A104240 -id:A104240 - cp's OEIS frontend

A103200 a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n) = a(n-4) + sqrt(60a(n-2)^2 + 60a(n-2) + 1) for n >= 5.

1, 2, 11, 19, 90, 153, 712, 1208, 5609, 9514, 44163, 74907, 347698, 589745, 2737424, 4643056, 21551697, 36554706, 169676155, 287794595, 1335857546, 2265802057, 10517184216, 17838621864, 82801616185, 140443172858, 651895745267, 1105706761003, 5132364345954

Offset: 1

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

nonn

The original version of this question was as follows: Let a(1) = 1, a(2) = 2, a(3) = 11, a(4) = 19; for n = 1..4 let b(n) = sqrt(60 a(n)^2 + 60 a(n) + 1); for n >= 5 let a(n) = a(n-4) + b(n-2), b(n) = sqrt(60 a(n)^2 + 60 a(n) +1). Bhanu and Deshpande ask for a proof that a(n) and b(n) are always integers. The b(n) sequence is A103201.

This sequence is also the interleaving of two sequences c and d that can be extended backwards: c(0) = c(1) = 0, c(n) = sqrt(60 c(n-1)^2 + 60 c(n-1) +1) + c(n-2) giving 0,0,1,11,90,712,5609,... d(0) = 1, d(1) = 0, d(n) = sqrt(60 d(n-1)^2 + 60 d(n-1) +1) + d(n-2) giving 1,0,2,19,153,1208,9514,... and interleaved: 0,1,0,0,1,2,11,19,90,153,712,1208,5609,9514,... lim_{n->infinity} a(n)/a(n-2) = 1/(4 - sqrt(15)), (1/(4-sqrt(15)))^n approaches an integer as n -> infinity. - Gerald McGarvey, Mar 29 2005

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.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-1,1).

Cf. A103201, A177187 (first differences).

I:=[1,2,11,19,90]; [n le 5 select I[n] else Self(n-1)+8*Self(n-2)-8*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Sep 28 2011

a[1]:=1: a[2]:=2:a[3]:=11: a[4]:=19: for n from 5 to 31 do a[n]:=a[n-4]+sqrt(60*a[n-2]^2+60*a[n-2]+1) od:seq(a[n],n=1..31); # Emeric Deutsch, Apr 13 2005

RecurrenceTable[{a[1]==1,a[2]==2,a[3]==11,a[4]==19,a[n]==a[n-4]+ Sqrt[60a[n-2]^2+60a[n-2]+1]},a[n],{n,40}] (* or *) LinearRecurrence[ {1,8,-8,-1,1},{1,2,11,19,90},40] (* Harvey P. Dale, Sep 27 2011 *)
CoefficientList[Series[-x*(1 + x + x^2)/((x - 1)*(x^4 - 8*x^2 + 1)), {x, 0, 40}], x] (* T. D. Noe, Jun 04 2012 *)

Comments from Pierre CAMI and Gerald McGarvey, Apr 20 2005: (Start)
Sequence satisfies a(0)=0, a(1)=1, a(2)=2, a(3)=11; for n > 3, a(n) = 8*a(n-2) - a(n-4) + 3.
G.f.: -x*(1 + x + x^2) / ( (x - 1)*(x^4 - 8*x^2 + 1) ). Note that the 3 = the sum of the coefficients in the numerator of the g.f., 8 appears in the denominator of the g.f. and 8 = 2*3 + 2. Similar relationships hold for other series defined as nonnegative n such that m*n^2 + m*n + 1 is a square, here m=60. Cf. A001652, A001570, A049629, A105038, A105040, A104240, A077288, A105036, A105037. (End)
a(2n) = (A105426(n)-1)/2, a(2n+1) = (A001090(n+2) - 5*A001090(n+1) - 1)/2. - Ralf Stephan, May 18 2007
a(1)=1, a(2)=2, a(3)=11, a(4)=19, a(5)=90, a(n) = a(n-1) + 8*a(n-2) - 8*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Sep 27 2011

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

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

A222390 Nonnegative integers m such that 10m(m+1)+1 is a square.

Original entry on oeis.org

0, 3, 15, 132, 588, 5031, 22347, 191064, 848616, 7255419, 32225079, 275514876, 1223704404, 10462309887, 46468542291, 397292260848, 1764580902672, 15086643602355, 67007605759263, 572895164628660, 2544524437949340, 21754929612286743, 96624921036315675

Offset: 1

Bruno Berselli, Feb 18 2013

a(n+1)/a(n) tends alternately to (7+2*sqrt(10))/3 and (13+4*sqrt(10))/3; a(n+2)/a(n) tends to A176398^2.

Subsequence of A014601.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. nonnegative integers m such that k*m*(m+1)+1 is a square: A001652 (k=2), A001921 (k=3), A001477 (k=4), A053606 (k=5), A105038 (k=6), A105040 (k=7), A053141 (k=8), this sequence (k=10), A105838 (k=11), A061278 (k=12), A104240 (k=13); A105063 (k=17), A222393 (k=18), A101180 (k=19), A077259 (k=20) [incomplete list].

Cf. A221875.

m:=22; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(3*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2))));

I:=[0,3,15,132,588]; [n le 5 select I[n] else Self(n-1) +38*Self(n-2)-38*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013

LinearRecurrence[{1, 38, -38, -1, 1}, {0, 3, 15, 132, 588}, 23]
CoefficientList[Series[3 x (1 + 4 x + x^2)/((1 - x) (1 - 6 x - x^2) (1 + 6 x - x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(-1/2+((5+(-1)^n*sqrt(10))*(3-sqrt(10))^(2*floor(n/2))+(5-(-1)^n*sqrt(10))*(3+sqrt(10))^(2*floor(n/2)))/20), n, 1, 23);

x='x+O('x^30); concat([0], Vec(3*x*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2)))) \\ G. C. Greubel, Jul 15 2018

G.f.: 3*x*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2)).
a(n) = a(-n+1) = a(n-1)+38*a(n-2)-38*a(n-3)-a(n-4)+a(n-5).
a(n) = -1/2+((5+t*(-1)^n)*(3-t)^(2*floor(n/2))+(5-t*(-1)^n)*(3+t)^(2*floor(n/2)))/20, where t=sqrt(10).
2*a(n)+1 = A221875(n).

A222393 Nonnegative integers m such that 18m(m+1)+1 is a square.

Original entry on oeis.org

0, 4, 12, 152, 424, 5180, 14420, 175984, 489872, 5978292, 16641244, 203085960, 565312440, 6898944364, 19203981732, 234361022432, 652370066464, 7961375818340, 22161378278060, 270452416801144, 752834491387592, 9187420795420572, 25574211328900084

Offset: 1

Bruno Berselli, Feb 19 2013

a(n+2)/a(n) tends to A156164.

a(n) is congruent to {0,2,4} (mod 5, 6 and 10).

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. nonnegative integers n such that k*n*(n+1)+1 is a square: A001652 (k=2), A001921 (k=3), A001477 (k=4), A053606 (k=5), A105038 (k=6), A105040 (k=7), A053141 (k=8), A222390 (k=10), A105838 (k=11), A061278 (k=12), A104240 (k=13); A105063 (k=17), this sequence (k=18), A101180 (k=19), A077259 (k=20) [incomplete list].

m:=22; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(4*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2))));

I:=[0,4,12,152,424]; [n le 5 select I[n] else Self(n-1)+34*Self(n-2)-34*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 4, 12, 152, 424}, 23]
CoefficientList[Series[4 x (1 + x)^2 / ((1 - x) (1 - 6 x + x^2) (1 + 6 x + x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(-1/2+((3+sqrt(2)*(-1)^n)*(3-2*sqrt(2))^(2*floor(n/2))+(3-sqrt(2)*(-1)^n)*(3+2*sqrt(2))^(2*floor(n/2)))/12), n, 1, 23);

x='x+O('x^30); concat([0], Vec(4*x*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)))) \\ G. C. Greubel, Jul 15 2018

G.f.: 4*x*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)).
a(n) = a(-n+1) = a(n-1)+34*a(n-2)-34*a(n-3)-a(n-4)+a(n-5).
a(n) = -1/2+((3+t*(-1)^n)*(3-2*t)^(2*floor(n/2))+(3-t*(-1)^n)*(3+2*t)^(2*floor(n/2)))/12, where t=sqrt(2).

A105046 a(n) = 1298*a(n-3) - a(n-6) - 648, for n>6, with a(0)=0, a(1)=1, a(2)=8, a(3)=145, a(4)=505, a(5)=9728, a(6)=187561.

Original entry on oeis.org

0, 1, 8, 145, 505, 9728, 187561, 654841, 12626288, 243453385, 849982465, 16388911448, 316002305521, 1103276584081, 21272794432568, 410170749112225, 1432052156154025, 27612070784561168, 532401316345361881, 1858802595411339721, 35840446605565962848

Offset: 0

Gerald McGarvey, Apr 03 2005

nonn

It appears this sequence gives all nonnegative m such that 13*m^2 - 13*m + 1 is a square and that a(n+1) = A104240(n) + 1. (A104240 is nonnegative n such that 13*n^2 + 13*n + 1 is a square.)

G. C. Greubel, Table of n, a(n) for n = 0..950
Index entries for linear recurrences with constant coefficients, signature (1,0,1298,-1298,0,-1,1).

Cf. A104240.

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+7*x+137*x^2-938*x^3+137*x^4+7*x^5+x^6) / ((1-x)*(1-11*x+x^2)*(1+11*x+120*x^2+11*x^3+x^4)) )); // G. C. Greubel, Mar 14 2023

CoefficientList[Series[x (1+7x+137x^2-938x^3+137x^4+7x^5+x^6)/((1-x) (1-11x+x^2)(1+11x+120x^2+11x^3+x^4)),{x,0,30}],x] (* or *)  LinearRecurrence[{1,0,1298,-1298,0,-1,1},{0,1,8,145,505,9728,187561, 654841},30] (* Harvey P. Dale, Jun 12 2012 *)

@CachedFunction
def a(n): # a = A105046
    if (n<8): return (0,1,8,145,505,9728,187561,654841)[n]
    else: return a(n-1) +1298*a(n-3) -1298*a(n-4) -a(n-6) +a(n-7)
[a(n) for n in range(41)] # G. C. Greubel, Mar 14 2023

a(n) = 1298*a(n-3) - a(n-6) - 648 for n > 6.

G.f.: x*(1+7*x+137*x^2-938*x^3+137*x^4+7*x^5+x^6) / ((1-x)*(1-11*x+x^2)*(1+11*x+120*x^2+11*x^3+x^4)). - R. J. Mathar, Sep 09 2008

More terms from Harvey P. Dale, Jun 12 2012

cp's OEIS Frontend

A103200 a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n) = a(n-4) + sqrt(60*a(n-2)^2 + 60*a(n-2) + 1) for n >= 5.

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

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A222390 Nonnegative integers m such that 10*m*(m+1)+1 is a square.

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A222393 Nonnegative integers m such that 18*m*(m+1)+1 is a square.

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A105046 a(n) = 1298*a(n-3) - a(n-6) - 648, for n>6, with a(0)=0, a(1)=1, a(2)=8, a(3)=145, a(4)=505, a(5)=9728, a(6)=187561.

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A103200 a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n) = a(n-4) + sqrt(60a(n-2)^2 + 60a(n-2) + 1) for n >= 5.

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

A222390 Nonnegative integers m such that 10m(m+1)+1 is a square.

A222393 Nonnegative integers m such that 18m(m+1)+1 is a square.