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 1-3 of 3 results.

A115391 a(0)=0; then a(4*k+1)=a(4*k)+(4*k+1)^2, a(4*k+2)=a(4*k+1)+(4*k+3)^2, a(4*k+3)=a(4*k+2)+(4*k+2)^2, a(4*k+4)=a(4*k+3)+(4*k+4)^2.

Original entry on oeis.org

0, 1, 10, 14, 30, 55, 104, 140, 204, 285, 406, 506, 650, 819, 1044, 1240, 1496, 1785, 2146, 2470, 2870, 3311, 3840, 4324, 4900, 5525, 6254, 6930, 7714, 8555, 9516, 10416, 11440, 12529, 13754, 14910, 16206, 17575, 19096, 20540, 22140, 23821, 25670, 27434, 29370
Offset: 0

Views

Author

Pierre CAMI, Mar 15 2006

Keywords

Comments

Probable answer to the riddle in A115603.
Partial sums of the squares of the terms of A116966.

Links

Crossrefs

Cf A000330, A004770

Programs

  • Mathematica
    LinearRecurrence[{2,-1,0,2,-4,2,0,-1,2,-1},{0,1,10,14,30,55,104,140,204,285,406},50] (* Harvey P. Dale, Jul 01 2020 *)

Formula

G.f.: x*(4*x^7-3*x^6+8*x^5+7*x^4+12*x^3-5*x^2+8*x+1) / ((x-1)^4*(x+1)^2*(x^2+1)^2). - Colin Barker, Jul 18 2013
a(n) = (2*n+1)*(2*n*(n+1)+3*(1+cos(n*Pi)-2*cos(n*Pi/2)))/12. - Luce ETIENNE, Feb 01 2017

Extensions

More terms from Stefan Steinerberger, Mar 31 2006
Entry revised by Don Reble, Apr 06 2006
More terms from Colin Barker, Jul 18 2013
Offset adapted to definition by Georg Fischer, Jun 18 2021

A162899 Partial sums of [A052938(n)^2].

Original entry on oeis.org

1, 10, 14, 30, 39, 64, 80, 116, 141, 190, 226, 290, 339, 420, 484, 584, 665, 786, 886, 1030, 1151, 1320, 1464, 1660, 1829, 2054, 2250, 2506, 2731, 3020, 3276, 3600, 3889, 4250, 4574, 4974, 5335, 5776, 6176, 6660, 7101, 7630, 8114, 8690, 9219, 9844, 10420
Offset: 0

Views

Author

Rick L. Shepherd, Jul 16 2009

Keywords

Comments

Another plausible solution, besides A115391 and A116955, to A115603: Each additional term of the partial sums here is the square of a number that alternately differs +2, -1, +2, -1, ..., from the previous number that is squared: a(3) = 30 = 1^2 + 3^2 + 2^2 + 4^2, where 1, 3, 2, 4 display this pattern.

Links

Crossrefs

Cf. A115603, A115391, A116955.

Programs

  • Mathematica
    LinearRecurrence[{2,1,-4,1,2,-1},{1,10,14,30,39,64},50] (* Harvey P. Dale, Sep 26 2020 *)
  • PARI
    a(n) = sum(k=0, n, (if(k%2==0, k+2, k+5)/2)^2)

Formula

a(n) = sum(k=0..n, A052938(n)^2).
a(n) = (60-36*(-1)^n+(109-9*(-1)^n)*n+24*n^2+2*n^3)/24. G.f.: (4*x^4-4*x^3-7*x^2+8*x+1) / ((x-1)^4*(x+1)^2). - Colin Barker, Jul 18 2013

A164765 Partial sums of [A080782^2].

Original entry on oeis.org

1, 10, 14, 30, 66, 91, 140, 221, 285, 385, 529, 650, 819, 1044, 1240, 1496, 1820, 2109, 2470, 2911, 3311, 3795, 4371, 4900, 5525, 6254, 6930, 7714, 8614, 9455, 10416, 11505, 12529, 13685, 14981, 16206, 17575, 19096, 20540, 22140, 23904, 25585
Offset: 1

Views

Author

Carl R. White, Aug 25 2009

Keywords

Comments

Yet another plausible solution to A115603.
The first differences of A115603 are all squares (assuming a prior term of 0), meaning that any sequence beginning 1,3,2,4 is sufficient to account for them; This solution chooses the permutation of integers A080782 = {1,3,2,4,6,5,7,9,8,...}
Ultimately that means this sequence is equal to A000330 for every two out of three consecutive terms, and is greater by 2n+1 where different.

Links

Crossrefs

Original puzzle: A115603; Used in this solution: A080782, A000330; Other solutions: A115391, A116955, A162899

Programs

  • Mathematica
    Accumulate[Array[#+Mod[#+1,3]&,70,0]^2] (* Harvey P. Dale, Mar 29 2013 *)
  • PARI
    Vec(x*(1 + 8*x - 5*x^2 + 10*x^3 + 4*x^4 - x^5 + x^6) / ((1 - x)^4*(1 + x + x^2)^2) + O(x^40)) \\ Colin Barker, Aug 03 2020

Formula

a(n) = ( n(n+1) + 6 - 8*sin^2(Pi*(n+1)/3) )*(2n+1)/6.
a(n) = Sum_{k=0..n} A080782(k)^2.
From Colin Barker, Aug 03 2020: (Start)
G.f.: x*(1 + 8*x - 5*x^2 + 10*x^3 + 4*x^4 - x^5 + x^6) / ((1 - x)^4*(1 + x + x^2)^2).
a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 4*a(n-4) + 2*a(n-5) - a(n-6) + 2*a(n-7) - a(n-8) for n>8.
(End)
Showing 1-3 of 3 results.

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