A001542 -id:A001542 - cp's OEIS frontend

A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350

Offset: 1

Benoit Cloitre, May 10 2003

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have
  a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
  a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ).
It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)
Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 1..2608
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Bisections are A001542 and A002315.

Cf. A084069, A084070, A133566, A079496, A001541, A001653, A008844, A055792, A049629, A108412, A143608.

a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
From Paul Barry, Jun 06 2006: (Start)
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007
Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007
From Peter Bala, Mar 23 2018: (Start)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
sqrt(1 + 2*a(2*n)^2) = A001541(n).
1 + 2*a(2*n)^2 = A055792(n+1).
a(2*n) - a(2*n-1) = A001653(n).
(1 + a(2*n+1)^2)/2 = A008844(n). (End)
a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021

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

Original entry on oeis.org

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

A077444 Numbers k such that (k^2 + 4)/2 is a square.

Original entry on oeis.org

2, 14, 82, 478, 2786, 16238, 94642, 551614, 3215042, 18738638, 109216786, 636562078, 3710155682, 21624372014, 126036076402, 734592086398, 4281516441986, 24954506565518, 145445522951122, 847718631141214, 4940866263896162, 28797478952235758, 167844007449518386

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003
Equivalently, 2*n^2 + 8 is a square.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 2 + n^2/2. - Ctibor O. Zizka, Nov 09 2009
The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013
a((p+1)/2) == 2 (mod p) where p is an odd prime. - Altug Alkan, Mar 17 2016

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Amiram Eldar, Table of n, a(n) for n = 1..1306
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Tanya Khovanova, Recursive Sequences
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
J. J. O'Connor and E. F. Robertson, Pell's Equation. [Broken link]
Eric Weisstein's World of Mathematics, Pell Equation.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

(A077445(n))^2 - 2*a(n) = 8.
First differences of A001541.
Pairwise sums of A001542.
Bisection of A002203 and A080039.
Cf. A001653.

[n: n in [0..10^8] | IsSquare((n^2 + 4) div 2)]; // Vincenzo Librandi, Jun 20 2015

LinearRecurrence[{6,-1},{2,14},30] (* Harvey P. Dale, Jul 25 2018 *)

for(n=1,20,q=(1+sqrt(2))^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec(2*x*(1+x)/(1-6*x+x^2) + O(x^100)) \\ Altug Alkan, Mar 17 2016

a(n) = (((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + ((3 + 2*sqrt(2))^(n-1) - (3 - 2*sqrt(2))^(n-1))) / (2*sqrt(2)).
a(n) = 2*A002315(n-1).
Recurrence: a(n) = 6*a(n-1) - a(n-2), starting 2, 14.
Offset 0, with a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = a^((2n+1)/2) - b^((2n+1)/2). a(n) = 2*(A001109(n+1) + A001109(n)) = (A003499(n+1) - A003499(n))/2 = 2*sqrt(A001108(2n+1)) = sqrt(A003499(2n+1)-2). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Limit_{n->oo} a(n)/a(n-1) = 5.82842712474619009760... = 3 + 2*sqrt(2). See A156035.
From R. J. Mathar, Nov 16 2007: (Start)
G.f.: 2*x*(1+x)/(1-6*x+x^2).
a(n) = 2*(7*A001109(n) - A001109(n+1)). (End)
a(n) = (1+sqrt(2))^(2*n-1) - (1+sqrt(2))^(1-2*n). - Gerson Washiski Barbosa, Sep 19 2010
a(n) = floor((1 + sqrt(2))^(2*n-1)). - Thomas Ordowski, Jun 07 2013
a(n) = sqrt(2*A075870(n)^2-4). - Derek Orr, Jun 18 2015
a(n) = 2*sqrt((2*A001653(n)^2)-1). - César Aguilera, Jul 13 2023
E.g.f.: 2*(1 + exp(3*x)*(sqrt(2)*sinh(2*sqrt(2)*x) - cosh(2*sqrt(2)*x))). - Stefano Spezia, Aug 27 2024

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Joseph L. Pe, Jan 07 2003

Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176.
Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016
From Sergey Pavlov, Mar 14 2017: (Start)
The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

From Benoit Cloitre, Jan 19 2003: (Start)
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
a(n) = A008844(n) - 1. (End)
From R. J. Mathar, Sep 21 2011: (Start)
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = 24*A029546(n-2). (End)
a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020
From Amiram Eldar, Dec 02 2024: (Start)
a(n) = A001542(n-1)*A001542(n).
Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Edited by Bruno Berselli, Nov 17 2016

A173958 Number A(n,k) of spanning trees in C_k X P_n; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 2, 1, 3, 12, 1, 4, 75, 70, 1, 5, 384, 1728, 408, 1, 6, 1805, 31500, 39675, 2378, 1, 7, 8100, 508805, 2558976, 910803, 13860, 1, 8, 35287, 7741440, 140503005, 207746836, 20908800, 80782, 1, 9, 150528, 113742727, 7138643400, 38720000000, 16864848000, 479991603, 470832, 1

Offset: 1

Alois P. Heinz, Nov 26 2010

Every row and every column of the array is a divisibility sequence, i.e., the terms satisfy the property that if n divides m then a(n) divides a(m) provided a(n) > 0. This follows from the representation of the elements of the array as a resultant. - Peter Bala, May 01 2014

			Square array A(n,k) begins:
  1,    2,      3,         4,           5,  ...
  1,   12,     75,       384,        1805,  ...
  1,   70,   1728,     31500,      508805,  ...
  1,  408,  39675,   2558976,   140503005,  ...
  1, 2378, 910803, 207746836, 38720000000,  ...

Alois P. Heinz, Antidiagonals n = 1..60, flattened
Germain Kreweras, Complexité et circuits Eulériens dans les sommes tensorielles de graphes, J. Combin. Theory, B 24 (1978), 202-212. See p. 210. - From _N. J. A. Sloane_, May 27 2012
Eric Weisstein's World of Mathematics, Cycle Graph
Eric Weisstein's World of Mathematics, Path Graph
Wikipedia, Kirchhoff's theorem

Columns k=1-11 give: A000012, A001542, A003690, A003753, A003733, A158880, A158898, A210812, A174001, A210813, A174089.
Rows n=1-2 give: A000027, A006235.
Main diagonal gives A252767.
Cf. A156308.

with(LinearAlgebra):
A:= proc(n, m) local M, i, j;
     if m=1 then 1 else
      M:= Matrix(n*m, shape=symmetric);
      for i to n do
        for j to m-1 do M[m*(i-1)+j, m*(i-1)+j+1]:=-1 od;
        M[m*(i-1)+1, m*i]:= M[m*(i-1)+1, m*i]-1
      od;
      for i to n-1 do
        for j to m do M[m*(i-1)+j, m*i+j]:=-1 od
      od;
      for i to n*m do
        M[i,i]:= -add(M[i,j], j=1..n*m)
      od;
      Determinant(DeleteColumn(DeleteRow(M, 1), 1))
     fi
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..9);
# Crude Maple program from N. J. A. Sloane, May 27 2012:
Digits:=200;
T:=(m,n)->round(Re(evalf(simplify(expand(
m*mul(mul( 4*sin(h*Pi/m)^2+4*sin(k*Pi/(2*n))^2, h=1..m-1), k=1..n-1))))));
# Alternative program using the resultant:
for n from 1 to 10 do seq(k*resultant(simplify((2*(ChebyshevT(k,(x + 2)/2) - 1))/x), simplify(ChebyshevU(n-1,1 - x/2)), x), k = 1 .. 10) end do; # Peter Bala, May 01 2014

t[m_, n_] := m*Product[Product[4*Sin[h*Pi/m]^2 + 4*Sin[k*Pi/(2*n)]^2, {h, 1, m-1}], {k, 1, n-1}]; Table[t[m, n-m+1] // Round, {n, 1, 9}, {m, n, 1, -1}] // Flatten (* Jean-François Alcover, Dec 05 2013, after N. J. A. Sloane *)

A(n,k) = m*Prod(Prod( 4*sin(h*Pi/m)^2+4*sin(k*Pi/(2*n))^2, h=1..m-1), k=1..n-1) [Kreweras]. - From N. J. A. Sloane, May 27 2012

Let T(n,x) and U(n,x) denote the Chebyshev polynomials of the first and second kind respectively. Let R(n,x) = 2*( T(n,(x + 2)/2) - 1 )/x (the row polynomials of A156308). Then the (n,k)-th element of the array equals k times the resultant (R(k,x), U(n-1,(2 - x)/2)). - Peter Bala, May 01 2014 [Corrected by Pontus von Brömssen, Apr 08 2025]

A079496 a(0) = a(1) = 1; thereafter a(2n+1) = 2a(2n) - a(2n-1), a(2n) = 4a(2n-1) - a(2n-2).

Original entry on oeis.org

1, 1, 3, 5, 17, 29, 99, 169, 577, 985, 3363, 5741, 19601, 33461, 114243, 195025, 665857, 1136689, 3880899, 6625109, 22619537, 38613965, 131836323, 225058681, 768398401, 1311738121, 4478554083, 7645370045, 26102926097, 44560482149, 152139002499, 259717522849, 886731088897

Offset: 0

Benoit Cloitre, Jan 20 2003

a(1)=1, a(n) is the smallest integer > a(n-1) such that sqrt(2)*a(n) is closer and > to an integer than sqrt(2)*a(n-1) (i.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(2)*a(n)) < frac(sqrt(2)*a(n-1))).
n such that floor(sqrt(2)*n^2) = n*floor(sqrt(2)*n).
The sequence 1,1,3,5,17,... has g.f. (1+x-3x^2-x^3)/(1-6x^2+x^4); a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*2^(n-k-floor((n+1)/2)); a(n) = -(sqrt(2)-1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4) - (sqrt(2)+1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4); a(2n) = A001541(n) = A001333(2n); a(2n+1) = A001653(n) = A000129(2n+1). - Paul Barry, Jan 22 2005
The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496. - Clark Kimberling, Aug 27 2008
From Richard Choulet, May 09 2010: (Start)
This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z.
The g.f is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n).
The general form of a(n) is given by:
a(2*m)=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)) and
a(2*m+1)=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)). (End)
The integer square roots of floor(n^2/2 + 1) or (A007590 + 1). - Richard R. Forberg, Aug 01 2013

			1 + x + 3*x^2 + 5*x^3 + 17*x^4 + 29*x^5 + 99*x^6 + 169*x^7 + 577*x^8 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 0..2608
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 47, 56.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv:1105.3399 [math.GM], 2011.
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Yujun Yang and Heping Zhang, Kirchhoff Index of linear hexagonal chains, Int. J. Quant. Chem. 108 (2008) 503-512, eq (3.3).
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A010914, A058580.

Cf. A000129, A084068.

H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -1):
a := n -> `if`(n < 3, [1, 1, 3][n+1], 2^(n - 1)*H(n, irem(n, 2), 1/2)):
seq(simplify(a(n)), n=0..26); # Peter Luschny, Sep 03 2019

a[1] = 1; a[2] = 3; a[3] = 5; a[n_] := a[n] = (a[n-1]*a[n-2] + 2) / a[n-3]; Table[a[n], {n, 1, 29}] (* Jean-François Alcover, Jul 17 2013, after Paul D. Hanna *)

{a(n) = n = abs(n); 2^((4-n)\2) * real( (10 + 7 * quadgen(8)) / 2 * (2 + quadgen(8))^(n-3) ) }  /* Michael Somos, Sep 03 2013 */

{a(n) = polcoeff( (1 + x - 3*x^2 - x^3) / (1 - 6*x^2 + x^4) + x * O(x^abs(n)), abs(n))} /* Michael Somos, Sep 03 2013 */

a(2n+1) - a(2n) = a(2n) - a(2n-1) = A001542(n).
a(2n+1) = ceiling((2+sqrt(2))/4*(3+2*sqrt(2))^n), a(2n) = ceiling(1/2*(3+2*sqrt(2))^n).
G.f.: (1 + x - 3*x^2 - x^3)/(1 - 6*x^2 + x^4).
a(n)*a(n+3) - a(n+1)*a(n+2) = 2. - Paul D. Hanna, Feb 22 2003
a(n) = 6*a(n-2) - a(n-4). - R. J. Mathar, Apr 04 2008
a(-n) = a(n) = A010914(n-3)*2^floor((4 - n)/2). - Michael Somos, Sep 03 2013
a(n) = (sqrt(2)*sqrt(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n))/(2+sqrt(2)+(-1)^n*(-2+sqrt(2))). - Gerry Martens, Jun 06 2015
a(n) = 2^(n - 1)*H(n, n mod 2, 1/2) for n >= 3 where H(n, a, b) = hypergeom([a - n/2, b - n/2], [1 - n], -1). - Peter Luschny, Sep 03 2019
a(n) == Pell(n)^(-1) (mod Pell(n+1)) where Pell(n) = A000129(n), use the identity a(n)*Pell(n) - A084068(n-1)*Pell(n+1) = 1, taken modulo Pell(n+1). - Gary W. Adamson, Nov 21 2023
E.g.f.: cosh(x)*(cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2)). - Stefano Spezia, Apr 21 2025

a(0)=1 added by Michael Somos, Sep 03 2013

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844

Offset: 0

Amarnath Murthy, Jun 08 2003

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Edited and extended by Robert G. Wilson v, Jun 15 2003

A031150 Appending a digit to n^2 gives another perfect square.

Original entry on oeis.org

1, 2, 4, 5, 6, 12, 18, 43, 80, 154, 191, 228, 456, 684, 1633, 3038, 5848, 7253, 8658, 17316, 25974, 62011, 115364, 222070, 275423, 328776, 657552, 986328, 2354785, 4380794, 8432812, 10458821, 12484830, 24969660, 37454490

Offset: 1

Patrick De Geest

Square root of 'Squares from A023110 with last digit removed'.

One could include an initial '0', and even list it with multiplicity 3 or 4, since 00, 01, 04 and 09 are all perfect squares: In analogy to corresponding sequences for other bases, this sequence could be defined as sqrt(floor[A023110/10]), see A204512 [base 8], A204517 (base 7), A204519 (base 6), A204521 [base 5], A001353 [base 3], A001542 [base 2]. (For bases 4 and 9, the corresponding sequence contains all integers.) - M. F. Hasler, Jan 16 2012

			5^2 = 25 and 16^2 = 256, so 5 is in the sequence.
115364^2 = 13308852496, 364813^2 = 133088524969.

R. K. Guy, Neg and Reg, preprint, Jan 2012.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Joshua Stucky, Pell's Equation and Truncated Squares, Number Theory Seminar, Kansas State University, Feb 19 2018.
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,38,0,0,0,0,0,0,-1).

Cf. A023110, A030686, A030687, A053784.

See A202303 for the resulting squares.

for i from 1 to 150000 do if (floor(sqrt(10 * i^2 + 9)) > floor(sqrt(10 * i^2))) then print(i) end if end do;

CoefficientList[Series[(x^10 + 2 x^9 + 4 x^8 + 5 x^7 + 18 x^6 + 12 x^5 + 6 x^4 + 5 x^3 + 4 x^2 + 2 x + 1)/(x^14 - 38 x^7 + 1), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 19 2013 *)
LinearRecurrence[{0,0,0,0,0,0,38,0,0,0,0,0,0,-1},{1,2,4,5,6,12,18,43,80,154,191,228,456,684},40] (* Harvey P. Dale, Jun 09 2017 *)

G.f.: x*(x^10+2*x^9+4*x^8+5*x^7+18*x^6+12*x^5+6*x^4+5*x^3+4*x^2+2*x+1) / (x^14-38*x^7+1). - Colin Barker, Jan 30 2013

A050795 Numbers n such that n^2 - 1 is expressible as the sum of two nonzero squares in at least one way.

Original entry on oeis.org

3, 9, 17, 19, 33, 35, 51, 73, 81, 99, 105, 129, 145, 147, 161, 163, 179, 195, 201, 233, 243, 273, 289, 291, 297, 339, 361, 387, 393, 451, 465, 467, 483, 489, 513, 521, 577, 579, 585, 611, 627, 649, 675, 721, 723, 739, 777, 801, 809, 819, 849, 883, 899, 915

Offset: 1

Patrick De Geest, Sep 15 1999

nonn

Analogous solutions exist for the sum of two identical squares z^2-1 = 2.r^2 (e.g. 99^2-1 = 2.70^2). Values of 'z' are the terms in sequence A001541, values of 'r' are the terms in sequence A001542.
Looking at a^2 + b^2 = c^2 - 1 modulo 4, we must have a and b even and c odd. Taking a = 2u, b = 2v and c = 2w - 1 and simplifying, we get u^2 + v^2 = w(w+1). - Franklin T. Adams-Watters, May 19 2008
If n is in this sequence, then so is n^(2^k), for all k >= 0. - Altug Alkan, Apr 13 2016

			E.g. 51^2 - 1 = 10^2 + 50^2 = 22^2 + 46^2 = 34^2 + 38^2.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
E.-B. Escott, Query 2521, L'Intermédiaire des Mathématiciens, 10 (1903), 285. [Contains errors]
Index entries for sequences related to sums of squares

Cf. A050796, A050797, A001541, A001542, A001333.

Cf. A140612, A002378.

t={}; Do[i=c=1; While[iJayanta Basu, Jun 01 2013 *)
Select[Range@ 1000, Length[PowersRepresentations[#^2 - 1, 2, 2] /. {0, } -> Nothing] > 0 &] (* _Michael De Vlieger, Apr 13 2016 *)

select( {is_A050795(n)=#qfbsolve(Qfb(1,0,1),n^2-1,2)}, [1..999]) \\ M. F. Hasler, Mar 07 2022

from itertools import islice, count
from sympy import factorint
def A050795_gen(startvalue=2): # generator of terms >= startvalue
    for k in count(max(startvalue,2)):
        if all(map(lambda d: d[0] % 4 != 3 or d[1] % 2 == 0, factorint(k**2-1).items())):
            yield k
A050795_list = list(islice(A050795_gen(),20)) # Chai Wah Wu, Mar 07 2022

a(n) = 2*A140612(n) + 1. - Franklin T. Adams-Watters, May 19 2008

{k : A025426(k^2-1)>0}. - R. J. Mathar, Mar 07 2022

A051009 Reduced denominators of Newton's iteration for sqrt(2).

Original entry on oeis.org

1, 2, 12, 408, 470832, 627013566048, 1111984844349868137938112, 3497379255757941172020851852070562919437964212608

Offset: 1

Eric W. Weisstein

(2^n)-th Pell numbers. - Sergio Falcon, Dec 04 2008

For n>1, Egyptian fraction expansion of 2-sqrt(2), i.e., 2-sqrt(2) = 1/2 + 1/12 + 1/408 + 1/470832 + ... - Simon Plouffe, Feb 22 2011

			G.f. = x + 2*x^2 + 12*x^3 + 408*x^4 + 470832*x^5 + ...

J. Conrad, Table of n, a(n) for n = 1..12
Neil J. Calkin, Eunice Y. S. Chan and Robert M. Corless, Computational Discovery with Newton Fractals, Bohemian Matrices, & Mandelbrot Polynomials, arXiv:2109.03765 [math.HO], 2021.
Eric Weisstein's World of Mathematics, Newton's Iteration
Eric Weisstein's World of Mathematics, Square Root
Eric Weisstein's World of Mathematics, Pythagoras's Constant

Cf. A000129, A001542, A001601, A051009, A098890.

Table[Simplify[Expand[(1/(2 Sqrt[2])) ((1 + Sqrt[2])^(2^n) - (1 - Sqrt[2])^(2^n))]], {n, 0, 7}] (* Artur Jasinski, Oct 10 2008 *)
Do[Print[Fibonacci[2^n,2]],{n,0,10}] (* Sergio Falcon, Dec 04 2008 *)

a(n) = A000129(2^n).
a(n) = 2*a(n-1)*A001601(n-1). - Joe Keane (jgk(AT)jgk.org), May 31 2002
sqrt(2) = 1 + 1/2 - Sum_{n>=3} (1/a(n)). - Donald S. McDonald, Jan 21 2003
For n>1, a(n) = 2*a(n-1)*sqrt(2*a(n-1)^2+1). - Mario Catalani (mario.catalani(AT)unito.it), May 27 2003
For n>0: a(n) = Sum_{r=0..2^(n-1)-1} binomial(2^n, 2*r+1)*2^r. - Mario Catalani (mario.catalani(AT)unito.it), May 30 2003
For n>=4, a(n) = A098890(n-2) - A098890(n-3). - Kieren MacMillan, Dec 19 2007
a(n+1) = (1/(2*sqrt(2)))*((1 + sqrt(2))^(2^n) - (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
For n>0, a(n) = sqrt((A001601(n)^2-1)/2). - Jose Hortal, Apr 14 2012
a(1)=1, a(2)=2, a(n) = 2 * a(n-1) * cos(2^(n-3) * arccos(3)). - Daniel Suteu, Dec 01 2016
0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 if n>0. - Michael Somos, Dec 01 2016
a(n) = A001542(2^(n-2)). - A.H.M. Smeets, May 28 2017

cp's OEIS Frontend

A084068 a(1) = 1, a(2) = 2; a(2*k) = 2*a(2*k-1) - a(2*k-2), a(2*k+1) = 4*a(2*k) - a(2*k-1).

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

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A077444 Numbers k such that (k^2 + 4)/2 is a square.

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A078522 Numbers k such that (k+1)*(2*k+1) is a perfect square.

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A173958 Number A(n,k) of spanning trees in C_k X P_n; square array A(n,k), n>=1, k>=1, read by antidiagonals.

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A079496 a(0) = a(1) = 1; thereafter a(2*n+1) = 2*a(2*n) - a(2*n-1), a(2*n) = 4*a(2*n-1) - a(2*n-2).

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A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

A079496 a(0) = a(1) = 1; thereafter a(2n+1) = 2a(2n) - a(2n-1), a(2n) = 4a(2n-1) - a(2n-2).