A103200 -id:A103200 - cp's OEIS frontend

A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2sqrt(8a(n-1)^2 + 8*a(n-1) + 1).

0, 2, 14, 84, 492, 2870, 16730, 97512, 568344, 3312554, 19306982, 112529340, 655869060, 3822685022, 22280241074, 129858761424, 756872327472, 4411375203410, 25711378892990, 149856898154532, 873430010034204, 5090723162050694, 29670908962269962, 172934730611569080

Offset: 0

Wolfdieter Lang

Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n).
The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
Also the indices of triangular numbers that are half other triangular numbers [a of T(a) such that 2T(a)=T(b)]. The T(a)'s are in A075528, the T(b)'s are in A029549 and the b's are in A001652. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002
Sequences A053141 (this entry), A016278, A077259, A077288 and A077398 are part of an infinite series of sequences. Each depends upon the polynomial p(n) = 4k*n^2 + 4k*n + 1, when 4k is not a perfect square. Equivalently, they each depend on the equation k*t(x)=t(z) where t(n) is the triangular number formula n(n+1)/2. The dependencies are these: they are the sequences of positive integers n such that p(n) is a perfect square and there exists a positive integer m such that k*t(n)=t(m). A053141 is for k=2, A016278 is for k=3, A077259 is for k=5. - Robert Phillips (bobanne(AT)bellsouth.net), Oct 11 2007, Nov 27 2007
Jason Holt observes that a pair drawn from a drawer with A053141(n)+1 red socks and A001652(n) - A053141(n) blue socks will as likely as not be matching reds: (A053141+1)*A053141/((A001652+1)*A001652) = 1/2, n>0. - Bill Gosper, Feb 07 2010
The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)=A011900(n) and c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009
Behera & Panda call these the balancers and A001109 are the balancing numbers. - Michel Marcus, Nov 07 2017

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
A. Behera and G. K. Panda, On the Square Roots of Triangular Numbers, Fib. Quart., 37 (1999), pp. 98-105.
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4.
aBa Mbirika, Janee Schrader, and Jürgen Spilker, Pell and Associated Pell Braid Sequences as GCDs of Sums of k Consecutive Pell, Balancing, and Related Numbers, J. Int. Seq. (2023) Vol. 26, Art. 23.6.4.
J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
G. K. Panda, Sequence balancing and cobalancing numbers, Fib. Q., Vol. 45, No. 3 (2007), 265-271. See p. 266.
Michael Penn, (co) balancing numbers, YouTube video, 2022.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Robert Phillips, A triangular number result, 2009.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
Burkard Polster, Nice merging together, Mathologer video (2015).
B. Polster and M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
A. Tekcan, M. Tayat, and M. E. Ozbek, The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001108, A001109, A001652, A001653.
Cf. A011900, A029549, A053142, A075528, A103200.
Partial sums of A001542.
Cf. A000194, A001541, A002024, A016278, A046090, A055997, A077259, A077288, A077398, A084068.

a053141 n = a053141_list !! n
a053141_list = 0 : 2 : map (+ 2)
   (zipWith (-) (map (* 6) (tail a053141_list)) a053141_list)
-- Reinhard Zumkeller, Jan 10 2012

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

A053141 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,2]) ;
    else
        6*procname(n-1)-procname(n-2)+2 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Join[{a=0,b=1}, Table[c=6*b-a+1; a=b; b=c, {n,60}]]*2 (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
a[n_] := Floor[1/8*(2+Sqrt[2])*(3+2*Sqrt[2])^n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 28 2013 *)
Table[(Fibonacci[2n + 1, 2] - 1)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

concat(0,Vec(2/(1-x)/(1-6*x+x^2)+O(x^30))) \\ Charles R Greathouse IV, May 14 2012

{x=1+sqrt(2); y=1-sqrt(2); P(n) = (x^n - y^n)/(x-y)};
a(n) = round((P(2*n+1) - 1)/2);
for(n=0, 30, print1(a(n), ", ")) \\ G. C. Greubel, Jul 15 2018

[(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # G. C. Greubel, Apr 27 2020

a(n) = (A001653(n)-1)/2 = 2*A053142(n) = A011900(n)-1. [Corrected by Pontus von Brömssen, Sep 11 2024]
a(n) = 6*a(n-1) - a(n-2) + 2, a(0) = 0, a(1) = 2.
G.f.: 2*x/((1-x)*(1-6*x+x^2)).
Let c(n) = A001109(n). Then a(n+1) = a(n)+2*c(n+1), a(0)=0. This gives a generating function (same as existing g.f.) leading to a closed form: a(n) = (1/8)*(-4+(2+sqrt(2))*(3+2*sqrt(2))^n + (2-sqrt(2))*(3-2*sqrt(2))^n). - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002
a(n) = 2*Sum_{k = 0..n} A001109(k). - Mario Catalani (mario.catalani(AT)unito.it), Mar 22 2003
For n>=1, a(n) = 2*Sum_{k=0..n-1} (n-k)*A001653(k). - Charlie Marion, Jul 01 2003
For n and j >= 1, A001109(j+1)*A001652(n) - A001109(j)*A001652(n-1) + a(j) = A001652(n+j). - Charlie Marion, Jul 07 2003
From Antonio Alberto Olivares, Jan 13 2004: (Start)
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
a(n) = -(1/2) - (1-sqrt(2))/(4*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(4*sqrt(2))*(3+2*sqrt(2))^n. (End)
a(n) = sqrt(2)*cosh((2*n+1)*log(1+sqrt(2)))/4 - 1/2 = (sqrt(1+4*A029549)-1)/2. - Bill Gosper, Feb 07 2010 [typo corrected by Vaclav Kotesovec, Feb 05 2016]
a(n+1) + A055997(n+1) = A001541(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
From Charlie Marion, Oct 18 2004: (Start)
For n>k, a(n-k-1) = A001541(n)*A001653(k)-A011900(n+k); e.g., 2 = 99*5 - 493.
For n<=k, a(k-n) = A001541(n)*A001653(k) - A011900(n+k); e.g., 2 = 3*29 - 85 + 2. (End)
a(n) = A084068(n)*A084068(n+1). - Kenneth J Ramsey, Aug 16 2007
Let G(n,m) = (2*m+1)*a(n)+ m and H(n,m) = (2*m+1)*b(n)+m where b(n) is from the sequence A001652 and let T(a) = a*(a+1)/2. Then T(G(n,m)) + T(m) = 2*T(H(n,m)). - Kenneth J Ramsey, Aug 16 2007
Let S(n) equal the average of two adjacent terms of G(n,m) as defined immediately above and B(n) be one half the difference of the same adjacent terms. Then for T(i) = triangular number i*(i+1)/2, T(S(n)) - T(m) = B(n)^2 (setting m = 0 gives the square triangular numbers). - Kenneth J Ramsey, Aug 16 2007
a(n) = A001108(n+1) - A001109(n+1). - Dylan Hamilton, Nov 25 2010
a(n) = (a(n-1)*(a(n-1) - 2))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - G. C. Greubel, Apr 27 2020
E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/4. - Stefano Spezia, Mar 16 2024
a(n) = A000194(A029549(n)) = A002024(A075528(n)). - Pontus von Brömssen, Sep 11 2024

Name corrected by Zak Seidov, Apr 11 2011

A105038 Nonnegative n such that 6n^2 + 6n + 1 is a square.

Original entry on oeis.org

0, 4, 44, 440, 4360, 43164, 427284, 4229680, 41869520, 414465524, 4102785724, 40613391720, 402031131480, 3979697923084, 39394948099364, 389969783070560, 3860302882606240, 38213059042991844, 378270287547312204, 3744489816430130200, 37066627876753989800

Offset: 0

Gerald McGarvey, Apr 03 2005

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A001652, A001570, A049629, A105040, A104240, A077288, A105036, A103200, A105037.

CoefficientList[Series[4x/(1-11x+11x^2-x^3),{x,0,30}],x] (* or *) LinearRecurrence[{11,-11,1},{0,4,44},30] (* Harvey P. Dale, Sep 29 2013 *)

for(n=0,427284,if(issquare(6*n*(n+1)+1),print1(n,",")))

Vec(4*x/(1-11*x+11*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2012

G.f.: 4*x/(1-11*x+11*x^2-x^3).
a(0)=0, a(1)=4, a(2)=44, a(n)=11*a(n-1)-11*a(n-2)+a(n-3). - Harvey P. Dale, Sep 29 2013
a(n) = (-6-(5-2*sqrt(6))^n*(-3+sqrt(6))+(3+sqrt(6))*(5+2*sqrt(6))^n)/12. - Colin Barker, Mar 05 2016
a(n) = (A072256(n+1) - 1)/2.

More terms from Vladeta Jovovic, Apr 05 2005

Incorrect conjectures deleted by N. J. A. Sloane, Nov 24 2010

A105045 a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=12; for n > 4, a(n) = 8*a(n-2) - a(n-4) - 3.

Original entry on oeis.org

0, 1, 2, 3, 12, 20, 91, 154, 713, 1209, 5610, 9515, 44164, 74908, 347699, 589746, 2737425, 4643057, 21551698, 36554707, 169676156, 287794596, 1335857547, 2265802058, 10517184217, 17838621865, 82801616186, 140443172859, 651895745268, 1105706761004, 5132364345955

Offset: 0

Gerald McGarvey, Apr 03 2005

It appears that this sequence gives all nonnegative m such that 60*m^2 - 60*m + 1 is a square and that for n > 3, a(n+1) = A103200(n) + 1.
From Paul Weisenhorn, Jun 30 2010: (Start)
Place b(n) red and a(n) blue balls in an urn, then draw 6 balls without replacement.
This gives binomial(b(n), 6) = binomial(b(n), 4) * binomial(a(n), 2), where b(n) = A179123(n). (End)

			For n=3, a(3)=3; b(3)=14; binomial(14,6)=3003; binomial(14,4)*binomial(3,2) = 1001*3 = 3003. - _Paul Weisenhorn_, Jun 30 2010

Muniru A Asiru, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-1,1).

Cf. A001090, A103200, A179123.

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+x-7*x^2+x^3+x^4)/((1-x)*(1-8*x^2+x^4)) )); // G. C. Greubel, Mar 14 2023

n:=1: for m from 1 to 2000 do w:=sqrt(1+60*m*(m-1)):
if (w=floor(w)) then a(n)=m: b(n)=(9+w)/2: inc(n): end if: end do # Paul Weisenhorn, Jun 30 2010

Join[{0},RecurrenceTable[{a[1]==1,a[2]==2,a[3]==3,a[4]==12,a[n] == 8a[n-2]-a[n-4]-3},a,{n,30}]] (* or *) LinearRecurrence[{1,8,-8,-1, 1}, {0,1,2,3,12,20}, 30] (* Harvey P. Dale, Nov 10 2011 *)

@CachedFunction
def a(n): # a = A105045
    if (n<6): return (0,1,2,3,12,20)[n]
    else: return a(n-1) +8*a(n-2) -8*a(n-3) -a(n-4) +a(n-5)
[a(n) for n in range(41)] # G. C. Greubel, Mar 14 2023

a(n) = 8*a(n-2) - a(n-4) - 3, for n > 4.
From Paul Weisenhorn, Jun 30 2010: (Start)
Let r=sqrt(15), then
a(n) = ((15+r)*(4+r)^((n-1)/2) + (15-r)*(4-r)^((n-1)/2) + 30)/60 when n is odd, and
a(n) = ((45+11*r)*(4+r)^((n-2)/2) + (45-11*r)*(4-r)^((n-2)/2) + 30)/60 when n is even. (End)
a(n) = a(n-1) + 8*a(n-2) - 8*a(n-3) - a(n-4) + a(n-5), a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=12, a(5)=20. - Harvey P. Dale, Nov 10 2011
G.f.: x*(1 +x -7*x^2 +x^3 +x^4)/((1-x)*(1-8*x^2+x^4)). - Colin Barker, Jan 01 2013

More terms from Colin Barker, Jan 01 2013

A103201 a(1) = 11, a(2) = 19, a(3) = 89, a(4) = 151; for n >= 5, a(n) = sqrt(a(n-4)^2 + 60a(n-2)^2 + 4a(n-2)sqrt(210 + 15a(n-4)^2)).

Original entry on oeis.org

11, 19, 89, 151, 701, 1189, 5519, 9361, 43451, 73699, 342089, 580231, 2693261, 4568149, 21203999, 35964961, 166938731, 283151539, 1314305849, 2229247351, 10347508061, 17550827269, 81465758639, 138177370801, 641378561051

Offset: 1

K. S. Bhanu and M. N. Deshpande, Mar 24 2005

nonn

K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, An interesting sequence of quadruples and related open problems, Institute of Sciences, Nagpur, India, Preprint, 2005.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-1).

This is the sequence b(n) defined in A103200. Bhanu and Deshpande ask for a proof that the terms of the sequence are always integers.

Cf. A103200.

a:=[11,19,89,151];; for n in [5..30] do a[n]:=8*a[n-2]-a[n-4]; od; a; # G. C. Greubel, May 24 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(11+19*x+x^2-x^3)/(1-8*x^2+x^4) )); // G. C. Greubel, May 24 2019

b[1]:=11:b[2]:=19:b[3]:=89:b[4]:=151: for n from 5 to 28 do b[n]:=sqrt(b[n-4]^2+60*b[n-2]^2+4*b[n-2]*sqrt(210+15*b[n-4]^2)) od:seq(b[n],n=1..28); # Emeric Deutsch, Apr 13 2005

LinearRecurrence[{0, 8, 0, -1}, {11, 19, 89, 151}, 30] (* Georg Fischer, May 24 2019 *)

my(x='x+O('x^30)); Vec(x*(11+19*x+x^2-x^3)/(1-8*x^2+x^4)) \\ G. C. Greubel, May 24 2019

a=(x*(11+19*x+x^2-x^3)/(1-8*x^2+x^4)).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, May 24 2019

G.f.: x*(11 + 19*x + x^2 - x^3)/(1 - 8*x^2 + x^4). - Georg Fischer, May 24 2019

More terms from Pierre CAMI and Emeric Deutsch, Apr 13 2005

A105063 a(1)=0, a(2)=0, a(3)=8, a(4)=24, a(n) = 32 + 66*a(n-2) - a(n-4) for n > 4.

Original entry on oeis.org

0, 0, 8, 24, 560, 1616, 36984, 106664, 2440416, 7038240, 161030504, 464417208, 10625572880, 30644497520, 701126779608, 2022072419144, 46263741881280, 133426135166016, 3052705837384904, 8804102848537944, 201432321525522416

Offset: 1

Pierre CAMI, Apr 05 2005

This sequence has the property 17*a(n)*(a(n) + 1) + 1 is a square.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,66,-66,-1,1).

Cf. A001090, A103200.

R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( 8*x^3*(1+x)^2/((1-x)*(1-66*x^2+x^4)) )); // G. C. Greubel, Mar 13 2023

LinearRecurrence[{1,66,-66,-1,1}, {0,0,8,24,560}, 40] (* G. C. Greubel, Mar 13 2023 *)

@CachedFunction
def a(n):
    if (n<6): return (0,0,0,8,24,560)[n]
    else: return a(n-1) +66*a(n-2) -66*a(n-3) -a(n-4) +a(n-5)
[a(n) for n in range(1,41)] # G. C. Greubel, Mar 13 2023

From R. J. Mathar, Nov 13 2009: (Start)
a(n) = a(n-1) +66*a(n-2) -66*a(n-3) -a(n-4) +a(n-5).
G.f.: 8*x^3*(1+x)^2/((1-x)*(1+8*x-x^2)*(1-8*x-x^2)). (End)
a(n) = (1/4)*(-32*[n=0] - 2 + i^n*((23 + 11*(-1)^n)*ChebyshevU(n, 4*I) - i*(187 + 89*(-1)^n)*ChebyshevU(n-1, 4*I))). - G. C. Greubel, Mar 13 2023

More terms from R. J. Mathar, Nov 13 2009

A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30a(i)^2 + 30a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4sqrt(30(a(n-2)^2) + 30*a(n-2) + 1).

Original entry on oeis.org

0, 0, 3, 7, 76, 164, 1679, 3611, 36872, 79288, 809515, 1740735, 17772468, 38216892, 390184791, 839030899, 8566292944, 18420462896, 188068259987, 404411152823, 4128935426780, 8878624899220, 90648511129183, 194925336630027, 1990138309415256, 4279478780961384, 43692394296006459

Offset: 1

Pierre CAMI, Mar 27 2005

nonn

By construction and recurrence, 30*a(n)^2 + 30*a(n) + 1 = j(n)^2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,22,-22,-1,1).

Cf. A053141, A103200.

m:=25; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))); // G. C. Greubel, Jul 15 2018

Rest[CoefficientList[Series[x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)), {x, 0, 50}], x]] (* G. C. Greubel, Jul 15 2018 *)
LinearRecurrence[{1,22,-22,-1,1},{0,0,3,7,76},40] (* Harvey P. Dale, Mar 05 2025 *)

x='x+O('x^30); concat([0,0], Vec(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))) \\ G. C. Greubel, Jul 15 2018

G.f.: x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009

Terms a(19) onward added by G. C. Greubel, Jul 15 2018

A103625 a(n) = 3 + 7a(n-2) + sqrt(1 + 48a(n-2) + 48*a(n-2)^2), with a(1) = 0, a(2) = 0, a(3) = 2.

Original entry on oeis.org

0, 0, 2, 4, 34, 62, 480, 870, 6692, 12124, 93214, 168872, 1298310, 2352090, 18083132, 32760394, 251865544, 456293432, 3508034490, 6355347660, 48860617322, 88518573814, 680540608024, 1232904685742, 9478707895020, 17172147026580, 132021369922262, 239177153686384

Offset: 1

Pierre CAMI, Mar 29 2005

Define j(n) = sqrt(48*a(n)^2 + 48*a(n) + 1), then j(n) is prime for n=3, 4, 5, 6, 7, 25, 28, 32, 35, 48, 65, 66, 88, 96, 113, 119, 151, 155, 182, 220, 231, 316, 488, 531, 599, 722, 1049, 1176, ...
For n > 1, first member of the Diophantine pair (m,k) that satisfies 12*(m^2 + m) = k^2 + k; a(n)=m. - Herbert Kociemba, May 12 2008
Former name: Define a(1)=0, a(2)=0, a(3)=2, a(4)=4, a(5)=34, a(6)=62, a(7)=480, a(8)=870 such that from i=1 to 8: 48*a(i)^2 + 48*a(i) + 1 = j(i)^2 with j(1)=1, j(2)=1, j(3)=17, j(4)=31, j(5)=239, j(6)=433, j(7)=3329, j(8)=6031. Then a(n) = a(n-8) + 28*sqrt(48*(a(n-4)^2) + 48*a(n-4) + 1). - G. C. Greubel, Mar 22 2024

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,14,-14,-1,1).

Cf. A053141, A103200, A103737, A125905.

R:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!(2*(x^2+x+1)/(1-x-14*x^2+14*x^3+x^4-x^5))); // G. C. Greubel, Jul 15 2018

a[1]=0; a[2]=0; a[3]=2; a[n_]:=a[n]= 3+7a[n-2]+Sqrt[1+48a[n-2]+48a[n-2]^2]; Table[a[n],{n,1,20}] (* Herbert Kociemba, May 12 2008 *)
Rest@CoefficientList[Series[2*x^3*(1+x+x^2)/(1-x-14*x^2+14*x^3+x^4-x^5), {x,0,30}], x] (* G. C. Greubel, Jul 15 2018 *)
LinearRecurrence[{1,14,-14,-1,1},{0,0,2,4,34},30] (* Harvey P. Dale, Jun 04 2021 *)

my(x='x+O('x^30)); concat([0,0], Vec(2*x^3*(1+x+x^2)/(1-x-14*x^2+14*x^3 + x^4-x^5))) \\ G. C. Greubel, Jul 15 2018

@CachedFunction
def b(n): return chebyshev_U(n, -2) # A125905
def A103625(n): return (1/8)*(-16*int(n==0) -4 +5*(-1)^n*(3*b(n) +11*b(n-1)) +5*b(n) +19*b(n-1))
[A103625(n) for n in range(1,41)] # G. C. Greubel, Mar 22 2024

G.f.: 2*x^3*(1+x+x^2)/((1-x)*(1-4*x+x^2)*(1+4*x+x^2)). - Ralf Stephan, May 18 2007
a(n) = (1/8)*(-16*[n=0] - 4 + 5*(-1)^n*(3*A125905(n) + 11*A125905(n-1)) + (5*A125905(n) + 19*A125905(n-1))), where A125905(n) = ChebyshevU(n, -2). - G. C. Greubel, Mar 22 2024
E.g.f.: (15*cosh(sqrt(3)*x)*(2*cosh(2*x) + sinh(2*x))/2 - sqrt(3)*(4*cosh(x) + sinh(x))*(cosh(x) + 4*sinh(x))*sinh(sqrt(3)*x) - 3*(4 + exp(x)))/6. - Stefano Spezia, Jun 02 2024

Terms a(17) onward added by G. C. Greubel, Jul 15 2018

Edited by G. C. Greubel, Mar 22 2024

cp's OEIS Frontend

A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).

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A105038 Nonnegative n such that 6*n^2 + 6*n + 1 is a square.

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A105045 a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=12; for n > 4, a(n) = 8*a(n-2) - a(n-4) - 3.

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A103201 a(1) = 11, a(2) = 19, a(3) = 89, a(4) = 151; for n >= 5, a(n) = sqrt(a(n-4)^2 + 60*a(n-2)^2 + 4*a(n-2)*sqrt(210 + 15*a(n-4)^2)).

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A105063 a(1)=0, a(2)=0, a(3)=8, a(4)=24, a(n) = 32 + 66*a(n-2) - a(n-4) for n > 4.

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A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30*a(i)^2 + 30*a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4*sqrt(30*(a(n-2)^2) + 30*a(n-2) + 1).

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A053141 a(0)=0, a(1)=2 then a(n) = a(n-2) + 2sqrt(8a(n-1)^2 + 8*a(n-1) + 1).

A105038 Nonnegative n such that 6n^2 + 6n + 1 is a square.

A103201 a(1) = 11, a(2) = 19, a(3) = 89, a(4) = 151; for n >= 5, a(n) = sqrt(a(n-4)^2 + 60a(n-2)^2 + 4a(n-2)sqrt(210 + 15a(n-4)^2)).

A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30a(i)^2 + 30a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4sqrt(30(a(n-2)^2) + 30*a(n-2) + 1).

A103625 a(n) = 3 + 7a(n-2) + sqrt(1 + 48a(n-2) + 48*a(n-2)^2), with a(1) = 0, a(2) = 0, a(3) = 2.

A103715 Define a(1)=0, a(2)=0, a(3)=1, a(4)=3, a(5)=18, a(6)=22, a(7)=119, a(8)=285. Then a(n) = a(n-8) + 4sqrt(420a(n-4)^2 + 420*a(n-4) + 1).

A105076 Numbers k such that 60k^2 + 60k + 1 is a square.