A025581 Triangle read by rows: T(n, k) = n-k, for 0 <= k <= n.

0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 2, 1, 0, 5, 4, 3, 2, 1, 0, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 8, 7, 6, 5, 4, 3, 2, 1, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3

Offset: 0

Author

David W. Wilson

Keywords

• nonn
• tabl
• easy
• nice

Comments

Decreasing integers m to 0 followed by decreasing integers m+1 to 0, etc.

The point with coordinates (x = A025581(n), y = A002262(n)) sweeps out the first quadrant by upwards antidiagonals. N. J. A. Sloane, Jul 17 2018

The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 0, k >= 0) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002

Riordan array (x/(1-x)^2, x). - Philippe Deléham, Feb 18 2012

a(n,k) = (A214604(n,k) - A214661(n,k)) / 2. - Reinhard Zumkeller, Jul 25 2012

Sequence B is called a reverse reluctant sequence of sequence A if B is a triangular array read by rows such that row number k lists the first k terms of the sequence A in reverse order. This sequence is the reverse reluctant sequence of sequence 0,1,2,3,..., the nonnegative integers A001477. - Boris Putievskiy, Dec 13 2012

A problem posed by François Viète (Vieta) in his book Zeteticorum liber quinque (1593), liber 2, problem 19 (quoted in the Alten et al. reference, on p. 292) is to find for a rectangle (a >= b >= 1) with given a^3 - b^3, name it C, and a*b, name it F, the difference a-b, name it x. This is a simple exercise which Viète found remarkable. It reduces to a standard cubic equation for x, namely x^3 + 3*F*x = C. Proof: Use the square of the diagonal d^2 = a^2 + b^2. Then (i) C = a^3 - b^3 = (a - b)*(a^2 + b^2 + a*b) = x*(d^2 + F). (ii) use the trivial relation d^2 = (a-b)^2 + 2*a*b = x^2 + 2*F, to eliminate d^2 in (i). End of the Proof. Here for positive integers a = n and b = k: (T(n, k)^2 + 3*A079904(n, k))*T(n, k) = A257238(n, k) (also true for n = k = 0). - Wolfdieter Lang, May 12 2015

See a comment on A051162 on the cubic equation for S = a+b in terms of Cplus = a^3 + b^3 and D = a - b. This equation leads to a - b = sqrt((4*Cplus -S^3)/(3*S)). - Wolfdieter Lang, May 15 2015

The entries correspond to the first of the 2 coordinates of the Cantor Pairs, specifically x=w-(CPKey-(w^2+w)/2), where w=floor((sqrt(8*CPKey+1)-1)/2) and CPKey=Cantor Pair key (A001477). The second of the coordinate pairs is A002262. - Bill McEachen, Sep 12 2015

Examples

			The triangle T(n, k) begins (note that one could use l <= k <= n, for any integer l, especially 1):
  n\k  0 1 2 3 4 5 6 7 8 9 10 ...
   0:  0
   1:  1 0
   2:  2 1 0
   3:  3 2 1 0
   4:  4 3 2 1 0
   5:  5 4 3 2 1 0
   6:  6 5 4 3 2 1 0
   7:  7 6 5 4 3 2 1 0
   8:  8 7 6 5 4 3 2 1 0
   9:  9 8 7 6 5 4 3 2 1 0
  10: 10 9 8 7 6 5 4 3 2 1 0
  ... [formatted by _Wolfdieter Lang_, May 12 2015]
		

References

• H.-W. Alten et al., 4000 Jahre Algebra, 2. Auflage, Springer, 2014, p. 203.

Links

• Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened
• Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
• Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
• Michael Somos, Sequences used for indexing triangular or square arrays
• Eric Weisstein's World of Mathematics, Pairing Function.

Crossrefs

Cf. A025669, A025676, A025683, A002262, A004736, A001477.

Cf. A141418 (partial sums per row).

Cf. A079904, A257238, A051162, A105125.

Programs

Haskell

a025581 n k = n - k
a025581_row n = [n, n-1 .. 0]
a025581_tabl = iterate (\xs@(x:_) -> (x + 1) : xs) [0]
-- Reinhard Zumkeller, Aug 04 2014, Jul 22 2012, Mar 07 2011

Magma

/* As triangle */ [[(n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Sep 13 2015

Maple

A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1): seq(A025581(n), n=0..100);

Mathematica

Flatten[NestList[Prepend[#, #[[1]]+1]&, {0}, 13]] (* Jean-François Alcover, May 17 2011 *)
With[{nn=20},Flatten[Table[Join[{0},Reverse[Range[i]]],{i,nn}]]] (* Harvey P. Dale, Dec 31 2014 *)
Table[Range[n,0,-1],{n,0,15}]//Flatten (* Harvey P. Dale, Aug 01 2020 *)

PARI

a(n)=binomial(1+floor(1/2+sqrt(2+2*n)),2)-(n+1) /* produces a(n) */

PARI

t1(n)=binomial(floor(3/2+sqrt(2+2*n)),2)-(n+1) /* A025581 */

PARI

t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262 */

PARI

apply( {A025581(n)=binomial(sqrtint(8*n+1)\/2+1,2)-n-1}, [0..90]) \\ M. F. Hasler, Dec 06 2019

Python

from math import isqrt, comb
def A025581(n): return comb((m:=isqrt(k:=n+1<<1))+(k>m*(m+1))+1,2)-n-1 # Chai Wah Wu, Nov 08 2024

Formula

T(n, k) = n-k, for 0 <= k <= n.

As a sequence: a(n) = (((trinv(n)-1)*(((1/2)*trinv(n))+1))-n), with trinv(n) = floor((1+sqrt(1+8*n))/2). Cf. A002262.

a(n) = A004736(n+1) - 1.

G.f. for T(n,k): y / ((1-x)^2 * (1-x*y)). - Ralf Stephan, Jan 25 2005

For the cubic equation satisfied by T(n, k) see the comment on a problem by Viète above. - Wolfdieter Lang, May 12 2015

G.f. for a(n): -(1-x)^(-2) + (1-x)^(-1) * Sum_{n>=0} (n+1)*x^(n*(n+1)/2). The sum is related to Jacobi theta functions. - Robert Israel, May 12 2015

T(n, k) = sqrt((4*A105125(n, k) - A051162(n, k)^3)/(3*A051162(n, k))). See a comment above. - Wolfdieter Lang, May 15 2015

a(n) = (1/2)*(t^2 + t - 2*n - 2), where t = floor(sqrt(2*n+1) + 1/2) = round(sqrt(2*n+1)). - Ridouane Oudra, Dec 01 2019

a(n) = ((1/2) * ceiling((-1 + sqrt(9 + 8 * n))/2) * ceiling((1 + sqrt(9 + 8 * n))/2)) - n - 1. - Ryan Jean, Apr 22 2022

Extensions

Typo in definition corrected by Arkadiusz Wesolowski, Nov 24 2011

Edited (part of name moved to first comment; definition of trinv added in formula) by Wolfdieter Lang, May 12 2015