A128217 -id:A128217 - cp's OEIS frontend

A128218 First differences of A128217.

1, 3, 1, 3, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 15, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

John W. Layman, Feb 19 2007

nonn

a(A130883(n-1)) = 2*n-1 and a(m) != 2*n-1 for m < A130883(n-1). - Reinhard Zumkeller, Jun 20 2015

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A128127.

Cf. A130883, A152271 (run lengths after initial term).

a128218 n = a128218_list !! (n-1)
a128218_list = zipWith (-) (tail a128217_list) a128217_list
-- Reinhard Zumkeller, Jun 20 2015

nsrQ[n_]:=Module[{sr=Sqrt[n]},Abs[First[sr-Nearest[{Floor[sr], Ceiling[ sr]}, sr]]]<1/4];Differences[Select[Range[0,250],nsrQ]] (* Harvey P. Dale, May 02 2012 *)

default(realprecision, 10000);
is_A128217(n) = ((abs(sqrt(n)-sqrtint(n))<(1/4)) || (abs(sqrt(n)-(1+sqrtint(n)))<(1/4)));
k=0; n=0; prevm=0; while(k<20000, n++; if(is_A128217(n), k++; write("b128218.txt", k, " ", (n-prevm)); prevm = n)); \\ Antti Karttunen, Jan 16 2025

Let A(1)={1}. Then, for k=2,3,4,..., form A(k) by appending to A(k-1) the term k-1 followed by k-1 1's, if k is even, or by appending to A(k-1) the term k followed by k-1 1's, if k is odd. {a(n)} appears to be the limit of {A(k)} as k->infinity.

Offset changed by Reinhard Zumkeller, Jun 20 2015

A063656 Numbers k such that the truncated square root of k is equal to the rounded square root of k.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 9, 10, 11, 12, 16, 17, 18, 19, 20, 25, 26, 27, 28, 29, 30, 36, 37, 38, 39, 40, 41, 42, 49, 50, 51, 52, 53, 54, 55, 56, 64, 65, 66, 67, 68, 69, 70, 71, 72, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 121

Offset: 0

Floor van Lamoen, Jul 24 2001

Also: take 1, skip 0, take 2, skip 1, take 3, skip 2, ...
The union of sets of numbers in closed intervals [k^2,k^2+k], k >= 0, intervals 0 to 1, 1 to 2, 4 to 6, 9 to 12 etc. - J. M. Bergot, Jun 27 2013
Conjecture: the following definition produces a(n) for n >= 1: a(1) = 1; for n > 1, smallest number > a(n-1) satisfying the condition that a(n) is a square if and only if n is a triangular number. - J. Lowell, May 13 2014
Thus a(2) = 2, because 2 is not a triangular number and not a square; a(3) != 3, because 3 is not a square but is a triangular number; a(3) = 4 is OK because 4 is a square and 3 is a triangular number; etc. [Examples supplied by N. J. A. Sloane, May 13 2014]

			The triangle begins as:
   0;
   1,  2;
   4,  5,  6;
   9, 10, 11, 12;
  16, 17, 18, 19, 20;
  25, 26, 27, 28, 29, 30;
  36, 37, 38, 39, 40, 41, 42;
  49, 50, 51, 52, 53, 54, 55, 56;
  ... - _Stefano Spezia_, Oct 19 2024

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Harry J. Smith)

Cf. A063657, A004201-A004202.
Cf. A128217, A217575.
Essentially partial sums of A051340.

a063656 n = a063656_list !! n
a063656_list = f 1 [0..] where
   f k xs = us ++ f (k + 1) (drop (k - 1) vs) where
                    (us, vs) = splitAt k xs
-- Reinhard Zumkeller, Jun 20 2015

Select[Range[121],Floor[Sqrt[#]]==Round[Sqrt[#]] &] (* Stefano Spezia, Oct 19 2024 *)

{ n=-1; for (m=0, 10^9, if (sqrt(m)%1 < .5, write("b063656.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 27 2009

As a triangle from Stefano Spezia, Oct 19 2024: (Start)
T(n,k) = n^2 + k with 0 <= k <= n.
G.f.: x*(1 + x + 2*y - 4*x*y + 3*x^3*y^2 - x^2*y*(2 + y))/((1 - x)^3*(1 - x*y)^3). (End)

A220098 Manhattan distances between 2n and 1 in the double spiral with positive integers and 1 at the center.

Original entry on oeis.org

1, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 3, 4, 5, 6, 7, 6, 5, 4, 5, 6, 7, 8, 7, 6, 5, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 6, 7, 8, 9, 10, 11, 10, 9, 8, 7, 6, 7, 8, 9, 10, 11, 12, 11, 10, 9, 8, 7, 6, 7, 8, 9, 10, 11, 12, 13

Offset: 1

Alex Ratushnyak, Dec 04 2012

Double spiral begins:
.
  82---84---86---88---90---92---94---96---98
   |
  80   51---53---55---57---59---61---63---65
   |    |                                  |
  78   49   26---28---30---32---34---36   67
   |    |    |                        |    |
  76   47   24   11---13---15---17   38   69
   |    |    |    |              |    |    |
  74   45   22    9    2----4   19   40   71
   |    |    |    |    |    |    |    |    |
  72   43   20    7    1    6   21   42   73
   |    |    |    |    |    |    |    |    |
  70   41   18    5----3    8   23   44   75
   |    |    |              |    |    |    |
  68   39   16---14---12---10   25   46   77
   |    |                        |    |    |
  66   37---35---33---31---29---27   48   79
   |                                  |    |
  64---62---60---58---56---54---52---50   81
                                           |
  99---97---95---93---91---89---87---85---83

			From _Philippe Deléham_, Mar 08 2013: (Start)
As a square array, this begins:
  1,  1,  2,  2,  3,  3,  4,  4,  5, ...
  2,  3,  3,  4,  4,  5,  5,  6,  6, ...
  2,  4,  5,  5,  6,  6,  7,  7,  8, ...
  3,  4,  6,  7,  7,  8,  8,  9,  9, ...
  3,  5,  6,  8,  9,  9, 10, 10, 11, ...
  4,  5,  7,  8, 10, 11, 11, 12, 12, ...
  4,  6,  7,  9, 10, 12, 13, 13, 14, ...
  5,  6,  8,  9, 11, 12, 14, 15, 15, ..., etc.
As a triangle, this begins:
  1
  2, 1
  2, 3, 2
  3, 4, 3, 2
  3, 4, 5, 4, 3
  4, 5, 6, 5, 4, 3, etc. (End)

Cf. A053615, A077043, A128217, A214526.

#include 
#define SIZE 20
int grid[SIZE][SIZE];
int direction[] = {0, -1,  1, 0, 0, 1, -1, 0};
main() {
  int i, j, x1, y1, x2, y2, stepSize;
  int direction1pos=0, direction2pos=4, val;
  x1 = y1 = x2 = y2 = SIZE/2;
  for (val=grid[y1][x1]=1, stepSize=0; ; ++stepSize) {
    if (x1<1 || x1>=SIZE-1 || x2<1 || x2>=SIZE-1) break;
    if (y1<1 || y1>=SIZE-1 || y2<1 || y2>=SIZE-1) break;
    for (i=stepSize|1; i; ++val,--i) {
      x1 += direction[direction1pos  ];
      y1 += direction[direction1pos+1];
      x2 += direction[direction2pos  ];
      y2 += direction[direction2pos+1];
      grid[y1][x1] = val*2;
      grid[y2][x2] = val*2+1;
      printf("%d, ",abs(x1-SIZE/2)+abs(y1-SIZE/2));
    }
    direction1pos = (direction1pos+2) & 7;
    direction2pos = (direction2pos+2) & 7;
  }
  for (i=0; i

step(v, m) = concat(v, vector(m, k, 1+v[#v-k+1]))
a(max_n) = {my(v=[0], k=1); while(#v < max_n+1, v=step(v,k); k++); v[2..max_n+1]} \\ Thomas Scheuerle, Jan 07 2025

A053615(n) = if(n<1, 0, sqrtint(n) - A053615(n - sqrtint(n)))
a(n) = A053615(floor( floor( (sqrtint(n*8) + 1)/2 )^2/2 ) + n) \\ Thomas Scheuerle, Jan 07 2025

abs( a(n) - a(n-1) ) = 1.
From Thomas Scheuerle, Jan 07 2025: (Start)
a(n*(n+1)/2 - k) = 1 + a(n*(n-1)/2 + k) with a(0) = 0 and for 0 <= k < n.
a(n) = A053615(A128217(n+1)). (End)

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A128218 First differences of A128217.

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A063656 Numbers k such that the truncated square root of k is equal to the rounded square root of k.

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A220098 Manhattan distances between 2n and 1 in the double spiral with positive integers and 1 at the center.

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C

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