A029549 -id:A029549 - cp's OEIS frontend

A123479 Coefficients of series giving the best rational approximations to sqrt(6).

20, 1980, 194040, 19013960, 1863174060, 182572043940, 17890197132080, 1753056746899920, 171781670999060100, 16832850701160989900, 1649447587042777950120, 161629030679491078121880, 15837995559003082877994140, 1551961935751622630965303860

Offset: 1

Gene Ward Smith, Sep 28 2006

The partial sums of the series 5/2 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(6), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2;2], [2;2,4,2], [2;2,4,2,4,2], [2;2,4,2,4,2,4,2] and so forth.

Sequence of numbers x=a(n) such 4*x+1 and 6*x+1 are both square, and their square roots are A138288(n) and A054320(n). - Paul Cleary, Jun 23 2014

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A123478, A123480, A029549, A123482.

LinearRecurrence[{99,-99,1},{0,20,1980},{2, 25}] (* Paul Cleary, Jun 23 2014 *)

Vec(-20*x/((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 99*a(n+2) - 99*a(n+1) + a(n).
a(n) = -5/24 + (( + 2*6^(1/2))/48)*(49 + 20*6^(1/2))^n + ((5 - 2*6^(1/2))/48)*(49 - 20*6^(1/2))^n.
G.f.: -20*x / ((x-1)*(x^2-98*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6T(m)=5T(k) for some k.

Original entry on oeis.org

0, 55, 26565, 12804330, 6171660550, 2974727580825, 1433812522297155, 691094661019647940, 333106192798948009980, 160556493834431921162475, 77387896922003387052303025, 37300805759911798127288895630

Offset: 0

Charlie Marion, Dec 20 2011

			6*0 = 5*0;
6*55 = 5*66;
6*26565 = 5*31878;
6*12804330 = 5*15365196.

Vincenzo Librandi, Table of n, a(n) for n = 0..229
Index entries for linear recurrences with constant coefficients, signature (483,-483,1).

Cf. A001652, A029549, A053141, A075528, A200993-A201008.

I:=[0, 55, 26565]; [n le 3 select I[n] else 483*Self(n-1)-483*Self(n-2)+Self(n-3): n in [1..15]]; // Vincenzo Librandi, Dec 22 2011

LinearRecurrence[{483,-483,1},{0,55,26565},30] (* Vincenzo Librandi, Dec 22 2011 *)

makelist(expand(((11-2*sqrt(30))^(2*n+1)+(11+2*sqrt(30))^(2*n+1)-22)/192), n, 0, 11); /* Bruno Berselli, Dec 21 2011 */

concat(0,Vec(55/(1-x)/(1-482*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 23 2011

For n > 1, a(n) = 482*a(n-1) - a(n-2) + 55. See A200993 for generalization.
From Bruno Berselli, Dec 21 2011: (Start)
G.f.: 55*x/((1-x)*(1-482*x+x^2)).
a(n) = a(-n-1) = 483*a(n-1)-483*a(n-2)+a(n-3).
a(n) = ((11-2*r)^(2*n+1)+(11+2*r)^(2*n+1)-22)/192, where r=sqrt(30). (End)

a(11) corrected by Bruno Berselli, Dec 21 2011

a(6) corrected by Vincenzo Librandi, Dec 22 2011

A123478 Coefficients of series giving the best rational approximations to sqrt(7).

Original entry on oeis.org

48, 12240, 3108960, 789663648, 200571457680, 50944360587120, 12939667017670848, 3286624478127808320, 834789677777445642480, 212033291530993065381648, 53855621259194461161296160, 13679115766543862141903843040, 3474441549080881789582414836048

Offset: 1

Gene Ward Smith, Sep 28 2006

The partial sums of the series 8/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(7), which constitute every fourth convergent of the continued fraction. The corresponding continued fractions are [2;1,1,1], [2;1,1,1,4,1,1,1], [2;1,1,1,4,1,1,1,4,1,1,1] and so forth.

Harvey P. Dale, Table of n, a(n) for n = 1..416
Index entries for linear recurrences with constant coefficients, signature (255,-255,1).

Cf. A123479, A123480, A029549, A123482.

LinearRecurrence[{255,-255,1},{48,12240,3108960},30] (* Harvey P. Dale, Nov 20 2016 *)

Vec(-48*x/((x-1)*(x^2-254*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 255 a(n+2) - 255 a(n+1) + a(n).
a(n) = -4/21 + (2/21+1/28*7^(1/2))*(127+48*7^(1/2))^n + (2/21-1/28*7^(1/2))*(127-48*7^(1/2))^n.
G.f.: -48*x / ((x-1)*(x^2-254*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

A123482 Coefficients of the series giving the best rational approximations to sqrt(11).

Original entry on oeis.org

60, 23940, 9528120, 3792167880, 1509273288180, 600686976527820, 239071907384784240, 95150018452167599760, 37869468272055319920300, 15071953222259565160679700, 5998599512991034878630600360, 2387427534217209622129818263640, 950190160018936438572789038328420

Offset: 1

Gene Ward Smith, Oct 02 2006

The partial sums of the series 10/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(11), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [3;3,6,3], [3;3,6,3,6,3], [3;3,6,3,6,3,6,3] and so forth.

G. C. Greubel, Table of n, a(n) for n = 1..380
Index entries for linear recurrences with constant coefficients, signature (399,-399,1).

Cf. A010468, A041014, A041015.

Cf. A123478, A123479, A123480, A029549.

CoefficientList[Series[-60*x/((x - 1)*(x^2 - 398*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)

Vec(-60*x/((x-1)*(x^2-398*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

a(n+3) = 399*a(n+2) - 399*a(n+1) + a(n).
a(n) = -5/33 + (5/66 + 1/44*11^(1/2))*(199 + 60*11^(1/2))^n + (5/66 - 1/44*11^(1/2))*(199 - 60*11^(1/2))^n.
G.f.: -60*x / ((x-1)*(x^2-398*x+1)). - Colin Barker, Jun 23 2014

More terms from Colin Barker, Jun 23 2014

A341895 Indices of triangular numbers that are ten times other triangular numbers.

Original entry on oeis.org

0, 4, 20, 39, 175, 779, 1500, 6664, 29600, 56979, 253075, 1124039, 2163720, 9610204, 42683900, 82164399, 364934695, 1620864179, 3120083460, 13857908224, 61550154920, 118481007099, 526235577835, 2337285022799, 4499158186320, 19983094049524, 88755280711460, 170849530073079, 758831338304095, 3370363382012699

Offset: 1

Vladimir Pletser, Feb 23 2021

Second member of the Diophantine pair (b(n), a(n)) that satisfies a(n)^2 + a(n) = 10*(b(n)^2 + b(n)) or T(a(n)) = 10*T(b(n)) where T(x) is the triangular number of x. The T(b)'s are in A068085 and the b's are in A341893.

Can be defined for negative n by setting a(-n) = -a(n+1) - 1 for all n in Z.

			a(2) = 4 is a term because its triangular number, T(a(2)) = 4*5 / 2 = 10 is ten times a triangular number.
a(4) = 38*a(1) - a(-2) + 18 = 38*0 - (-21) + 18 = 39, etc.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A341893, A068085, A166477 (n=10).

Cf. A336623, A336624, A336626, A336625, A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.

f := gfun:-rectoproc({a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)]) ; #

Rest@ CoefficientList[Series[x^2*(4 + 16*x + 19*x^2 - 16*x^3 - 4*x^4 - x^5)/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 30}], x] (* Michael De Vlieger, May 19 2022 *)

a(n) = 38*a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20.
G.f.: x^2*(4 + 16*x + 19*x^2 - 16*x^3 - 4*x^4 - x^5)/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7). - Stefano Spezia, Feb 24 2021
a(n) = (A198943(n) + 1)/2 - 1. - Hugo Pfoertner, Feb 26 2021

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

Original entry on oeis.org

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

A214838 Triangular numbers of the form k^2 + 2.

Original entry on oeis.org

3, 6, 66, 171, 2211, 5778, 75078, 196251, 2550411, 6666726, 86638866, 226472403, 2943171003, 7693394946, 99981175206, 261348955731, 3396416785971, 8878171099878, 115378189547778, 301596468440091, 3919462027838451, 10245401755863186, 133146330756959526, 348042063230908203

Offset: 1

Alex Ratushnyak, Mar 07 2013

Corresponding k values are in A077241.

Except 3, all terms are in A089982: in fact, a(2) = 3+3 and a(n) = (k-2)*(k-1)/2+(k+1)*(k+2)/2, where k = sqrt(a(n)-2) > 2 for n > 2. [Bruno Berselli, Mar 08 2013]

			2211 is in the sequence because 2211 = 47^2 + 2.

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A000217, A001110, A001219, A006454, A029549, A069017, A077241, A089982, A164055, A165892.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)))); // Bruno Berselli, Mar 08 2013

LinearRecurrence[{1, 34, -34, -1, 1}, {3, 6, 66, 171, 2211}, 25] (* Bruno Berselli, Mar 08 2013 *)

t[n]:=((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4$
makelist(expand(t[n]*(t[n]+1)/2), n, 1, 25); /* Bruno Berselli, Mar 08 2013 */

for(n=1, 10^9, t=n*(n+1)/2; if(issquare(t-2), print1(t,", "))); \\ Joerg Arndt, Mar 08 2013

import math
for i in range(2, 1<<32):
      t = i*(i+1)//2 - 2
      sr = int(math.sqrt(t))
      if sr*sr == t:
          print(f'{sr:10} {i:10} {t+2}')

G.f.: -3*x*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)). - Joerg Arndt, Mar 08 2013

a(n) = A000217(t), where t = ((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4. - Bruno Berselli, Mar 08 2013

A253651 Triangular numbers that are the product of a triangular number and a prime number.

Original entry on oeis.org

0, 3, 6, 15, 21, 45, 66, 78, 105, 190, 210, 231, 435, 465, 630, 861, 903, 1035, 1326, 2415, 2556, 2628, 3003, 3570, 4005, 4950, 5460, 5565, 5995, 7140, 8646, 8778, 9870, 12246, 16471, 16836, 17205, 17391, 17766, 20100, 22155, 26565, 26796, 28680, 28920, 30381, 32131, 33411, 33930, 36856

Offset: 1

Antonio Roldán, Jan 07 2015

nonn

			190 is in the sequence because it is triangular (190=19*20/2) and 190=10*19, with 10 triangular number and 19 prime number.

Harvey P. Dale, Table of n, a(n) for n = 1..259 (all terms up to and including the 10000th triangular number)

Cf. A029549 (T is 2*t), A076140 (T is 3*t), A225503 (first T to be prime(n)*t).

Cf. A188630, A253650, A253652, A253653.

N:= 10^5: # to get all terms <= N
Primes:= select(isprime, [2,seq(2*k+1,k=1..N/3)]):
select(t -> issqr(1+8*t), {seq(seq(a*(a+1)/2*p, a = 2 .. floor(sqrt(2*N/p))), p = Primes)});
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(%,list)); # Robert Israel, Jan 07 2015

Join[{0},Module[{nn=300,trs},trs=Accumulate[Range[nn]];Select[ trs,AnyTrue[ #/trs,PrimeQ]&]]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 16 2018 *)

{i=1; j=2;print1(0,", "); while(i<=10^5, k=1; p=2; c=0; while(k1, c=k); if(c>0, print1(i, ", ")); k+=p; p+=1); i+=j; j+=1)}

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

Original entry on oeis.org

0, 1, 6, 12, 55, 246, 474, 2107, 9360, 18018, 80029, 355452, 684228, 3039013, 13497834, 25982664, 115402483, 512562258, 986657022, 4382255359, 19463867988, 37466984190, 166410301177, 739114421304, 1422758742216, 6319209189385, 28066884141582, 54027365220036, 239963538895471, 1065802482958830, 2051617119619170

Offset: 1

Vladimir Pletser, Feb 23 2021

The indices of triangular numbers that are one-tenth of other triangular numbers [t of T(t) such that T(t)=T(u)/10].
First member of the Diophantine pair (t, u) that satisfies 10*(t^2 + t) = u^2 + u; a(n) = t.
The T(t)'s are in A068085 and the u's are in A341895.
Also, nonnegative t such that 40*t^2 + 40*t + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(4) = 12 is a term because its triangular number, (12*13) / 2 = 78 is one-tenth of 780, the triangular number of 39.
a(4) = 38 a(1) - a(-2) +18 = 0 - 6 +18 = 12 ;
a(5) = 38 a(2) - a(-1) + 18 = 38*1 - 1 +18 = 55.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
V. Pletser, Recurrent relations for triangular multiples of other triangular numbers, Indian J. Pure Appl. Math. 53 (2022) 782-791
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A068085, A341895.
Cf. A336623, A336624, A336626, A336625, A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.
Cf. A180003.

f := gfun:-rectoproc({a(-3) = 6, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 6, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ;

Rest@ CoefficientList[Series[(x^2*(1 + 5*x + 6*x^2 + 5*x^3 + x^4))/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 31}], x] (* Michael De Vlieger, May 19 2022 *)

a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A068085(n).
a(n) = 38 a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
G.f.: x^2*(1 + 4*x+x^2)*(1+x+x^2)/ ((1-x)*(1-38*x^3+x^6)). - Stefano Spezia, Feb 24 2021
a(n) = A180003(n) - 1. - Hugo Pfoertner, Feb 28 2021

A227027 Triangular numbers representable as x!/y! with y < x-1.

Original entry on oeis.org

6, 120, 210, 990, 7140, 185136, 242556, 2162160, 8239770, 258474216, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550

Offset: 1

Alex Ratushnyak, Jun 27 2013

nonn

Triangular numbers in A045619, except A045619(1)=0. The sequence is infinite because A029549 is a subsequence. According to Melissen's comment in A097571,  y > x-7.
The sequence of x's producing a(n): A227026.
a(2) and a(3) have two representations:
a(2) = 120 = 5*4*3*2 = 6*5*4.
a(3) = 210 = 7*6*5 = 15*14.

			990 is in the sequence since 990 = 11!/8! = 11*10*9 is a ratio of factorials and 990 = (44)(44 + 1)/2 is a triangular number.

Cf. A000217, A001219, A029549, A045619, A097571, A227026.

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A123479 Coefficients of series giving the best rational approximations to sqrt(6).

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A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6*T(m)=5*T(k) for some k.

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A123478 Coefficients of series giving the best rational approximations to sqrt(7).

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A123482 Coefficients of the series giving the best rational approximations to sqrt(11).

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A341895 Indices of triangular numbers that are ten times other triangular numbers.

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

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A214838 Triangular numbers of the form k^2 + 2.

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A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6T(m)=5T(k) for some k.