A115599 -id:A115599 - cp's OEIS frontend

A055997 Numbers k such that k*(k - 1)/2 is a square.

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

Offset: 1

Barry E. Williams, Jun 14 2000

Numbers k such that (k-th triangular number - k) is a square.
Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002
Number of closed walks of length 2k on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - Bradley Klee & Bill Gosper, Jul 22 2015
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.

Colin Barker, Table of n, a(n) for n = 1..100
Dario Alpern, a^4+b^3=c^2.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 46, 56.
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A007913, A001541.
A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
a(n) = A001108(n-1)+1.
A001110(n-1)=a(n)*(a(n)-1)/2.
Cf. A001652, A001653, A046090.
Identical to A115599, but with additional leading term.

I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
map(A,[$1..100]); # Robert Israel, Jul 22 2015

Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
(1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
LinearRecurrence[{7,-7,1},{1,2,9},30] (* Harvey P. Dale, Dec 06 2015 *)

Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */

t(n)=(1+sqrt(2))^(n-1);
for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ Hugo Pfoertner, Nov 29 2019

a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020

a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - Charlie Marion, Jul 21 2003; corrected by Michel Marcus, Apr 20 2020
a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004; corrected by Bernard Schott, Apr 20 2020
A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004
a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Feb 21 2006; corrected by Michel Marcus, Apr 20 2020
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with mA001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020

A233073 T(n,k)=Number of nXk 0..3 arrays with no element x(i,j) adjacent to value 3-x(i,j) horizontally, antidiagonally or vertically, top left element zero, and 1 appearing before 2 in row major order.

Original entry on oeis.org

1, 2, 2, 5, 9, 5, 14, 50, 50, 14, 41, 289, 582, 289, 41, 122, 1682, 6854, 6854, 1682, 122, 365, 9801, 80811, 164495, 80811, 9801, 365, 1094, 57122, 952869, 3957778, 3957778, 952869, 57122, 1094, 3281, 332929, 11235652, 95264272, 194998895, 95264272

Offset: 1

R. H. Hardin, Dec 03 2013

Table starts
....1........2...........5.............14................41
....2........9..........50............289..............1682
....5.......50.........582...........6854.............80811
...14......289........6854.........164495...........3957778
...41.....1682.......80811........3957778.........194998895
..122.....9801......952869.......95264272........9622519979
..365....57122....11235652.....2293174089......475027244071
.1094...332929...132484030....55201144642....23452561697310
.3281..1940450..1562171807..1328800293991..1157902539075279
.9842.11309769.18420188169.31986846738550.57168401542703366

			Some solutions for n=4 k=4
..0..0..1..0....0..0..1..1....0..0..0..0....0..1..0..1....0..0..1..1
..1..1..0..0....0..0..1..0....1..0..2..2....1..0..0..0....0..1..1..1
..1..1..0..2....0..1..1..1....1..0..2..2....1..0..0..1....0..1..0..0
..0..1..0..2....1..3..1..3....1..0..0..2....0..0..1..1....0..0..0..2

R. H. Hardin, Table of n, a(n) for n = 1..264

Column 1 is A007051(n-1)

Column 2 is A115599

Empirical for column k:
k=1: a(n) = 4*a(n-1) -3*a(n-2)
k=2: a(n) = 7*a(n-1) -7*a(n-2) +a(n-3)
k=3: a(n) = 14*a(n-1) -28*a(n-2) +24*a(n-3) -11*a(n-4) +2*a(n-5) for n>6
k=4: [order 11] for n>12
k=5: [order 21] for n>23
k=6: [order 58] for n>60

A158464 Number of distinct squares in row n of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1

Offset: 0

Reinhard Zumkeller, Mar 19 2009

nonn

It seems that some subset of terms in A055997 (A115599) gives the positions of 3's. E.g., we have a(9) = a(50) = a(289) = a(9801) = 3, but on the other hand, a(1682) = a(57122) = 2. - Antti Karttunen, Nov 03 2017

			a(8) = #{1} = 1;
a(9) = #{1,9,36} = 3.

Antti Karttunen, Table of n, a(n) for n = 0..11111
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000290, A055997 (A115599).

A158464 := proc(n)
    local sqset,k ;
    sqset := {} ;
    for k from 0 to n do
        P := binomial(n,k) ;
        if issqr(P) then
            sqset := sqset union {P} ;
        end if;
    end do:
    nops(sqset) ;
end proc:
seq(A158464(n),n=0..120) ; # R. J. Mathar, Jul 09 2016

CountDistinct /@ Table[Sqrt@ Binomial[n, k] /. k_ /; ! IntegerQ@ k -> Nothing, {n, 0, 104}, {k, 0, n}] (* Michael De Vlieger, Nov 03 2017 *)

A158464(n) = sum(k=0,n\2,issquare(binomial(n,k))); \\ Antti Karttunen, Nov 03 2017

a(n) = Sum_{k=0..floor(n/2)} A010052(A007318(n,k));

a(A000290(n)) > 1 for n > 1.

More terms from Antti Karttunen, Nov 03 2017

cp's OEIS Frontend

A055997 Numbers k such that k*(k - 1)/2 is a square.

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A233073 T(n,k)=Number of nXk 0..3 arrays with no element x(i,j) adjacent to value 3-x(i,j) horizontally, antidiagonally or vertically, top left element zero, and 1 appearing before 2 in row major order.

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A158464 Number of distinct squares in row n of Pascal's triangle.

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Maple

Mathematica

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