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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.

A206772 Table T(n,k)=max{4*n+k-4,n+4*k-4} n, k > 0, read by antidiagonals.

Original entry on oeis.org

1, 5, 5, 9, 6, 9, 13, 10, 10, 13, 17, 14, 11, 14, 17, 21, 18, 15, 15, 18, 21, 25, 22, 19, 16, 19, 22, 25, 29, 26, 23, 20, 20, 23, 26, 29, 33, 30, 27, 24, 21, 24, 27, 30, 33, 37, 34, 31, 28, 25, 25, 28, 31, 34, 37, 41, 38, 35, 32, 29, 26, 29, 32, 35, 38, 41, 45
Offset: 1

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Author

Boris Putievskiy, Jan 15 2013

Keywords

Comments

In general, let m be natural number. Table T(n,k)=max{m*n+k-m,n+m*k-m}. For m=1 the result is A002024, for m=2 the result is A204004, for m=3 the result is A204008. This sequence is result for m=4.

Examples

			The start of the sequence as table for general case:
  1........m+1..2*m+1..3*m+1..4*m+1..5*m+1..6*m+1 ...
  m+1......m+2..2*m+2..3*m+2..4*m+2..5*m+2..6*m+2 ...
  2*m+1..2*m+2..2*m+3..3*m+3..4*m+3..5*m+3..6*m+3 ...
  3*m+1..3*m+2..3*m+3..3*m+4..4*m+4..5*m+4..6*m+4 ...
  4*m+1..4*m+2..4*m+3..4*m+4..4*m+5..5*m+5..6*m+5 ...
  5*m+1..5*m+2..5*m+3..5*m+4..5*m+5..5*m+6..6*m+6 ...
  6*m+1..6*m+2..6*m+3..6*m+4..6*m+5..6*m+6..6*m+7 ...
  . . .
The start of the sequence as triangle array read by rows for general case:
  1;
  m+1,     m+1;
  2*m+1,   m+2, 2*m+1;
  3*m+1, 2*m+2, 2*m+2, 3*m+1;
  4*m+1, 3*m+2, 2*m+3, 3*m+2, 4*m+1;
  5*m+1, 4*m+2, 3*m+3, 2*m+4, 3*m+3, 4*m+2; 5*m+1;
  6*m+1, 5*m+2, 4*m+3, 3*m+4, 2*m+5, 3*m+4, 4*m+3; 5*m+2, 6*m+1;
  . . .
Row number r contains r numbers: r*m+1, (r-1)*m+2, ... (r-1)*m+2, r*m+1.
		

Crossrefs

Programs

  • Python
    t=int((math.sqrt(8*n-7)-1)/2)
    result=4*(t+1)+3*max(t*(t+1)/2-n,n-(t*t+3*t+4)/2)

Formula

For the general case
a(n) = m*A002024(n) + (m-1)*max{-A002260(n),-A004736(n)}.
a(n) = m*(t+1) + (m-1)*max{t*(t+1)/2-n,n-(t*t+3*t+4)/2}
where t=floor((-1+sqrt(8*n-7))/2).
For m=4
a(n) = 4*(t+1) + 3*max{t*(t+1)/2-n,n-(t*t+3*t+4)/2}
where t=floor((-1+sqrt(8*n-7))/2).