A008844 -id:A008844 - cp's OEIS frontend

A165518 Perfect squares (A000290) that can be expressed as the sum of four consecutive triangular numbers (A000217).

4, 100, 3364, 114244, 3880900, 131836324, 4478554084, 152139002500, 5168247530884, 175568277047524, 5964153172084900, 202605639573839044, 6882627592338442564, 233806732499933208100, 7942546277405390632804, 269812766699283348307204, 9165691521498228451812100, 311363698964240484013304164

Offset: 1

Ant King, Sep 28 2009

As T(n) + T(n+1) = (n+1)^2 and T(n+2) + T(n+3) = (n+3)^2, it follows that the equation T(n) + T(n+1) + T(n+2) + T(n+3) = s^2 becomes (n+1)^2 + (n+3)^2 = s^2. Hence the solutions to this equation correspond to those Pythagorean triples with shorter legs that differ by two, such as 6^2 + 8^2 = 10^2.

Terms are the squares of the hypotenuses of Pythagorean triangles where other two sides are m and m+2, excepting the initial 4. See A075870. - Richard R. Forberg, Aug 15 2013

			As the third perfect square that can be expressed as the sum of four consecutive triangular numbers is 3364 = T(39) + T(40) + T(41) + T(42), we have a(3)=3364.
The first term, 4, equals T(-1) + T(0) + T(1) + T(2).

Harvey P. Dale, Table of n, a(n) for n = 1..600
Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A000217, A165516 (squares that can be expressed as the sum of three consecutive triangular numbers), A029549, A075870.

I:=[4,100,3364]; [n le 3 select I[n] else 35*Self(n-1) - 35*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Oct 21 2018

A165518:=n->(1/2)*(2+(3+2*sqrt(2))^(2*n+1)+(3-2*sqrt(2))^(2*n+1)); seq(A165518(k), k=1..20); # Wesley Ivan Hurt, Oct 24 2013

TriangularNumber[n_]:=1/2 n (n+1); data=Select[Range[10^7],IntegerQ[Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1]+TriangularNumber[ #+2]+TriangularNumber[ #+3]]] &];2(#^2+4#+5)&/@data
t={4, 100}; Do[AppendTo[t, 34 t[[-1]] - t[[-2]] - 32], {20}]; t
LinearRecurrence[{35,-35,1},{4,100,3364},20] (* Harvey P. Dale, May 22 2012 *)

x='x+O('x^50); Vec(4*x*(1-10*x+x^2)/((1-x)*(1-34*x+x^2))) \\ G. C. Greubel, Oct 21 2018

a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 32.
a(n) = (2 + (3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1))/2.
a(n) = ceiling((1/2)*(2 + (3+2*sqrt(2))^(2n+1))).
G.f.: 4*x*(x^2-10*x+1)/((1-x)*(x^2-34*x+1)).
a(n) = 4*A008844(n-1). - R. J. Mathar, Dec 14 2010
a(n) = A075870(n)^2.  - Richard R. Forberg, Aug 15 2013

Extended by T. D. Noe, Dec 09 2010

A156161 a(n) = 34*a(n-1)-a(n-2)-2312 for n > 2; a(1)=289, a(2)=7225.

Original entry on oeis.org

289, 7225, 243049, 8254129, 280395025, 9525174409, 323575532569, 10992042930625, 373405884106369, 12684808016683609, 430910066683134025, 14638257459209870929, 497269843546452475249, 16892536423120174285225

Offset: 1

Klaus Brockhaus, Feb 09 2009

nonn

lim_{n -> infinity} a(n)/a(n-1) = (17+12*sqrt(2)).

			a(3) = 34*a(2)-a(1)-2312 = 34*7225-289-2312 = 243049.

Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Second trisection of A156159.
Equals 289*A008844. - Klaus Brockhaus, Sep 23 2009
Cf. A156164 (decimal expansion of (17+12*sqrt(2))).

RecurrenceTable[{a[1]==289,a[2]==7225,a[n]==34a[n-1]-a[n-2]-2312},a,{n,20}] (* or *) LinearRecurrence[{35,-35,1},{289,7225,243049},20] (* Harvey P. Dale, Dec 11 2013 *)

{m=14; v=concat([289, 7225], vector(m-2)); for(n=3, m, v[n]=34*v[n-1]-v[n-2]-2312); v}

a(n) = (578+(867-578*sqrt(2))*(17+12*sqrt(2))^n+(867+578*sqrt(2))*(17-12*sqrt(2))^n)/8.
G.f.: x*(289-2890*x+289*x^2)/((1-x)*(1-34*x+x^2)). [corrected by Klaus Brockhaus, Sep 23 2009]
a(1)=289, a(2)=7225, a(3)=243049, a(n) = 35*a(n-1)-35*a(n-2)+a(n-3). - Harvey P. Dale, Dec 11 2013

A363319 Squares (A000290) and centered squares (A001844) sorted, including duplicates.

Original entry on oeis.org

1, 1, 4, 5, 9, 13, 16, 25, 25, 36, 41, 49, 61, 64, 81, 85, 100, 113, 121, 144, 145, 169, 181, 196, 221, 225, 256, 265, 289, 313, 324, 361, 365, 400, 421, 441, 481, 484, 529, 545, 576, 613, 625, 676, 685, 729, 761, 784, 841, 841, 900, 925, 961, 1013, 1024

Offset: 1

Clark Kimberling, May 27 2023

This sequence consists of the numbers in A363267 arranged in nondecreasing order, including duplicates, which are given by A008844 = (1, 25, 841, 28561, ...).

Cf. A000290, A001844, A363267, A363282.

c[1] = 1; c[2] = 1;
c[n_] := If[OddQ[n], c[n - 2] + n, 2 c[n - 1] - n + 1]
u = Table[c[n], {n, 1, 120}]  (* A363267 *)
s = Sort[u] (* this sequence *)

Incorrect recurrence removed by Georg Fischer, Aug 14 2023

A383734 Numbers k such that 2+k and 2*k are squares.

Original entry on oeis.org

2, 98, 3362, 114242, 3880898, 131836322, 4478554082, 152139002498, 5168247530882, 175568277047522, 5964153172084898, 202605639573839042, 6882627592338442562, 233806732499933208098, 7942546277405390632802, 269812766699283348307202, 9165691521498228451812098

Offset: 1

Emilio Martín, May 07 2025

The limit of a(n+1)/a(n) is 33.97056... = 17+12*sqrt(2) = (3+2*sqrt(2))^2 (see A156164).

			98 is a term becouse 98+2=100 is a square and 98*2=196 is a square.

Eric Weisstein's World of Mathematics, NSW numbers.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A002315, A088165, A008843, A008844, A075870, A031396, A156164.
Cf. A382209 (10+k and 10*k are squares).
Cf. A245226 (m such that k+m and k*m are squares).

LinearRecurrence[{35, -35, 1}, {2, 98, 3362}, 20] (* Amiram Eldar, May 07 2025 *)

from itertools import islice
def A383734_gen(): # generator of terms
    x, y = 1, 7
    while True:
        yield 2*x**2
        x, y = y, 6*y - x
A383734_list = list(islice(A383734_gen(), 100))

a(n) = (1/2) * ((3+2*sqrt(2))^(2*n-1) + (3-2*sqrt(2))^(1-2*n)) - 1.
a(n) = -2*sqrt(2)*sinh(n*log(17+12*sqrt(2))) + 3*cosh(n*log(17+12*sqrt(2))) - 1.
a(n) = 2*A002315(n-1)^2.
a(n) = A075870(n)^2 - 2.
a(n) = 34*a(n-1) - a(n-2) + 32.
G.f.: 2 * (1 + 14*x + x^2) / ((1 - x)*(1 - 34*x + x^2)). - Stefano Spezia, May 08 2025

A008845 Numbers k such that k+1 and k/2+1 are squares.

Original entry on oeis.org

0, 48, 1680, 57120, 1940448, 65918160, 2239277040, 76069501248, 2584123765440, 87784138523760, 2982076586042448, 101302819786919520, 3441313796169221280, 116903366249966604048, 3971273138702695316400, 134906383349641674153600, 4582845760749114225906048

Offset: 0

N. J. A. Sloane

			48+1 = 49 = 7^2 and 48/2+1 = 24+1 = 25 = 5^2.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Colin Barker, Table of n, a(n) for n = 0..650
Henry Ernest Dudeney, Amusements in Mathematics, 1917. See problem 114, "Curious numbers".
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

a:=[0,48,1680];; for n in [4..20] do a[n]:=35*a[n-1]-35*a[n-2] +a[n-3]; od; a; # G. C. Greubel, Sep 13 2019

I:=[0,48]; [n le 2 select I[n] else  34*Self(n-1) - Self(n-2)+48: n in [1..20]]; // Vincenzo Librandi, Mar 03 2016

seq(coeff(series(48*x/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 0..20); # G. C. Greubel, Sep 13 2019

LinearRecurrence[{35,-35,1},{0,48,1680},20] (* Harvey P. Dale, May 24 2014 *)

concat(0, Vec(48*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Mar 02 2016

def A008845_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(48*x/((1-x)*(1-34*x+x^2))).list()
A008845_list(20) # G. C. Greubel, Sep 13 2019

a(n) = 2*(A008844(n)-1) = 16*A075528(n) = 48*A029546(n). - corrected by Sean A. Irvine, Apr 07 2018
a(0)=0, a(1)=48, a(2)=1680, a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). - Harvey P. Dale, May 24 2014
From Colin Barker, Mar 02 2016: (Start)
a(n) = (-6+(3-2*sqrt(2))*(17+12*sqrt(2))^(-n)+(3+2*sqrt(2))*(17+12*sqrt(2))^n)/4.
G.f.: 48*x / ((1-x)*(1-34*x+x^2)).
(End)
a(n) = 34*a(n-1) - a(n-2) + 48. - Vincenzo Librandi, Mar 03 2016

A009759 Expansion of (3 - 21x + 4x^2)/((x-1)(x^2 - 6x + 1)).

Original entry on oeis.org

-3, 0, 17, 116, 693, 4056, 23657, 137900, 803757, 4684656, 27304193, 159140516, 927538917, 5406093000, 31509019097, 183648021596, 1070379110493, 6238626641376, 36361380737777, 211929657785300, 1235216565974037

Offset: 0

N. J. A. Sloane, Nick Baxter (nick(AT)visigenic.com)

sign

Numbers k such that 2*k^2 + 14*k + 25 is a square. - James R. Buddenhagen, Jul 19 2008

If k satisfies the previous condition, i.e., P(k) = 2*k^2 + 14*k + 25 is a square, then there exists an integer m such that (k+3)^2 + (k+4)^2 = m^2; finding such numbers k is equivalent to finding Pythagorean triples (X,X+1,Z) with X in A001652 (1st formula). For k >= 0, P(k) = A001844(k+3), and the successive values of P(k) that are squares are in A008844. - Bernard Schott, Mar 24 2019

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, New-York, 1964, pp. 122-124.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001652 (X values), A001653 (Z values), A001844, A008844.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))) ) // G. C. Greubel, Feb 12 2018

CoefficientList[Series[(3-21x+4x^2)/((x-1)(x^2-6x+1)),{x,0,30}],x] (* or *) LinearRecurrence[{7,-7,1},{-3,0,17},30] (* Harvey P. Dale, Dec 12 2016 *)

my(x='x+O('x^30)); Vec((3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))) \\ G. C. Greubel, Feb 12 2018

((3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 04 2019

a(n) = A001652(n) - 3.
a(n) = ( (1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 14 )/4.
a(n) - a(n-1) = A001541(n), n > 0. - R. J. Mathar, Apr 23 2009
a(n) = (Q(2*n+1) -14)/4 = (4*P(n)*P(n+1) + (-1)^n - 7)/2, where P(n) = A000129(n) (Pell) and Q(n) =  A002203(n) (Pell-Lucas). - G. C. Greubel, Apr 04 2019

G.f. and Binet formula corrected by R. J. Mathar, Aug 24 2016

A251924 Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to a hexagonal number H(m) for some m.

Original entry on oeis.org

0, 34, 1188, 40390, 1372104, 46611178, 1583407980, 53789260174, 1827251437968, 62072759630770, 2108646576008244, 71631910824649558, 2433376321462076760, 82663163018885960314, 2808114166320660573948, 95393218491883573553950, 3240561314557720840260384

Offset: 1

Colin Barker, Dec 11 2014

Also nonnegative integers x in the solutions to 2*x^2-4*y^2+4*x+2*y+2 = 0, the corresponding values of y being A008844.

First bisection of A076708. [Bruno Berselli, Dec 11 2014]

			34 is in the sequence because T(34)+T(35) = 595+630 = 1225 = H(25).

Colin Barker, Table of n, a(n) for n = 1..653
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A000384, A008844, A076708.

LinearRecurrence[{35,-35,1},{0,34,1188},20] (* Harvey P. Dale, Feb 04 2019 *)

concat(0, Vec(2*x^2*(x-17)/((x-1)*(x^2-34*x+1)) + O(x^100)))

a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
G.f.: 2*x^2*(x-17) / ((x-1)*(x^2-34*x+1)).
a(n) = (-8-(4+3*sqrt(2))*(17+12*sqrt(2))^(-n)+(-4+3*sqrt(2))*(17+12*sqrt(2))^n)/8. - Colin Barker, Mar 02 2016

A251925 Prime powers p^k (k>=2) of the form (n^2+1)/2.

Original entry on oeis.org

25, 841, 28561, 32959081, 1119638521, 1985636569351347658201, 3051519929713402294221039791281, 4689566069222821420312720463003656425961, 183840368926047361112315395593676258316051401, 17020879736268069268391497343746883355223007561030622302744641179601

Offset: 1

Joerg Arndt, Dec 11 2014

nonn

The corresponding n are a subsequence of A001333; see example.

			The first few terms correspond to
7^2 + 1 = 2 * 5^2 = 2 * 25,
41^2 + 1 = 2 * 29^2 = 2 * 841,
239^2 + 1 = 2 * 13^4 = 2 * 28561,
8119^2 + 1 = 2 * 5741^2 = 2 * 32959081,
47321^2 + 1 = 2 * 33461^2 = 2 * 1119638521,
63018038201^2+1 = 2 * 44560482149^2 = 2 * 1985636569351347658201.

Paolo Xausa, Table of n, a(n) for n = 1..19
Joerg Arndt, Arctan relations for Pi (the author's starting point for this sequence).

Cf. A027861 (primes of the form (n^2+1)/2), A001333, A008844 (primes and composites with k=2).

With[{r=Range[100]},Select[((ChebyshevT[r,I]/I^r)^2+1)/2,!PrimeQ[#]&&PrimePowerQ[#]&]] (* Paolo Xausa, Nov 13 2023, after Joerg Arndt *)

forstep(n=1,10^9,2, t=(n^2+1)/2; if( !isprime(t) && isprimepower(t), print1(t,", ")));

/* much more efficient: */
{b(n) = polchebyshev(n, 1, I) / I^n}
for(n=1,10^3,t=(b(n)^2+1)/2;if(!isprime(t)&&isprimepower(t),print1(t,", ")));

A276916 Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.

Original entry on oeis.org

1, 5, 41, 61, 145, 181, 313, 365, 545, 613, 841, 925, 1201, 1301, 1625, 1741, 2113, 2245, 2665, 2813, 3281, 3445, 3961, 4141, 4705, 4901, 5513, 5725, 6385, 6613, 7321, 7565, 8321, 8581, 9385, 9661, 10513, 10805, 11705, 12013, 12961, 13285, 14281, 14621, 15665

Offset: 0

Daniel Poveda Parrilla, Sep 27 2016

All terms of this sequence are centered square numbers. Graphically, each term of the sequence is made of four squares, eight triangles and a central element.

a(A220185(n+1)) = A008844(2n) = A079291(4n+1), which is a square of a Pell number.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008844, A079291, A220185, A276914.

[4*n*(2*n+(-1)^n)+1 : n in [0..60]]; // Wesley Ivan Hurt, Sep 27 2016

A276916:=n->4*n*(2*n+(-1)^n)+1: seq(A276916(n), n=0..60); # Wesley Ivan Hurt, Sep 27 2016

Table[4 n (2 n + (-1)^n) + 1, {n, 0, 44}] (* or *)
CoefficientList[Series[(1 +4x +34x^2 +12x^3 +13x^4)/((1-x)^3*(1+x)^2), {x, 0, 44}], x] (* Michael De Vlieger, Sep 28 2016 *)

Vec((1+4*x+34*x^2+12*x^3+13*x^4)/((1-x)^3*(1+x)^2) + O(x^50)) \\ Colin Barker, Sep 27 2016

[4*n*(2*n+(-1)^n) +1 for n in (0..60)] # G. C. Greubel, Aug 19 2022

a(n) = 4*n*(2*n + (-1)^n) + 1.
a(n) = 4*n*(2*n + 1) + 1 for n even.
a(n) = 4*n*(2*n - 1) + 1 for n odd.
a(n) is sum of two squares; a(n) = k^2 + (k+1)^2 where k = 2n-(n mod 2). - David A. Corneth, Sep 27 2016
From Colin Barker, Sep 27 2016: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4.
G.f.: (1+4*x+34*x^2+12*x^3+13*x^4) / ((1-x)^3*(1+x)^2). (End)
E.g.f.: (1+8*x+8*x^2)*exp(x) - 4*x*exp(-x). - G. C. Greubel, Aug 19 2022

A307492 Numbers that are both centered square and square pyramidal.

Original entry on oeis.org

1, 5, 42925, 1026745

Offset: 1

Ilya Gutkovskiy, Apr 10 2019

If it exists, a(5) > 10^29. - Bert Dobbelaere, Apr 12 2019

Eric Weisstein's World of Mathematics, Centered Square Number
Eric Weisstein's World of Mathematics, Square Pyramidal Number

Intersection of A000330 and A001844.

Cf. A008844, A039596, A131750, A307491.

csQ[n_] := IntegerQ[Sqrt[2*n-1]]; Select[Table[n(n+1)(2n+1)/6, {n, 0, 1000}], csQ] (* Amiram Eldar, Apr 11 2019 *)

cp's OEIS Frontend

A165518 Perfect squares (A000290) that can be expressed as the sum of four consecutive triangular numbers (A000217).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A156161 a(n) = 34*a(n-1)-a(n-2)-2312 for n > 2; a(1)=289, a(2)=7225.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A363319 Squares (A000290) and centered squares (A001844) sorted, including duplicates.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Extensions

A383734 Numbers k such that 2+k and 2*k are squares.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Formula

A008845 Numbers k such that k+1 and k/2+1 are squares.

Views

Author

Keywords

Examples

References

Links

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A009759 Expansion of (3 - 21*x + 4*x^2)/((x-1)*(x^2 - 6*x + 1)).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

Extensions

A251924 Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to a hexagonal number H(m) for some m.

A009759 Expansion of (3 - 21x + 4x^2)/((x-1)(x^2 - 6x + 1)).