents in the expansion of Sum_{n >= 0} x^n * Product_{k = 1..n+1} (1 - x^k) = 1 - x^3 - x^5 + x^10 + x^13 - x^20 - x^24 + + - - .... (End)
Sum_{n>=1} 1/a(n) = 159/98 - 2*Pi/(7*sqrt(3)). - Amiram Eldar, Jul 26 2024
E.g.f.: (x*(19 + 3*x)*cosh(x) + (5 + 17*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Dec 07 2024

Extensions

Definition simplified by R. J. Mathar, Sep 25 2009

A165719 Integers of the form k*(k+9)/8.

Original entry on oeis.org

14, 17, 45, 50, 92, 99, 155, 164, 234, 245, 329, 342, 440, 455, 567, 584, 710, 729, 869, 890, 1044, 1067, 1235, 1260, 1442, 1469, 1665, 1694, 1904, 1935, 2159, 2192, 2430, 2465, 2717, 2754, 3020, 3059, 3339, 3380, 3674, 3717, 4025, 4070, 4392, 4439, 4775
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Keywords

Comments

Only one term is a prime number (17). Are all others composite?
There is no prime other than 17 in the first 1 million terms. - Harvey P. Dale, Jan 07 2020
Integers of the form k+k*(k+1)/8 = k+A000217(k)/4; for k see A047521, for A000217(k)/4 see A154260.

Examples

			for k = 1,2,..., k(k+9)/8 is 5/4, 11/4, 9/2, 13/2, 35/4, 45/4, 14, 17,.. and the integer values out of these become the sequence.

Links

Crossrefs

Cf. A000217, A047521, A154260, A165717, A165718.

Programs

  • Mathematica
    q=4;s=0;lst={};Do[s+=((n+q)/q);If[IntegerQ[s],AppendTo[lst,s]],{n,6!}];lst
Select[Table[(n(n+9))/8,{n,200}],IntegerQ] (* or *) Rest[Flatten[Table[ {9n+8n^2,14+23n+8n^2},{n,0,30}]]] (* or *) LinearRecurrence[{1,2,-2,-1,1},{14,17,45,50,92},60] (* Harvey P. Dale, Jan 07 2020 *)

Formula

From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = 2*n^2 + 6*n + 9/4 - 3*(-1)^n*(2*n+3)/4.
G.f.: x*(-14-3*x+x^3)/((1+x)^2 * (x-1)^3 ). (End)
Sum_{n>=1} 1/a(n) = 89/81 - (sqrt(2)+1)*Pi/9. - Amiram Eldar, Jul 26 2024

Extensions

Definition simplified by R. J. Mathar, Sep 25 2009

A165720 Integers of the form k*(k+11)/10.

Original entry on oeis.org

6, 8, 18, 21, 35, 39, 57, 62, 84, 90, 116, 123, 153, 161, 195, 204, 242, 252, 294, 305, 351, 363, 413, 426, 480, 494, 552, 567, 629, 645, 711, 728, 798, 816, 890, 909, 987, 1007, 1089, 1110, 1196, 1218, 1308, 1331, 1425, 1449, 1547, 1572, 1674, 1700, 1806
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Keywords

Comments

Integers of the form k + k*(k+1)/10 = k + A000217(k)/5. For k see A047208, for A000217(k)/5 see A057569. - R. J. Mathar, Sep 25 2009
Are all terms composite numbers?
Yes. They are alternately of the form (h+2)*(5*h-1)/2 and h*(5*h+11)/2, with h>0. - Bruno Berselli, Dec 22 2016

Links

Crossrefs

Cf. A000217, A047208, A057569, A165717, A165718, A165719, A279895.

Programs

  • Mathematica
    Select[k = Range[0, 130]; k (k + 11)/10, IntegerQ] (* Bruno Berselli, Dec 22 2016 *)

Formula

From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = 5*(2*n^2 + 10*n + 3)/16 - 3*(-1)^n*(5 + 2*n)/16.
G.f.: x*(-6 - 2*x + 2*x^2 + x^3) / ((1 + x)^2*(x - 1)^3). (End)
Sum_{n>=1} 1/a(n) = 514/363 - 2*Pi*sqrt(1+2/sqrt(5))/11. - Amiram Eldar, Jul 26 2024

Extensions

Definition simplified by R. J. Mathar, Sep 25 2009
Corrected A-number in my comment - R. J. Mathar, Oct 30 2009

A222964 Numbers k such that 25*k+36 is a square.

Original entry on oeis.org

0, 13, 37, 76, 124, 189, 261, 352, 448, 565, 685, 828, 972, 1141, 1309, 1504, 1696, 1917, 2133, 2380, 2620, 2893, 3157, 3456, 3744, 4069, 4381, 4732, 5068, 5445, 5805, 6208, 6592, 7021, 7429, 7884, 8316, 8797, 9253, 9760, 10240, 10773, 11277, 11836, 12364, 12949, 13501, 14112, 14688
Offset: 1

Views

Author

Vincenzo Librandi, Apr 07 2013

Keywords

Comments

Also, numbers of the form 25m^2+12*m, where m = 0,-1,1,-2,2,-3,3,... - Bruno Berselli, Apr 07 2013

Links

Crossrefs

Cf. numbers n such that k^2*n+(k+1)^2 is a square: A028552 (k=2), A218864 (k=3), A165717 (k=4).
Cf. numbers of the form k^2*m^2+floor(k^2/2)*m, where m=0,-1,1,-2,2,-3,3,...: A002378 (k=2), A185039 (k=3), A033996 (k=4), this sequence (k=5), A163758 (k=6).

Programs

  • Magma
    [n: n in [0..15000] | IsSquare(25*n+36)];
  • Magma
    I:=[0, 13, 37, 76, 124]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]];
  • Magma
    [0] cat [25*m^2+12*m where m is n*t: t in [-1, 1], n in [1..20]]; // Bruno Berselli, Apr 07 2013
  • Mathematica
    Select[Range[0, 10000], IntegerQ[Sqrt[25 # + 36]]&] (* or *) CoefficientList[Series[x (13 + 24 x + 13 x^2)/((1+x)^2(1-x)^3), {x, 0, 40}], x]
LinearRecurrence[{1,2,-2,-1,1},{0,13,37,76,124},50] (* Harvey P. Dale, Jan 23 2025 *)

Formula

G.f.: x^2*(13+24*x+13*x^2)/((1+x)^2*(1-x)^3).
a(n) = (50*n*(n-1)+(2*n-1)*(-1)^n+1)/8.
a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5).
Sum_{n>=2} 1/a(n) = 25/144 - tan(Pi/50)*Pi/12. - Amiram Eldar, Feb 16 2023

A165721 Integers of the form k*(k+13)/12.

Original entry on oeis.org

4, 14, 22, 25, 35, 55, 69, 74, 90, 120, 140, 147, 169, 209, 235, 244, 272, 322, 354, 365, 399, 459, 497, 510, 550, 620, 664, 679, 725, 805, 855, 872, 924, 1014, 1070, 1089, 1147, 1247, 1309, 1330, 1394, 1504, 1572, 1595, 1665, 1785, 1859, 1884, 1960, 2090
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Keywords

Comments

Integers of the form k+k*(k+1)/12 = k+A000217(k)/6 (see A069497). - R. J. Mathar, Sep 25 2009
Are all terms composite numbers?
Contribution from Zak Seidov, Sep 25 2009: (Start)
Integers of form n(13+n)/12, n=0,1,2,...
Each four terms of the sequence are composite numbers of forms:
{(4+3 m) (1+4 m), (2+3 m) (7+4 m), (2+m) (11+12 m), m (13+12 m)}, m=0,1,2,...
m=0: {4,14,22,25}; m=1: {35,55,69,74}; m=2: {90,120,140,147}, etc. (End)

Links

Crossrefs

Cf. A000217, A069497, A165717, A165718, A165719, A165720.

Programs

  • Mathematica
    q=6;s=0;lst={};Do[s+=((n+q)/q);If[IntegerQ[s],AppendTo[lst,s]],{n,6!}];lst
Select[Table[k (k+13)/12,{k,200}],IntegerQ] (* or *) LinearRecurrence[ {3,-5,7,-7,5,-3,1},{4,14,22,25,35,55,69},50] (* Harvey P. Dale, Jan 30 2013 *)

Formula

From R. J. Mathar, Sep 25 2009: (Start)
a(n) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7).
G.f.: x*(-4-2*x-x^3+x^5)/((x^2+1)^2*(x-1)^3). (End)
Sum_{n>=1} 1/a(n) = 712/507 - (3 + 4*sqrt(3))*Pi/39. - Amiram Eldar, Jul 26 2024

Extensions

Definition simplified by R. J. Mathar, Sep 25 2009
Showing 1-5 of 5 results.

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