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.

Previous Showing 21-27 of 27 results.

A191740 Dispersion of A047220, (numbers >1 and congruent to 0 or 1 or 3 mod 5), by antidiagonals.

Original entry on oeis.org

1, 3, 2, 6, 5, 4, 11, 10, 8, 7, 20, 18, 15, 13, 9, 35, 31, 26, 23, 16, 12, 60, 53, 45, 40, 28, 21, 14, 101, 90, 76, 68, 48, 36, 25, 17, 170, 151, 128, 115, 81, 61, 43, 30, 19, 285, 253, 215, 193, 136, 103, 73, 51, 33, 22, 476, 423, 360, 323, 228, 173, 123
Offset: 1

Views

Author

Clark Kimberling, Jun 14 2011

Keywords

Comments

For a background discussion of dispersions and their fractal sequences, see A191426. For dispersions of congruence sequences mod 3, mod 4, or mod 5, see A191655, A191663, A191667, A191702.
...
Suppose that {2,3,4,5,6} is partitioned as {x1, x2} and {x3,x4,x5}. Let S be the increasing sequence of numbers >1 and congruent to x1 or x2 mod 5, and let T be the increasing sequence of numbers >1 and congruent to x3 or x4 or x5 mod 5. There are 10 sequences in S, each matched by a (nearly) complementary sequence in T. Each of the 20 sequences generates a dispersion, as listed here:
...
A191722=dispersion of A008851 (0, 1 mod 5 and >1)
A191723=dispersion of A047215 (0, 2 mod 5 and >1)
A191724=dispersion of A047218 (0, 3 mod 5 and >1)
A191725=dispersion of A047208 (0, 4 mod 5 and >1)
A191726=dispersion of A047216 (1, 2 mod 5 and >1)
A191727=dispersion of A047219 (1, 3 mod 5 and >1)
A191728=dispersion of A047209 (1, 4 mod 5 and >1)
A191729=dispersion of A047221 (2, 3 mod 5 and >1)
A191730=dispersion of A047211 (2, 4 mod 5 and >1)
A191731=dispersion of A047204 (3, 4 mod 5 and >1)
...
A191732=dispersion of A047202 (2,3,4 mod 5 and >1)
A191733=dispersion of A047206 (1,3,4 mod 5 and >1)
A191734=dispersion of A032793 (1,2,4 mod 5 and >1)
A191735=dispersion of A047223 (1,2,3 mod 5 and >1)
A191736=dispersion of A047205 (0,3,4 mod 5 and >1)
A191737=dispersion of A047212 (0,2,4 mod 5 and >1)
A191738=dispersion of A047222 (0,2,3 mod 5 and >1)
A191739=dispersion of A008854 (0,1,4 mod 5 and >1)
A191740=dispersion of A047220 (0,1,3 mod 5 and >1)
A191741=dispersion of A047217 (0,1,2 mod 5 and >1)
...
For further information about these 20 dispersions, see A191722.
...
Regarding the dispersions A191722-A191741, there are general formulas for sequences of the type "(a or b mod m)" and "(a or b or c mod m)" used in the relevant Mathematica programs.

Examples

			Northwest corner:
1....3....6....11...20
2....5....10...18...31
4....8....15...26...45
7....13...23...40...68
9....16...28...48...81
12...21...36...61...103
		

Crossrefs

Programs

  • Mathematica
    (* Program generates the dispersion array t of the increasing sequence f[n] *)
    r = 40; r1 = 12;  c = 40; c1 = 12;
    a=3; b=5; c2=6; m[n_]:=If[Mod[n,3]==0,1,0];
    f[n_]:=a*m[n+2]+b*m[n+1]+c2*m[n]+5*Floor[(n-1)/3]
    Table[f[n], {n, 1, 30}]  (* A047220 *)
    mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
    rows = {NestList[f, 1, c]};
    Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
    t[i_, j_] := rows[[i, j]];
    TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]]
      (* A191740 *)
    Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191740  *)

A191741 Dispersion of A047217, (numbers >1 and congruent to 0 or 1 or 2 mod 5), by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 6, 4, 10, 11, 7, 8, 17, 20, 12, 15, 9, 30, 35, 21, 26, 16, 13, 51, 60, 36, 45, 27, 22, 14, 86, 101, 61, 76, 46, 37, 25, 18, 145, 170, 102, 127, 77, 62, 42, 31, 19, 242, 285, 171, 212, 130, 105, 71, 52, 32, 23, 405, 476, 286, 355, 217, 176, 120
Offset: 1

Views

Author

Clark Kimberling, Jun 14 2011

Keywords

Comments

For a background discussion of dispersions and their fractal sequences, see A191426. For dispersions of congruence sequences mod 3, mod 4, or mod 5, see A191655, A191663, A191667, A191702.
...
Suppose that {2,3,4,5,6} is partitioned as {x1, x2} and {x3,x4,x5}. Let S be the increasing sequence of numbers >1 and congruent to x1 or x2 mod 5, and let T be the increasing sequence of numbers >1 and congruent to x3 or x4 or x5 mod 5. There are 10 sequences in S, each matched by a (nearly) complementary sequence in T. Each of the 20 sequences generates a dispersion, as listed here:
...
A191722=dispersion of A008851 (0, 1 mod 5 and >1)
A191723=dispersion of A047215 (0, 2 mod 5 and >1)
A191724=dispersion of A047218 (0, 3 mod 5 and >1)
A191725=dispersion of A047208 (0, 4 mod 5 and >1)
A191726=dispersion of A047216 (1, 2 mod 5 and >1)
A191727=dispersion of A047219 (1, 3 mod 5 and >1)
A191728=dispersion of A047209 (1, 4 mod 5 and >1)
A191729=dispersion of A047221 (2, 3 mod 5 and >1)
A191730=dispersion of A047211 (2, 4 mod 5 and >1)
A191731=dispersion of A047204 (3, 4 mod 5 and >1)
...
A191732=dispersion of A047202 (2,3,4 mod 5 and >1)
A191733=dispersion of A047206 (1,3,4 mod 5 and >1)
A191734=dispersion of A032793 (1,2,4 mod 5 and >1)
A191735=dispersion of A047223 (1,2,3 mod 5 and >1)
A191736=dispersion of A047205 (0,3,4 mod 5 and >1)
A191737=dispersion of A047212 (0,2,4 mod 5 and >1)
A191738=dispersion of A047222 (0,2,3 mod 5 and >1)
A191739=dispersion of A008854 (0,1,4 mod 5 and >1)
A191740=dispersion of A047220 (0,1,3 mod 5 and >1)
A191741=dispersion of A047217 (0,1,2 mod 5 and >1)
...
For further information about these 20 dispersions, see A191722.
...
Regarding the dispersions A191722-A191741, there are general formulas for sequences of the type "(a or b mod m)" and "(a or b or c mod m)" used in the relevant Mathematica programs.

Examples

			Northwest corner:
1....2....5....10...17
3....6....11...20...35
4....7....12...21...36
8....15...26...45...76
9....16...27...46...77
13...22...37...62...105
		

Crossrefs

Programs

  • Mathematica
    (* Program generates the dispersion array t of the increasing sequence f[n] *)
    r = 40; r1 = 12;  c = 40; c1 = 12;
    a=2; b=5; c2=6; m[n_]:=If[Mod[n,3]==0,1,0];
    f[n_]:=a*m[n+2]+b*m[n+1]+c2*m[n]+5*Floor[(n-1)/3]
    Table[f[n], {n, 1, 30}]  (* A047217 *)
    mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
    rows = {NestList[f, 1, c]};
    Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
    t[i_, j_] := rows[[i, j]];
    TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191741 *)
    Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191741  *)

A143977 Rectangular array R by antidiagonals: label each unit square in the first quadrant lattice by its northeast vertex (x,y) and mark squares having |x-y| == 0 (mod 3); then R(m,n) is the number of marked squares in the rectangle [0,m] X [0,n].

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 4, 4, 4, 4, 2, 3, 4, 5, 6, 5, 4, 3, 3, 5, 6, 7, 7, 6, 5, 3, 3, 6, 7, 8, 9, 8, 7, 6, 3, 4, 6, 8, 10, 10, 10, 10, 8, 6, 4, 4, 7, 9, 11, 12, 12, 12, 11, 9, 7, 4, 4, 8, 10, 12, 14, 14, 14, 14, 12, 10, 8, 4, 5, 8, 11, 14, 15, 16, 17, 16, 15, 14, 11, 8, 5
Offset: 1

Views

Author

Clark Kimberling, Sep 06 2008

Keywords

Comments

Rows numbered 3,6,9,12,15,... are, except for initial terms, multiples of (1,2,3,4,5,6,7,...) = A000027.

Examples

			Northwest corner:
  1  1  1  2  2  2  3
  1  2  2  3  4  4  5
  1  2  3  4  5  6  7
  2  3  4  6  7  8 10
  2  4  5  7  9 10 12
		

Crossrefs

Rows and columns: A002264, A004523, A000027, A004772, A047212, et al.

Programs

  • Mathematica
    T[m_,n_]:=Ceiling[m n/3];Flatten[Table[T[m-n+1,n],{m,13},{n,m}]] (* Stefano Spezia, Oct 27 2022 *)

Formula

R(m,n) = ceiling(m*n/3). [Corrected by Stefano Spezia, Oct 27 2022]

A270109 a(n) = n^3 + (n+1)*(n+2).

Original entry on oeis.org

2, 7, 20, 47, 94, 167, 272, 415, 602, 839, 1132, 1487, 1910, 2407, 2984, 3647, 4402, 5255, 6212, 7279, 8462, 9767, 11200, 12767, 14474, 16327, 18332, 20495, 22822, 25319, 27992, 30847, 33890, 37127, 40564, 44207, 48062, 52135, 56432, 60959, 65722, 70727, 75980, 81487, 87254
Offset: 0

Views

Author

Bruno Berselli, Mar 11 2016, at the suggestion of Giuseppe Amoruso in BASE Cinque forum

Keywords

Comments

For n>1, many consecutive terms of the sequence are generated by floor(sqrt(n^2 + 2)^3) + n^2 + 2.
It appears that this is a subsequence of A000037 (the nonsquares).
The primes in the sequence belong to A045326.
Inverse binomial transform is 2, 5, 8, 6, 0, 0, 0, ... (0 continued).

Crossrefs

Subsequence of A001651, A047212.
Cf. A027444: numbers of the form n^3+n*(n+1); A085490: numbers of the form n^3+(n-1)*n.
Cf. A008865: numbers of the form n+(n+1)*(n+2); A130883: numbers of the form n^2+(n+1)*(n+2).

Programs

  • Magma
    [n^3+(n+1)*(n+2): n in [0..50]];
  • Mathematica
    Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]
  • Maxima
    makelist(n^3+(n+1)*(n+2), n, 0, 50);
    
  • PARI
    vector(50, n, n--; n^3+(n+1)*(n+2))
    
  • Sage
    [n^3+(n+1)*(n+2) for n in (0..50)]
    

Formula

O.g.f.: (2 - x + 4*x^2 + x^3)/(1 - x)^4.
E.g.f.: (2 + x)*(1 + x)^2*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
a(n+h) - a(n) + a(n-h) = n^3 + n^2 + (6*h^2+3)*n + (2*h^2+2) for any h. This identity becomes a(n) = n^3 + n^2 + 3*n + 2 if h=0.
a(h*a(n) + n) = (h*a(n))^3 + (3*n+1)*(h*a(n))^2 + (3*n^2+2*n+3)*(h*a(n)) + a(n) for any h, therefore a(h*a(n) + n) is always a multiple of a(n).
a(n) + a(-n) = 2*A059100(n) = A255843(n).
a(n) - a(-n) = 4*A229183(n).

A238738 Expansion of (1 + 2*x + 2*x^2)/(1 - x - 2*x^3 + 2*x^4 + x^6 - x^7).

Original entry on oeis.org

1, 3, 5, 7, 11, 15, 18, 24, 30, 34, 42, 50, 55, 65, 75, 81, 93, 105, 112, 126, 140, 148, 164, 180, 189, 207, 225, 235, 255, 275, 286, 308, 330, 342, 366, 390, 403, 429, 455, 469, 497, 525, 540, 570, 600, 616, 648, 680, 697, 731, 765, 783, 819, 855, 874
Offset: 0

Views

Author

Bruno Berselli, Mar 04 2014

Keywords

Comments

Subsequence of A008732: a(n) = A008732(A047212(n+1)).
See also Deléham's example in A008732: these numbers are in the first (A000566), third (A005475) and fifth (A028895) column.

Examples

			G.f.: 1 + 3*x + 5*x^2 + 7*x^3 + 11*x^4 + 15*x^5 + 18*x^6 + 24*x^7 + ...
		

Crossrefs

Cf. A000212 (see illustration above), A000217, A008732, A211538.

Programs

  • Magma
    m:=60; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+2*x^2)/(1-x-2*x^3+2*x^4+x^6-x^7)));
    
  • Mathematica
    CoefficientList[Series[(1 + 2 x + 2 x^2)/(1 - x - 2 x^3 + 2 x^4 + x^6 - x^7), {x, 0, 60}], x]
  • Maxima
    makelist(coeff(taylor((1+2*x+2*x^2)/(1-x-2*x^3+2*x^4+x^6-x^7), x, 0, n), x, n), n, 0, 60);
    
  • PARI
    Vec((1+2*x+2*x^2)/(1-x-2*x^3+2*x^4+x^6-x^7)+O(x^60))
    
  • Sage
    m = 60; L. = PowerSeriesRing(ZZ, m); f = (1+2*x+2*x^2)/(1-x-2*x^3+2*x^4+x^6-x^7); print(f.coefficients())

Formula

G.f.: (1 + 2*x + 2*x^2) / ((1 - x)^3*(1 + x + x^2)^2).
a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7), with n>6.
a(3k) = k*(5*k + 7)/2 + 1 (A000566);
a(3k+1) = k*(5*k + 11)/2 + 3 (A005475);
a(3k+2) = k*(5*k + 15)/2 + 5 (A028895).
a(n) = (floor(n/3)+1)*(4*n-7*floor(n/3)+2)/2. [Luce ETIENNE, Jun 14 2014]

A260307 a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-6) - a(n-7) - a(n-8) + a(n-9) with a(0) - a(8) as shown below.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 7, 10, 9, 13, 10, 15, 12, 17, 14, 20, 15, 22, 17, 24, 19, 27, 20, 29, 22, 31, 24, 34, 25, 36, 27, 38, 29, 41, 30, 43, 32, 45, 34, 48, 35, 50, 37, 52, 39, 55, 40, 57, 42, 59, 44, 62, 45, 64, 47, 66, 49, 69, 50, 71, 52, 73, 54, 76, 55, 78
Offset: 0

Views

Author

Paul Curtz, Nov 22 2015

Keywords

Comments

A260708 difference table rows have the same nine-step recurrence:
0, 1, 3, 6, 10, 16, 21, 29, 36, 46, 55, 65, 78, 93, ...
1, 2, 3, 4, 6, 5, 8, 7, 10, 9, 13, 10, 15, 12, ... = a(n)
1, 1, 1, 2, -1, 3, -1, 3, -1, 4, -3, 5, -3, 5, ... = b(n)
0, 0, 1, -3, 4, -4, 4, -4, 5, -7, 8, -8, 8, -8, ... (see A042965(n)).
(b(2n) + b(2n+1) = A052901(n+2).)

Crossrefs

Cf. A004767, A010718, A042965, A047212, A047282, A052901, A152467, A260160 (eight-step recurrence), A260699 (nine-step recurrence), A260708.

Programs

  • Magma
    I:=[1,2,3,4,6,5,8,7];[n le 8 select I[n] else Self(n-2) + Self(n-6) - Self(n-8): n in [1..70]]; // Vincenzo Librandi, Dec 26 2015
  • Mathematica
    RecurrenceTable[{a[n] == a[n-2] + a[n-6] - a[n-8], a[0]=1, a[1]=2, a[2]=3, a[3]=4, a[4]=6, a[5]=5, a[6]=8, a[7]=7}, a, {n,0,100}] (* G. C. Greubel, Nov 23 2015 *)
  • PARI
    Vec((x^6+x^5+3*x^4+2*x^3+2*x^2+2*x+1)/((x-1)^2*(x+1)^2*(x^2-x+1)*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Nov 22 2015
    
  • PARI
    vector(100, n, n--; n + (-1)^n *((n+2)\6) + 1) \\ Altug Alkan, Nov 24 2015
    

Formula

a(2n) = A047282(n). a(2n+1) = A047212(n+1).
a(n) = A260708(n+1) - A260708(n).
a(n+6) = a(n) + period of length 2: repeat 7, 5.
a(2n) + a(2n+1) = 3 + 4*n.
a(n) = n + 1 + (-1)^n*A152467(n+2).
From Colin Barker, Nov 22 2015: (Start)
a(n) = a(n-2) + a(n-6) - a(n-8) for n>7.
G.f.: (x^6+x^5+3*x^4+2*x^3+2*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^2-x+1)*(x^2+x+1)).
(End)

A093303 a(n) = a(n-1)*(2n-1) + 2n with a(0)=0.

Original entry on oeis.org

0, 2, 10, 56, 400, 3610, 39722, 516400, 7746016, 131682290, 2501963530, 52541234152, 1208448385520, 30211209638026, 815702660226730, 23655377146575200, 733316691543831232, 24199450820946430690, 846980778733125074186
Offset: 0

Views

Author

Emrehan Halici (emrehan(AT)halici.com.tr), Apr 24 2004

Keywords

Comments

Obviously, a(n) is always an even number. a(2) and a(6) are even semiprimes. - Altug Alkan, Dec 07 2015

Crossrefs

Cf. A005843.

Programs

  • Mathematica
    Flatten[{0,Table[n!*Binomial[2*n-1,n]/2^(n-1)*Sum[2^k*k/(k!*Binomial[2*k-1,k]), {k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Oct 28 2012 *)
  • PARI
    a(n) = if(n==0, 0, n!*binomial(2*n-1,n)/2^(n-1) * sum(k=1, n, 2^k*k/(k!*binomial(2*k-1,k)))) \\ Altug Alkan, Dec 07 2015
    
  • PARI
    a(n) = if(n==0, 0, a(n-1)*(2*n-1) + 2*n); \\ Altug Alkan, Dec 07 2015

Formula

a(n) = n!*C(2*n-1,n)/2^(n-1) * Sum_{k=1..n} 2^k*k/(k!*C(2*k-1,k)), for n>0. - Vaclav Kotesovec, Oct 28 2012
From Altug Alkan, Dec 07 2015: (Start)
a(A047212(k)) mod 10 = 0.
a(A016861(k)) mod 10 = 2.
a(A016885(k)) mod 10 = 6. (End)
a(n) ~ (sqrt(2) + 2*sqrt(Pi)*exp(1/2)*erf(1/sqrt(2))) * 2^n * n^n / exp(n). - Vaclav Kotesovec, Dec 18 2015

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 24 2004
Previous Showing 21-27 of 27 results.