K. S. Bhanu - cp's OEIS frontend

User: K. S. Bhanu

K. S. Bhanu has authored 4 sequences.

A151794 a(1)=2, a(2)=4, a(3)=6; a(n+3) = a(n+2)+ 2*a(n), n>=1.

2, 4, 6, 14, 26, 54, 106, 214, 426, 854, 1706, 3414, 6826, 13654, 27306, 54614, 109226, 218454, 436906, 873814, 1747626, 3495254, 6990506, 13981014, 27962026, 55924054, 111848106, 223696214, 447392426, 894784854, 1789569706, 3579139414, 7158278826, 14316557654

Offset: 1

K. S. Bhanu (bhanu_105(AT)yahoo.com), Jun 21 2009

Consider the following coin tossing experiment. Let n >= 1 be a predetermined integer. We toss an unbiased coin sequentially. For each outcome, we score two points for a head (H) and one point for a tail (T). The coin is tossed until the total score reaches n or jumps from n-1 to n+1. The results of the tosses are written in a linear array. Then the probability of non-occurrence of double heads (HH) is given by p(n) = a(n) / 2^n, n>=1.

Bhanu K. S, Deshpande M. N. & Cholkar C. P. (2006): Coin tossing -Some Surprising Results, International Journal of Mathematical Education In Science and Technology, Vol.37, No.1, pp.115-119.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2).

Join[{2},LinearRecurrence[{1,2},{4,6},40]] (* Harvey P. Dale, Oct 19 2012 *)

Vec(2*x*(-x+x^2-1)/((1+x)*(2*x-1)) + O(x^100)) \\ Colin Barker, Jun 12 2015

G.f.: 2*x*(-x+x^2-1)/((1+x)*(2*x-1)).
a(n) = A084214(n), n>1.
a(n) = A168648(n-2), n>2.
a(n) = 2*A048573(n-2), n>1.
a(n) = (4*(-1)^n+5*2^n)/6 for n>1. - Colin Barker, Jun 12 2015

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

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.

Original entry on oeis.org

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

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

Original entry on oeis.org

1, 8, 56, 385, 2640, 18096, 124033, 850136, 5826920, 39938305, 273741216, 1876250208, 12860010241, 88143821480, 604146740120, 4140883359361, 28382036775408, 194533374068496, 1333351581704065, 9138927697859960

Offset: 1

K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, Apr 06 2004

a(n) such that 9*(T(a(n)-1) + T(a(n+1)-1)) = 7*(T(a(n) + a(n+1) - 1)), where T(i) denotes the i-th triangular number.
Partial sums of Chebyshev sequence S(n,7) = U(n,7/2) = A004187(n+1). - Wolfdieter Lang, Aug 31 2004
From Klaus Purath, Aug 06 2025: (Start)
Numbers k such that both 3*k + 1 and 15*k + 1 are perfect squares. Also the sum of two consecutive terms is a square.
Take any recurrence (r) of the form (3,-1) with initial value 0 followed by an arbitrary positive integer i. Then the product of two consecutive terms of r divided by 3*i^2 gives the current sequence. (End)

			G.f. = x + 8*x^2 + 56*x^3 + 385*x^4 + 2640*x^5 + 18096*x^6 + ... - _Michael Somos_, Jan 23 2025

Michael De Vlieger, Table of n, a(n) for n = 1..1197
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).

Cf. A000032, A004187, A058038, A081079.

A092521:= func< n | (Lucas(4*n+2) -3)/15 >; // G. C. Greubel, Jun 12 2025

a[1] = 1; a[2] = 8; a[3] = 56; a[n_] := a[n] = 8 a[n - 1] - 8 a[n - 2] + a[n - 3]; Table[ a[n], {n, 20}] (* Robert G. Wilson v, Apr 08 2004 *)
Table[(LucasL[4n+2]-3)/15, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
LinearRecurrence[{8,-8,1},{1,8,56},30] (* Harvey P. Dale, Dec 27 2015 *)

Vec(x/((1-x)*(1-7*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 29 2015

def A092521(n): return (lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

G.f.: x/(1 - 8*x + 8*x^2 - x^3) = x/((1 - x)*(1 - 7*x + x^2)).
a(n) = 7*a(n-1) - a(n-2) + 1, n>=2, a(0):=0, a(1)=1.
a(n) = (S(n, 7)-S(n-1, 7) -1)/5, n>=1, with S(n, 7) = U(n, 7/2) = A004187(n+1).
a(n) = A058038(n)/3.
a(n) = (1/3)*Sum_{k=0..n} Fibonacci(4*k). - Gary Detlefs, Dec 07 2010
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = A081079(n)/15.
E.g.f.: (1/15)*( exp(7*x/2)*( 3*cosh(p*x) + sqrt(5)*sinh(p*x) ) - 3*exp(x) ), where p = 3*sqrt(5)/2. (End)

Edited and extended by Robert G. Wilson v, Apr 08 2004

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User: K. S. Bhanu

A151794 a(1)=2, a(2)=4, a(3)=6; a(n+3) = a(n+2)+ 2*a(n), n>=1.

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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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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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A092521 a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

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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 + 60a(n-2)^2 + 4a(n-2)sqrt(210 + 15a(n-4)^2)).

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.

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.