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 1-4 of 4 results.

A070262 5th diagonal of triangle defined in A051537.

Original entry on oeis.org

5, 3, 21, 2, 45, 15, 77, 6, 117, 35, 165, 12, 221, 63, 285, 20, 357, 99, 437, 30, 525, 143, 621, 42, 725, 195, 837, 56, 957, 255, 1085, 72, 1221, 323, 1365, 90, 1517, 399, 1677, 110, 1845, 483, 2021, 132, 2205, 575, 2397, 156, 2597, 675, 2805, 182, 3021, 783
Offset: 1

Views

Author

Amarnath Murthy, May 09 2002

Keywords

Links

Crossrefs

Cf. A061037. [From R. J. Mathar, Sep 29 2008]
Cf. A010873, A051537.
Cf. A000466, A002378, A003185, A085027.

Programs

  • Magma
    [LCM(n + 4, n)/GCD(n + 4, n): n in [1..50]]; // G. C. Greubel, Sep 20 2018
  • Mathematica
    Table[ LCM[i + 4, i] / GCD[i + 4, i], {i, 1, 60}]
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{5,3,21,2,45,15,77,6,117,35,165,12},90] (* Harvey P. Dale, Jul 13 2019 *)
  • PARI
    Vec(x*(5 + 3*x + 21*x^2 + 2*x^3 + 30*x^4 + 6*x^5 + 14*x^6 - 3*x^8 - x^9 - 3*x^10) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3) + O(x^60)) \\ Colin Barker, Mar 27 2017
  • PARI
    a(n) = lcm(n+4,n)/gcd(n+4,n); \\ Altug Alkan, Sep 20 2018

Formula

a(n) = lcm(n + 4, n) / gcd(n + 4, n).
From Colin Barker, Mar 27 2017: (Start)
G.f.: x*(5 + 3*x + 21*x^2 + 2*x^3 + 30*x^4 + 6*x^5 + 14*x^6 - 3*x^8 - x^9 - 3*x^10) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n>12. (End)
From Luce ETIENNE, May 10 2018: (Start)
a(n) = n*(n+4)*4^((5*(n mod 4)^3 - 24*(n mod 4)^2 + 31*(n mod 4)-12)/6).
a(n) = n*(n+4)*(37-27*cos(n*Pi)-6*cos(n*Pi/2))/64. (End)
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 11/6.
Sum_{n>=1} (-1)^n/a(n) = 7/6.
Sum_{k=1..n} a(k) ~ (37/192) * n^3. (End)

Extensions

Edited by Robert G. Wilson v, May 10 2002

A085027 a(n) = (4*n+3)*(4*n+7).

Original entry on oeis.org

21, 77, 165, 285, 437, 621, 837, 1085, 1365, 1677, 2021, 2397, 2805, 3245, 3717, 4221, 4757, 5325, 5925, 6557, 7221, 7917, 8645, 9405, 10197, 11021, 11877, 12765, 13685, 14637, 15621, 16637, 17685, 18765, 19877, 21021, 22197, 23405, 24645, 25917, 27221, 28557, 29925, 31325, 32757, 34221
Offset: 0

Views

Author

Gary W. Adamson, Jun 19 2003

Keywords

Comments

1 = 3/7 + Sum_{n>=1} 16/a(n) = 3/7 + 16/77 + 16/165 + 16/285...+...; with partial sums: 3/7, 7/11, 11/15, 15/19, 19/23, ...(4n+3)/(4n+7), ... ==> 1.
With A003185(n) = (4*n+1)*(4*n+5), a bisection of A078371(n) which is a bisection of A061037(n+2).
A quadrisection of A061037(n+2). After A002378(n), A003185(n) and A000466(n+1). - Paul Curtz, Mar 30 2011

Examples

			21 = (3)(7), 77 = (7)(11), 165 = (11)(15), 285 = (15)(19), 437 = (19)(23)...

Links

Crossrefs

Cf. A000466, A002378, A003185, A061037, A078371.

Programs

Formula

a(n) = 16*n^2+40*n+21. - Vincenzo Librandi, Aug 13 2011
From Colin Barker, Jul 11 2012: (Start)
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
G.f.: (21+14*x-3*x^2)/(1-x)^3. (End)
E.g.f.: (21 +56*x +16*x^2)*exp(x). - G. C. Greubel, Sep 20 2018
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=0} 1/a(n) = 1/12.
Sum_{n>=0} (-1)^n/a(n) = Pi/(8*sqrt(2)) + log(sqrt(2)-1)/(4*sqrt(2)) - 1/12. (End)

A128139 Triangle read by rows: matrix product A004736 * A128132.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 5, 4, 1, 5, 7, 7, 5, 1, 6, 9, 10, 9, 6, 1, 7, 11, 13, 13, 11, 7, 1, 8, 13, 16, 17, 16, 13, 8, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10
Offset: 0

Views

Author

Gary W. Adamson, Feb 16 2007

Keywords

Comments

A077028 with the final term in each row omitted.
Interchanging the factors in the matrix product leads to A128140 = A128132 * A004736.
From Gary W. Adamson, Jul 01 2012: (Start)
Alternatively, antidiagonals of an array A(n,k) of sequences with arithmetic progressions as follows:
1, 2, 3, 4, 5, 6, ...
1, 3, 5, 7, 9, 11, ...
1, 4, 7, 10, 13, 16, ...
1, 5, 9, 13, 17, 21, ...
... (End)
From Gary W. Adamson, Jul 02 2012: (Start)
A summation generalization for Sum_{k>=1} 1/(A(n,k)*A(n,k+1)) (formulas copied from A002378, A000466, A085001, A003185):
1 = 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4) + ...
1 = 2/(1)*(3) + 2/(3)*(5) + 2/(5)*(7) + ...
1 = 3/(1)*(4) + 3/(4)*(7) + 3/(7)*(10) + ...
1 = 4/(1)*(5) + 4/(5)*(9) + 4/(9)*(13) + ...
...
As a summation of terms equating to a definite integral:
Integral_{0..1} dx/(1+x) = ... 1 - 1/2 + 1/3 - 1/4 + ... = log(2).
Integral_{0..1} dx/(1+x^2) = 1 - 1/3 + 1/5 - 1/7 + ... = Pi/4 (see A157142)
Integral_{0..1} dx/(1+x^3) = 1 - 1/4 + 1/7 - 1/10 + ... (see A016777)
Integral_{0..1} dx/(1+x^4) = 1 - 1/5 + 1/9 - 1/13 + ... (see A016813). (End)

Examples

			First few rows of the triangle:
  1;
  1,  2;
  1,  3,  3;
  1,  4,  5,  4;
  1,  5,  7,  7,  5;
  1,  6,  9, 10,  9,  6;
  1,  7, 11, 13, 13, 11,  7;
  1,  8, 13, 16, 17, 16, 13,  8;
  1,  9, 15, 19, 21, 21, 19, 15,  9;
  1, 10, 17, 22, 25, 26, 25, 22, 17, 10;
  ...

Crossrefs

Cf. A004736, A128132, A128140, A004006 (row sums).

Formula

A004736 * A128132 as infinite lower triangular matrices.
T(n,k) = k*(1+n-k)+1 = 1 + A094053(n+1,1+n-k). - R. J. Mathar, Jul 09 2012

A173039 Odd numerators of the fractions (1/4-1/n^2), n>= -2.

Original entry on oeis.org

-3, -1, -3, 5, 3, 21, 45, 15, 77, 117, 35, 165, 221, 63, 285, 357, 99, 437, 525, 143, 621, 725, 195, 837, 957, 255, 1085, 1221, 323, 1365, 1517, 399, 1677, 1845, 483, 2021, 2205, 575, 2397, 2597, 675
Offset: 1

Views

Author

Paul Curtz, Nov 21 2010

Keywords

Comments

Odd numbers in A061037, which is extended to negative n, and uses -1 to represent -1/0 at n=0. Trisections are apparently A003185, A000466 and A085027.

Links

Formula

Conjecture: a(n)= +3*a(n-3) -3*a(n-6) +a(n-9). G.f.: ( 3+x+3*x^2-14*x^3-6*x^4-30*x^5-21*x^6-3*x^7-5*x^8 ) / ( (x-1)^3*(1+x+x^2)^3 ). - R. J. Mathar, Dec 02 2010
Showing 1-4 of 4 results.

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