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 1-6 of 6 results.

A219190 Numbers of the form k*(5*k+1), where k = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

0, 4, 6, 18, 22, 42, 48, 76, 84, 120, 130, 174, 186, 238, 252, 312, 328, 396, 414, 490, 510, 594, 616, 708, 732, 832, 858, 966, 994, 1110, 1140, 1264, 1296, 1428, 1462, 1602, 1638, 1786, 1824, 1980, 2020, 2184, 2226, 2398, 2442, 2622, 2668, 2856, 2904, 3100
Offset: 1

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Author

Bruno Berselli, Nov 14 2012

Keywords

Comments

Equivalently, numbers m such that 20*m+1 is a square.
Also, integer values of h*(h+1)/5.
More generally, for the numbers of the form n*(k*n+1) with n in A001057, we have:
. generating function (offset 1): x^2*(k-1+2*x+(k-1)*x^2)/((1+x)^2*(1-x)^3);
. n-th term: b(n) = (2*k*n*(n-1)+(k-2)*(-1)^n*(2*n-1)+k-2)/8;
. first differences: (n-1)*((-1)^n*(k-2)+k)/2;
. b(2n+1)-b(2n) = 2*n (independent from k);
. (4*k)*b(n)+1 = (2*k*n+(k-2)*(-1)^n-k)^2/4.

Crossrefs

Subsequence of A011858.
Cf. A090771: square roots of 20*a(n)+1 (see the first comment).
Cf. numbers of the form n*(k*n+1) with n in A001057: k=0, A001057; k=1, A110660; k=2, A000217; k=3, A152749; k=4, A074378; k=5, this sequence; k=6, A036498; k=7, A219191; k=8, A154260.
Cf. similar sequences listed in A219257.

Programs

  • Magma
    k:=5; f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]];
    
  • Magma
    I:=[0,4,6,18,22]; [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]]; // Vincenzo Librandi, Aug 18 2013
  • Mathematica
    Rest[Flatten[{# (5 # - 1), # (5 # + 1)} & /@ Range[0, 25]]]
    CoefficientList[Series[2 x (2 + x + 2 x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
    LinearRecurrence[{1,2,-2,-1,1},{0,4,6,18,22},50] (* Harvey P. Dale, Jan 21 2015 *)

Formula

G.f.: 2*x^2*(2 + x + 2*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (10*n*(n-1) + 3*(-1)^n*(2*n - 1) + 3)/8.
a(n) = 2*A057569(n) = A008851(n+1)*A047208(n)/5.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Jan 21 2015
Sum_{n>=2} 1/a(n) = 5 - sqrt(1+2/sqrt(5))*Pi. - Amiram Eldar, Mar 15 2022
a(n) = A132356(n-1)/2, n >= 1. - Bernard Schott, Mar 15 2022

A008738 a(n) = floor((n^2 + 1)/5).

Original entry on oeis.org

0, 0, 1, 2, 3, 5, 7, 10, 13, 16, 20, 24, 29, 34, 39, 45, 51, 58, 65, 72, 80, 88, 97, 106, 115, 125, 135, 146, 157, 168, 180, 192, 205, 218, 231, 245, 259, 274, 289, 304, 320, 336, 353, 370, 387, 405, 423, 442, 461, 480, 500, 520, 541, 562, 583, 605, 627, 650, 673
Offset: 0

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Author

Keywords

Comments

Without initial zeros, Molien series for 3-dimensional group [2+,n] = 2*(n/2).

Crossrefs

Cf. A011858. Partial sums of A288156.

Programs

  • GAP
    List([0..60], n-> Int((n^2 + 1)/5)); # G. C. Greubel, Aug 03 2019
  • Magma
    [(n^2+1) div 5: n in [0..60]]; // Bruno Berselli, Oct 28 2011
    
  • Mathematica
    Floor[(Range[0,60]^2 + 1)/5] (* G. C. Greubel, Aug 03 2019 *)
  • PARI
    a(n)=(n^2+1)\5;
    
  • Sage
    [floor((n^2+1)/5) for n in (0..60)] # G. C. Greubel, Aug 03 2019
    

Formula

G.f.: x^2*(1+x^3)/((1-x)^2*(1-x^5)) = x^2*(1+x)*(1-x+x^2)/( (1-x)^3 *(1+x+x^2+x^3+x^4) ).
a(n+2)= A249020(n) + A249020(n-1). - R. J. Mathar, Aug 11 2021

Extensions

More terms from Philip Mummert (s1280900(AT)cedarville.edu), May 10 2000

A194200 [sum{(k*e) : 1<=k<=n}], where [ ]=floor, ( )=fractional part.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 6, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 12, 13, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 22, 22, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 34, 34, 35, 36, 37
Offset: 1

Views

Author

Clark Kimberling, Aug 19 2011

Keywords

Comments

The defining [sum] is equivalent to
...
a(n)=[n(n+1)r/2]-sum{[k*r] : 1<=k<=n},
...
where []=floor and r=sqrt(2). Let s(n) denote the n-th partial sum of the sequence a; then the difference sequence d defined by d(n)=s(n+1)-s(n) gives the runlengths of a.
...
Examples:
...
r...........a........s....
1/2......A002265...A001972
1/3......A002264...A001840
2/3......A002264...A001840
1/4......A194220...A194221
1/5......A194222...A118015
2/5......A057354...A011858
3/5......A194222...A118015
4/5......A057354...A011858
1/6......A194223...A194224
3/7......A057357...A194229
1/8......A194235...A194236
3/8......A194237...A194238
sqrt(2)..A194161...A194162
sqrt(3)..A194163...A194164
sqrt(5)..A194169...A194170
sqrt(6)..A194173...A194174
tau......A194165...A194166; tau=(1+sqrt(5))/2
e........A194200...A194201
2e.......A194202...A194203
e/2......A194204...A194205
pi.......A194206...A194207

Examples

			a(5)=[(e)+(2e)+(3e)+4(e)+5(e)]
    =[.718+.436+.154+.873+.591]
    =[2.77423]=2.
		

Crossrefs

Cf. A194201.

Programs

  • Mathematica
    r = E;
    a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]
    Table[a[n], {n, 1, 90}]  (* A194200 *)
    s[n_] := Sum[a[k], {k, 1, n}]
    Table[s[n], {n, 1, 100}] (* A194201 *)

A211434 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w+2x+5y=0.

Original entry on oeis.org

1, 1, 5, 9, 17, 25, 33, 45, 57, 73, 89, 105, 125, 145, 169, 193, 217, 245, 273, 305, 337, 369, 405, 441, 481, 521, 561, 605, 649, 697, 745, 793, 845, 897, 953, 1009, 1065, 1125, 1185, 1249, 1313, 1377, 1445, 1513, 1585, 1657, 1729, 1805, 1881
Offset: 0

Views

Author

Clark Kimberling, Apr 11 2012

Keywords

Comments

For a guide to related sequences, see A211422.
Also, a(n) is the number of ordered pairs (w,x) with both terms in {-n,...,0,...,n} and w+2x divisible by 5. If (w,x) is such a pair it is easy to see that (-x,w), (-w,-x), and (x,-w) also are such pairs. If both w and x are nonzero these four pairs lie one in each quadrant. If one of w or x is zero, the other must be a multiple of 5. This means that a(n) equals 4*A211523(n) (the nonzero pairs) plus 4*floor(n/5) + 1 (pairs with w or x equal to zero). Since the sequences A211523(n), floor(n/5), and the constant sequence all satisfy the recurrence conjectured in the formula section, a(n) must also satisfy the recurrence, so this proves the conjecture. - Pontus von Brömssen, Jan 17 2020

Crossrefs

Programs

  • Magma
    a:=[]; for n in [0..50] do m:=0; for i, j in [-n..n] do if (i+2*j) mod 5 eq 0  then m:=m+1; end if; end for; Append(~a, m); end for; a; // Marius A. Burtea, Jan 19 2020
    
  • Magma
    R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1 - x + 4*x^2 + 4*x^4 - x^5 + x^6) / ((1 - x)^3*(1 + x + x^2 + x^3 + x^4)))); // Marius A. Burtea, Jan 19 2020
  • Mathematica
    t[n_] := t[n] = Flatten[Table[w + 2 x + 5 y, {w, -n, n}, {x, -n, n}, {y, -n, n}]]
    c[n_] := Count[t[n], 0]
    t = Table[c[n], {n, 0, 70}]  (* A211434 *)
    (t - 1)/4                    (* A011858 *)

Formula

Conjectures from Colin Barker, May 15 2017: (Start)
G.f.: (1 - x + 4*x^2 + 4*x^4 - x^5 + x^6) / ((1 - x)^3*(1 + x + x^2 + x^3 + x^4)).
a(n) = 2*a(n-1) - a(n-2) + a(n-5) - 2*a(n-6) + a(n-7) for n>6.
(End)
a(n) = (4*n*(n+1) + c(n))/5, where c(n) is 5 if n is 0 or 4 (mod 5), -3 if n is 1 or 3 (mod 5), and 1 if n is 2 (mod 5). - Pontus von Brömssen, Jan 17 2020

A227353 Number of lattice points in the closed region bounded by the graphs of y = 3*x/5, x = n, and y = 0, excluding points on the x-axis.

Original entry on oeis.org

0, 1, 2, 4, 7, 10, 14, 18, 23, 29, 35, 42, 49, 57, 66, 75, 85, 95, 106, 118, 130, 143, 156, 170, 185, 200, 216, 232, 249, 267, 285, 304, 323, 343, 364, 385, 407, 429, 452, 476, 500, 525, 550, 576, 603, 630, 658, 686, 715, 745, 775, 806, 837, 869, 902, 935
Offset: 1

Views

Author

Clark Kimberling, Jul 08 2013

Keywords

Comments

See A227347.

Examples

			a(1) = floor(3/5) = 0; a(2) = floor(6/5) = 1; a(3) = a(2) + floor(9/5) = 2; a(4) = a(3) + floor(12/5) = 4.
		

Crossrefs

Cf. A057355 (first differences).

Programs

  • Mathematica
    z = 150; r = 3/5; k = 1; a[n_] := Sum[Floor[r*x^k], {x, 1, n}]; t = Table[a[n], {n, 1, z}]
  • PARI
    a(n) = (3*n^2-n)\10; \\ Kevin Ryde, Mar 15 2022
    
  • Python
    a = lambda n: n*(3*n-1)//10 # Gennady Eremin, Mar 20 2022

Formula

a(n) = 2*a(n-1) - a(n-2) + a(n-5) - 2*a(n-6) + a(n-7).
G.f.: (x*(1 + x^2 + x^3))/((1 - x)^3*(1 + x + x^2 + x^3 + x^4)).
According to Wolfram Alpha, a(n) = floor(Re(E(n^2|Pi))) where E(x|m) is the incomplete elliptic integral of the second kind. - Kritsada Moomuang, Jan 28 2022
a(n) = a(n-1) + floor(3*n/5), n > 1. - Gennady Eremin, Mar 15 2022
a(n) = floor(n*(3*n-1)/10). - Kevin Ryde, Mar 15 2022

A376080 a(n) is the highest degree of the rational function in the recursive substitution {y1, y2} -> {y2, (y2 + 1)/(y1*y2)} after n steps.

Original entry on oeis.org

1, 1, 1, 2, 4, 5, 7, 10, 13, 16, 20, 25, 29, 34, 40, 46, 52, 59, 67, 74, 82, 91, 100, 109, 119, 130, 140, 151, 163, 175, 187, 200, 214, 227, 241, 256, 271, 286, 302, 319, 335, 352, 370, 388, 406, 425, 445, 464, 484, 505, 526, 547, 569, 592, 614, 637, 661, 685, 709, 734, 760, 785, 811, 838, 865
Offset: 0

Views

Author

Thomas Scheuerle, Sep 09 2024

Keywords

Comments

An example where the degree of the n-th iterate of a rational map exhibits polynomial growth. Also an example for exponential growth was given in the thesis from Khaled Hamad by A011782.

Crossrefs

Cf. A011782 (highest degree of the rational function in the substitution: {y1, y2} -> {y2, y2 + y1/y2}).

Programs

  • Mathematica
    A376080[n_] := Ceiling[(3*n*(n - 1) + 8)/14];
    Array[A376080, 100, 0] (* Paolo Xausa, Sep 23 2024 *)
  • PARI
    r(v) = [v[2], (v[2]+1)/(v[1]*v[2])];
    a(n) = {my(v = [x,x]); if(n < 2, 1,for(k=0, n-2, v = r(v)); poldegree(numerator(v[2])))};

Formula

G.f.: (1 - x + x^3 + x^4 - x^5 + x^6 + x^8)/(1 - 2*x + x^2 - x^7 + 2*x^8 - x^9).
a(n) = ceiling((3*n^2 - 3*n + 8)/14).
a(n) = 2*a(n-1) - a(n-2) + a(n-7) - 2*a(n-8) + a(n-9).
a(n) = a(n-7) + 3*(n-7) + 9.
(2*a(n+6) - a(n+5) - 2*a(n-1) + a(n-2) - 9)/3 = n.
Showing 1-6 of 6 results.