A047447 -id:A047447 - cp's OEIS frontend

A274651 Triangle read by rows: T(n,k), (1<=k<=n), in which each term is the least positive integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

1, 2, 3, 4, 1, 2, 3, 5, 4, 6, 6, 2, 1, 3, 4, 5, 4, 6, 2, 7, 8, 7, 8, 3, 1, 6, 5, 9, 9, 6, 10, 5, 8, 3, 11, 7, 8, 11, 9, 4, 1, 7, 10, 6, 5, 12, 7, 13, 8, 2, 9, 4, 11, 10, 14, 10, 9, 5, 12, 3, 1, 2, 13, 7, 8, 11, 11, 12, 8, 13, 5, 4, 3, 10, 9, 15, 14, 16, 13, 10, 11, 7, 9, 2, 1, 12, 8, 5, 17, 15, 18

Offset: 1

Omar E. Pol, Jul 02 2016

Analog of A269526, but note that this is a right triangle.
The same rule applied to an equilateral triangle gives A269526.
We construct the triangle by reading from left to right in each row, starting with T(1,1) = 1.
Presumably every diagonal and every column is also a permutation of the positive integers, but the proof does not seem so straightforward. Of course, neither the rows nor the antidiagonals are permutations of the positive integers, since they are finite in length.
Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the positive integers is true: see the link in A274650 (duplicated below). - N. J. A. Sloane, Jun 07 2017
It appears that the numbers generally appear for the first time in or near the right border of the triangle.
Theorem 1: the middle diagonal gives A000012 (the all 1's sequence).
Theorem 2: all 1's are in the middle diagonal.
For the proofs of the theorems 1 and 2 see the proofs of the theorems 1 and 2 of A274650 since both sequences are essentially the same.
From Bob Selcoe, Feb 15 2017: (Start)
The columns and diagonal are permutations of the natural numbers. The proofs are essentially the same as the proofs given for the columns and rows (respectively) in A269526.
All coefficients j <= 4 eventually populate in a repeating pattern toward the "middle diagonal" (i.e., relatively near the 1's); this is because we can build the triangle by j in ascending order; that is, we can start by placing all the 1's in the proper cells, then add the 2's, 3's, 4's, 5's, etc. So for i >= 0: since the 1's appear at T(1+2i, 1+i), the 2's appear at T(2+8i, 1+4i), T(3+8i, 3+4i), T(5+8i, 2+4i) and T(6+8i, 4+4i). Accordingly, after the first five 3's appear (at T(2,2), T(4,1), T(5,4), T(7,3) and T(8,6)), the remaining 3's appear at T(11+8i, 5+4i), T(12+8i, 7+4i), T(16+8i, 8+4i) and T(17+8i, 10+4i). Similarly for 4's, after the first 21 appearances, 4's appear at T(44+8i, 21+4i), T(45+8i, 24+4i), T(47+8i, 23+4i) and T(48+8i, 26+4i). So starting at T(41,21), this 16-coefficient pattern repeats at T(41+8i, 21+4i):
n/k  21 22 23 24 25 26
41    1  3
42    2
43    3  1  2
44    4     3
45       2  1  4
46             2
47          4  1
48             3     4
where the next 1 appears at T(49,25), and the pattern repeats at that point from the top left (so T(49,26) = 3, T(50,25) = 2, etc.).
Conjecture: as n gets sufficiently large, all coefficients j>4 will appear in a repeating pattern, populating all rows and diagonals around smaller j's near the "middle diagonal" (while I can offer no formal proof, it appears very likely that this is the case). (End)
From Hartmut F. W. Hoft, Jun 12 2017: (Start)
T(2k+1,k+1) = 1, for all k>=0, and T(n,{n/2,(n+3)/2,(n-1)/2,(n+2)/2}) = 2, for all n>=1 with mod(n,8) = {2,3,5,6} respectively, and no 1's or 2's occur in other positions.
Proof by (recursive) picture:
Positions in the triangle that are empty and those containing the dots of the guiding diagonals contain numbers larger than two.
  n\k 1 2 3 4 5 6 7 8  10  12  14  16  18  20  22  24
   1 |1
   2 |2 .
   3 |  1 2
   4 |      .
   5 |  2 1   .
   6 |      2   .
   7 |      1     .
   8 |____________.
   9 |       |1       .
  10 |       |2 .       .
  11 |       |  1 2       .
  12 |       |      .       .
  13 |       |  2 1   .       .
  14 |       |      2   .       .
  15 |       |      1     .       .
  16 |_____|______________._____.
  17 |       |       |1       .       .
  18 |       |       |2 .       .       .
  19 |       |       |  1 2       .       .
  20 |       |       |      .       .       .
  21 |       |       |  2 1   .       .       .
  22 |       |       |      2   .       .       .
  23 |       |       |      1     .       .       .
  24 |_____|_______|____._______._____._______.
      1 2 3 4 5 6 7 8      12      16      20      24
Consider the center of the triangle. In each octave of rows the columns in the first central quatrain contain a 1 and a 2, and the diagonals in the second central quatrain contain a 1 and a 2. Therefore, no 1's or 2's can occur in the respective downward quatrains of leading columns and trailing diagonals.
The sequence of rows containing 2's is A047447 (n mod 8 = {2,3,5,6}), those containing only 2's is A016825 (n mod 8 = {2,6}), those containing both 1's and 2's is A047621 (n mod 8 = {3,5}), those containing only 1's is A047522 (n mod 8 = {1,7}), and those containing neither 1's nor 2's is A008586 (n mod 8 = {0,4}).
(End)

			Triangle begins:
   1;
   2,  3;
   4,  1,  2;
   3,  5,  4,  6;
   6,  2,  1,  3,  4;
   5,  4,  6,  2,  7,  8;
   7,  8,  3,  1,  6,  5,  9;
   9,  6, 10,  5,  8,  3, 11,  7;
   8, 11,  9,  4,  1,  7, 10,  6,  5;
  12,  7, 13,  8,  2,  9,  4, 11, 10, 14;
  10,  9,  5, 12,  3,  1,  2, 13,  7,  8, 11;
  11, 12,  8, 13,  5,  4,  3, 10,  9, 15, 14, 16;
  13, 10, 11,  7,  9,  2,  1, 12,  8,  5, 17, 15, 18;
  ...
From _Omar E. Pol_, Jun 07 2017: (Start)
The triangle may be reformatted as an isosceles triangle so that the all 1's sequence (A000012) appears in the central column (but note that this is NOT the way the triangle is constructed!):
.
.                  1;
.                2,  3;
.              4,  1,  2;
.            3,  5,  4,  6;
.          6,  2,  1,  3,  4;
.        5,  4,  6,  2,  7,  8;
.      7,  8,  3,  1,  6,  5,  9;
.    9,  6, 10,  5,  8,  3, 11,  7;
.  8, 11,  9,  4,  1,  7, 10,  6,  5;
...
(End)

Michel Marcus, Table of n, a(n) for n = 1..20100
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
N. J. A. Sloane, Notes on A274650 and Proof by Non-Attacking Queens

Cf. A001844 (indices of the 1's).
Cf. A000012 (middle diagonal).
Every diagonal and every column of the right triangle is a permutation of A000027.
Cf. A274650 is the same triangle but with every entry minus 1.
Other sequences of the same family are A269526, A274528, A274820, A274821, A286297, A288530, A288531.
Sequences mentioned in N. J. A. Sloane's proof are A000170, A274616 and A287864.
Cf. A008586, A016825, A047447, A047522, A047621. - Hartmut F. W. Hoft, Jun 12 2017

f[1,1] = 1; (* for 1 < n and 1 <= k <= n *)
f[n_,k_] := f[n,k] = Module[{vals=Sort[Join[Map[f[n, #]&, Range[1, k-1]], Map[f[#, k]&, Range[k, n-1]], Map[f[n-k+#, #]&, Range[1, k-1]], Map[f[n-#, k+#]&, Range[1, Floor[(n-k)/2]]]]], c}, c=Complement[Range[1, Last[vals]], vals]; If[c=={}, Last[vals]+1, First[c]]]
(* computation of rows 1 ... n of triangle *)
a274651[n_] := Prepend[Table[f[i, j], {i, 2, n}, {j, 1, i}], {1}]
Flatten[a274651[13]] (* data *)
TableForm[a274651[13]] (* triangle *)
(* Hartmut F. W. Hoft, Jun 12 2017 *)

T(n,k) = A274650(n-1,k-1) + 1.

A188582 Decimal expansion of sqrt(2) - 1.

Original entry on oeis.org

4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7, 2, 7, 3, 5, 0, 1

Offset: 0

Robert G. Wilson v, Apr 04 2011

"In his Book 'The Theory of Poker,' David Sklansky coined the phrase 'Fundamental Theorem of Poker,' a tongue-in-cheek reference to the Fundamental Theorem of Algebra and Fundamental Theorem of Calculus from introductory texts on those two subjects. The constant [sqrt(2) - 1] appears so often in poker analysis that we will in the same vein go so far as to call it 'the golden mean of poker,' and we call it 'r' for short. We will see this value in a number of important results throughout this book." [Chen and Ankenman]
If a triangle has sides whose lengths form a harmonic progression in the ratio 1/(1 - d) : 1 : 1/(1 + d) then the triangle inequality condition requires that d be in the range 1 - sqrt(2) < d < sqrt(2) - 1. - Frank M Jackson, Oct 01 2013
This constant is the 6th smallest radius r < 1 for which a compact packing of the plane exists, with disks of radius 1 and r. - Jean-François Alcover, Sep 02 2014, after Steven Finch
This constant is also the largest argument of the arctangent function in the Viète-like formula for Pi given by Pi/2^(k+1) = arctan(sqrt(2 - a_(k-1))/a_k), where the index k >= 2 and the nested radicals are defined by recurrence using the relations a_k = sqrt(2 + a_(k-1)), a_1 = sqrt(2). When k = 2 the argument of the arctangent function sqrt(2 - a_1)/a_2 = sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2)) = sqrt(2) - 1 is largest. Consequently, at k = 2 the Viète-like formula for Pi can be written as Pi/8 = arctan(sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2))) = arctan(sqrt(2) - 1) (after Abrarov-Quine, see the article). - Sanjar Abrarov, Jan 07 2017
If r and R are respectively the inradius and the circumradius of a triangle, then the ratio r/R <= 1/2 (Euler inequality), and this maximum value 1/2 is obtained when the triangle is equilateral. Now, for a right triangle, the ratio r/R <= this constant = sqrt(2) - 1, and this maximum value sqrt(2) - 1 is obtained when the right triangle is isosceles. This is the answer to the question 1 of the Olympiade Mathématique Belge Maxi in 2008. - Bernard Schott, Sep 07 2022

			0.414213562373095048801688724209698078569671875376948073...

B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
Bill Chen and Jerrod Ankenman, The Mathematics of Poker, Chpt 14 - You Don't Have To Guess: No-Limit Bet Sizing, p. 153, ConJelCo, LLC, Pittsburgh PA 2006.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, pp. 396 and 486.

G. C. Greubel, Table of n, a(n) for n = 0..10000
S. M. Abrarov and B. M. Quine, A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals, figshare, 4509014, (2017).
Steven R. Finch, Errata and Addenda to Mathematical Constants, p. 62.
Olympiade Mathématique Belge, OMB 2008, Finale Maxi, Question 1.
Index entries for algebraic numbers, degree 2
Index to sequences related to Olympiads.

Cf. A002193, A014176, A020807, A120731, A182168 (sin(Pi/8)), A144981 (cos(Pi/8)).

Sqrt(2) - 1; // G. C. Greubel, Jan 31 2018

Mathematica
```
RealDigits[ Sqrt[2] - 1, 10, 111][[1]]
```

sqrt(2) - 1 \\ G. C. Greubel, Jan 31 2018

Equals exp(asinh(cos(Pi))) = exp(asinh(-1)). - Geoffrey Caveney, Apr 23 2014
Equals tan(Pi/8) = A182168 / A144981 = 1 / A014176. - Bernard Schott, Apr 12 2022
From Antonio Graciá Llorente, Mar 15 2024: (Start)
Equals Product_{k >= 0} ((8*k - 1)*(8*k + 9))/((8*k - 5)*(8*k + 13)).
Equals Product_{k >= 1} A047554(k)/A047447(k). (End)
From  Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 2 for k >= 0.
For example, taking k = 0 and k = 1 yields
Equals 1/(2 + (1*3)/(4 + (5*7)/(4 + (9*11)/(4 + (13*15)/(4 + ... + (4*n + 1)*(4*n + 3)/(4 + ...)))))) and
Equals (21/5) * 1/(10 + (1*3)/(20 + (5*7)/(20 + (9*11)/(20 + (13*15)/(20 + ... + (4*n + 1)*(4*n + 3)/(20 + ...)))))). (End)
Tan(arctan(c) + arctan(c^3)) = 1/2. - Gary W. Adamson, Apr 04 2024

cp's OEIS Frontend

A274651 Triangle read by rows: T(n,k), (1<=k<=n), in which each term is the least positive integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

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