The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A025581 Triangle T(n, k) = n-k, 0 <= k <= n. 146
 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 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 triangle array read by rows: row number k lists first k elements 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 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 Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012. Eric Weisstein's World of Mathematics, Pairing Function. 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. 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 EXAMPLE 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] 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]&, {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 *) PROG (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 (Haskell) a025581 n k = n - k a025581_row n = [n, n-1 .. 0] a025581_tabl = iterate (\xs@(x:_) -> (x + 1) : xs)  -- 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 CROSSREFS A004736(n+1) = 1+a(n). Cf. A025669, A025676, A025683, A002262, A004736, A001477. Cf. A141418 (partial sums per row). Cf. A079904, A257238, A051162, A105125. Sequence in context: A257571 A219649 A292160 * A025669 A025676 A025683 Adjacent sequences:  A025578 A025579 A025580 * A025582 A025583 A025584 KEYWORD nonn,tabl,easy,nice AUTHOR EXTENSIONS Typo in definition corrected by Arkadiusz Wesolowski, Nov 24 2011 Edited: part of name moved to first comment; added definition of trinv in formula. - Wolfdieter Lang, May 12 2015 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified February 17 18:14 EST 2020. Contains 332005 sequences. (Running on oeis4.)