A001009 -id:A001009 - cp's OEIS frontend

A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597, 244566641436218640

Offset: 0

Eric W. Weisstein

Solution to a*(a-1) = 2b*(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n).
n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - Benoit Cloitre, Apr 15 2003
Place a(n) balls in an urn, of which b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - Paul Weisenhorn, Aug 03 2010
Find base x in which repdigit yy has a square that is repdigit zzzz, corresponding to Diophantine equation zzzz_x = (yy_x)^2; then, solution z = a(n) with x = A002315(n) and y = A001653(n+1) for n >= 1 (see Maurice Protat reference). - Bernard Schott, Dec 21 2022

			For n=4: a(4)=697; b(4)=493; 2*binomial(493,2)=485112=binomial(697,2). - _Paul Weisenhorn_, Aug 03 2010

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
Maurice Protat, Des Olympiades à l'Agrégation, De zzzz_x = (yy_x)^2 à Pell-Fermat, Problème 23, pp. 52-54, Ellipses, Paris, 1997.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Ron Knott, Pythagorean Triples and Online Calculators
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Other 2 sides are A001652 and A001653.
Cf. A001009, A001541, A001542, A011900, A046729, A053141, A084159, A084703, A156035.
See comments in A301383.
Cf. A001109, A001653, A002024, A002315, A029549.

a046090 n = a046090_list !! n
a046090_list = 1 : 4 : map (subtract 2)
   (zipWith (-) (map (* 6) (tail a046090_list)) a046090_list)
-- Reinhard Zumkeller, Jan 10 2012

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018

Digits:=100: seq(round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^(n-1))/2)/2, n=0..20); # Paul Weisenhorn, Aug 03 2010

Join[{1},#+1&/@With[{c=3+2Sqrt[2]},NestList[Floor[c #]+3&,3,20]]] (* Harvey P. Dale, Aug 19 2011 *)
LinearRecurrence[{7,-7,1},{1,4,21},25] (* Harvey P. Dale, Apr 13 2012 *)
a[n_] := (2-ChebyshevT[n, 3]+ChebyshevT[n+1, 3])/4; Array[a, 21, 0] (* Jean-François Alcover, Jul 10 2016, adapted from PARI *)

a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)),x,3))/4

x='x+O('x^30); Vec((1-3*x)/((1-6*x+x^2)*(1-x))) \\ G. C. Greubel, Jul 15 2018

a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by Michel Marcus, Dec 23 2022]
a(n) = 6*a(n-1) - a(n-2) - 2, n >= 2, a(0) = 1, a(1) = 4.
a(n) = (A(n+1) - 3*A(n) + 2)/4 with A(n) = A001653(n).
A001652(n) = -a(-1-n).
From Barry E. Williams, May 03 2000: (Start)
G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)).
a(n) = partial sums of A001541(n). (End)
From Charlie Marion, Jul 01 2003: (Start)
A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1).
Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. (End)
a(n) = 1/2 + ((1-2^(1/2))/4)*(3 - 2^(3/2))^n + ((1+2^(1/2))/4)*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003
2*a(n) = 2*A084159(n) + 1 + (-1)^(n+1) = 2*A046729(n) + 1 - (-1)^(n+1). - Lekraj Beedassy, Jul 16 2004
a(n) = A001109(n+1) - A053141(n). - Manuel Valdivia, Apr 03 2010
From Paul Weisenhorn, Aug 03 2010: (Start)
a(n+1) = round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^n)/2);
b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1).
(End)
a(n)*(a(n)-1)/2 = b(n)*b(n+1) and 2*a(n) - 1 = b(n) + b(n+1), where b(n) = A001109. - Kenneth J Ramsey, Apr 24 2011
T(a(n)) = A011900(n)^2 + A001109(n), where T(n) is the n-th triangular number. See also A001653. - Charlie Marion, Apr 25 2011
a(0)=1, a(1)=4, a(2)=21, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Harvey P. Dale, Apr 13 2012
Limit_{n->oo} a(n+1)/a(n) = 3 + 2*sqrt(2) = A156035. - Ilya Gutkovskiy, Jul 10 2016
a(n) = A001652(n)+1. - Dimitri Papadopoulos, Jul 06 2017
a(n) = (A002315(n) + 1)/2. - Bernard Schott, Dec 21 2022
E.g.f.: (exp(x) + exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - Stefano Spezia, Mar 16 2024
a(n) = A002024(A029549(n))+1. - Pontus von Brömssen, Sep 11 2024

Additional comments from Wolfdieter Lang

Comment moved to A001653 by Claude Morin, Sep 22 2023

A001623 Number of 3 X n reduced (normalized) Latin rectangles.

Original entry on oeis.org

1, 4, 46, 1064, 35792, 1673792, 103443808, 8154999232, 798030483328, 94866122760704, 13460459852344064, 2246551018310998016, 435626600453967929344, 97108406689489312301056, 24658059294992101453262848, 7075100096781964808223653888, 2277710095706779480096994066432, 817555425148510266964075644059648

Offset: 3

N. J. A. Sloane

A Latin rectangle [L_{n,k}] is called normalized [N_{n,k}] if the first row is (0,1, . . . , n-1), and reduced [R_{n,k}] if the first row is (0,1, . . . , n-1) and the first column is (0,1, . . . , k-1). Then L_{n,k} = n! N_{n,k} = (n! (n-1)! /(n-k)!) R_{n,k}.

			G.f. = x^3 + 4*x^4 + 46*x^5 + 1064*x^6 + 35792*x^7 + 1673792*x^8 + ...

S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Robert Israel, Table of n, a(n) for n = 3..254
S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127. [Annotated scanned copy]
D. S. Stones, The many formulas for the number of Latin rectangles, Electron. J. Combin. 17 (2010), A1.
D. S. Stones and I. M. Wanless, Divisors of the number of Latin rectangles, J. Combin. Theory Ser. A 117 (2010), 204-215.
R. J. Stones, S. Lin, X. Liu, G. Wang, On Computing the Number of Latin Rectangles, Graphs and Combinatorics (2016) 32:1187-1202; DOI 10.1007/s00373-015-1643-1.
Index entries for sequences related to Latin squares and rectangles

Cf. A001009.

f:= n-> add(n*factorial(n-3)*factorial(i)*simplify(hypergeom([3*i+3, -n+i], [], 1/2))/(2^(-n+i)*factorial(n-i)),i=0..n):
map(f, [$3..30]); # Robert Israel, Nov 07 2016

Table[Sum[  n (n - 3)! (-1)^j 2^(n -i-j) i!/(n-i-j)! Binomial[3 i + j + 2, j], {i, 0, n}, {j, 0, n - i} ], {n, 3, 25}] (* Wouter Meeussen, Oct 27 2013 *)

A001623 = n->n*(n-3)!*sum(i=0,n,sum(j=0,n-i,(-1)^j*binomial(3*i+j+2,j)<<(n-i-j)/(n-i-j)!)*i!) \\ - M. F. Hasler, Oct 27 2013

a(n) ~ (n-1)!^2/exp(3) ~ 2*Pi*n^(2*n-1)/exp(2*n+3). - Vaclav Kotesovec, Sep 08 2016

Better description Jul 15 1995
Mathematica program, more terms, better definition, comment and Stones link from Wouter Meeussen, Oct 27 2013
Minor corrections by M. F. Hasler, Oct 27 2013

A000573 Number of 4 X n normalized Latin rectangles.

Original entry on oeis.org

4, 56, 6552, 1293216, 420909504, 207624560256, 147174521059584, 143968880078466048, 188237563987982390784, 320510030393570671051776, 695457005987768649183581184, 1888143905499961681708381310976, 6314083806394358817244705266941952, 25655084790196439186603345691314159616

Offset: 4

Brendan McKay and Eric Rogoyski

S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127.

Sheng Lin, Xiaoguang Liu and Douglas S. Stones, Gang Wang, Table of n, K(4,n) for n=4..150
P. G. Doyle, The number of Latin rectangles, arXiv:math/0703896v1 [math.CO], 2007.
B. D. McKay and E. Rogoyski, Latin squares of order ten, Electron. J. Combinatorics, 2 (1995) #N3.
Douglas Stones, Doyle's formula for the number of reduced 6xn Latin rectangles
Douglas Stones, Enumeration Of Latin Squares And Rectangles
D. S. Stones, The many formulas for the number of Latin rectangles, Electron. J. Combin 17 (2010), A1.
D. S. Stones and I. M. Wanless, Divisors of the number of Latin rectangles, J. Combin. Theory Ser. A 117 (2010), 204-215.
R. J. Stones, S. Lin, X. Liu, G. Wang, On Computing the Number of Latin Rectangles, Graphs and Combinatorics (2016) 32:1187-1202; DOI 10.1007/s00373-015-1643-1.
Index entries for sequences related to Latin squares and rectangles

Cf. A003170, A001009.

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A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

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A001623 Number of 3 X n reduced (normalized) Latin rectangles.

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A000573 Number of 4 X n normalized Latin rectangles.

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A000576 a(n) is the number of (n-2) X n normalized Latin rectangles.

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A162544 Triangle giving number L(k,n) of isotopy classes of Latin rectangles.

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A162545 Triangle giving number L(k,n) of normalized Latin rectangles which are lexicographically minimum of its isotopy class.

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