cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A006061 Star numbers (A003154) that are squares.

Original entry on oeis.org

1, 121, 11881, 1164241, 114083761, 11179044361, 1095432263641, 107341182792481, 10518340481399521, 1030690025994360601, 100997104206965939401, 9896685522256667700721, 969774184076946468731281
Offset: 1

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Author

Keywords

Examples

			a(2)=121 because this is the 2nd star number (A003154) that is a square.
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 121, p. 42, Ellipses, Paris 2008.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A007667 is 3*a(n)+2, sqrt(a(n)) is A054320.
Cf. A003154.

Programs

  • GAP
    a:=[1,121,11881];; for n in [4..20] do a[n]:=99*a[n-1]-99*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1+22*x+x^2)/((1-x)*(1-98*x+x^2)) )); // G. C. Greubel, Jul 23 2019
    
  • Maple
    Digits := 1000:q := seq(floor(evalf(( (5+2*sqrt(6))^n*(sqrt(6)-2)-(5-2*sqrt(6))^n*(sqrt(6)+2))^2/16)),n=1..100);
    A006061:=-(1+22*z+z**2)/(z-1)/(z**2-98*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    CoefficientList[Series[(1+22*x+x^2)/((1-x)*(1-98*x+x^2)), {x,0,20}], x] (* or *) LinearRecurrence[{99,-99,1}, {1,121,11881}, 20] (* G. C. Greubel, Jul 23 2019 *)
  • PARI
    my(x='x+O('x^20)); Vec((1+22*x+x^2)/((1-x)*(1-98*x+x^2))) \\ G. C. Greubel, Jul 23 2019
    
  • Sage
    ((1+22*x+x^2)/((1-x)*(1-98*x+x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019
    

Formula

A007667 = 3*square star numbers (A006061) + 2.
a(n) = denominator of kappa(sqrt(6)/A054320(n)) where kappa(x) is the sum of successive remainders by computing the Euclidean algorithm for (1, x). - Thomas Baruchel, Nov 29 2003
From Ignacio Larrosa Cañestro, Feb 27 2000: (Start)
a(n) = 99*(a(n-1) - a(n-2)) + a(n-3).
a(n) = (5 - 2*sqrt(6))/8*(sqrt(3) + sqrt(2))^(4*n) + (5 + 2*sqrt(6))/8*(sqrt(3) - sqrt(2))^(4*n) - 1/4. (End)
a(n) = 98*a(n-1) - a(n-2) + 24. - Lekraj Beedassy, Jul 14 2008

Extensions

More terms from Eric W. Weisstein and Sascha Kurz, Mar 24 2002

A054318 a(n)-th star number (A003154) is a square.

Original entry on oeis.org

1, 5, 45, 441, 4361, 43165, 427285, 4229681, 41869521, 414465525, 4102785725, 40613391721, 402031131481, 3979697923085, 39394948099365, 389969783070561, 3860302882606241, 38213059042991845, 378270287547312205
Offset: 1

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Author

Keywords

Comments

A two-way infinite sequence which is palindromic.
Also indices of centered hexagonal numbers (A003215) which are also centered square numbers (A001844). - Colin Barker, Jan 02 2015
Also positive integers y in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0. - Colin Barker, Jan 02 2015

Examples

			a(2) = 5 because the 5th Star number (A003154) 121=11^2 is the 2nd that is a square.
		

Crossrefs

A031138 is 3*a(n)-2. Cf. A003154, A006061, A182432, A211955.
Quintisection of column k=2 of A233427.

Programs

  • GAP
    a:=[1,5,45];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) )); // G. C. Greubel, Jul 23 2019
    
  • Mathematica
    CoefficientList[Series[x(1-6x+x^2)/((1-x)(1-10x+x^2)), {x,0,30}], x] (* Michael De Vlieger, Aug 11 2016 *)
    LinearRecurrence[{11,-11,1},{1,5,45},30] (* Harvey P. Dale, Nov 05 2016 *)
  • PARI
    a(n)=if(n<1,a(1-n),1/2+subst(poltchebi(n)+poltchebi(n-1),x,5)/12)
    
  • PARI
    Vec(x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jan 02 2015
    
  • Sage
    (x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019
    

Formula

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 1/2 + (3 - sqrt(6))/12*(5 + 2*sqrt(6))^n + (3 + sqrt(6))/12*(5 - 2*sqrt(6))^n.
From Michael Somos, Mar 18 2003: (Start)
G.f.: x*(1-6*x+x^2)/((1-x)*(1-10*x+x^2)).
12*a(n)*a(n-1) + 4 = (a(n) + a(n-1) + 2)^2.
a(n) = a(1-n) = 10*a(n-1) - a(n-2) - 4.
a(n) = 12*a(n-1)^2/(a(n-1) + a(n-2)) - a(n-1).
a(n) = (a(n-1) + 4)*a(n-1)/a(n-2). (End)
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + (1/2)*Sum_{k = 1..n} 8^k*binomial(n+k,2*k).
a(n+1) = R(n,4), where R(n,x) is the n-th row polynomial of A211955.
a(n+1) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(3) and T(n,x) the Chebyshev polynomial of the first kind.
Sum {k>=0} 1/a(k) = sqrt(3/2). (End)
A003154(a(n)) = A006061(n). - Zak Seidov, Oct 22 2012
a(n) = (4*a(n-1) + a(n-1)^2) / a(n-2), n >= 3. - Seiichi Manyama, Aug 11 2016
2*a(n) = 1+A072256(n). - R. J. Mathar, Feb 07 2022

Extensions

More terms from James Sellers, Mar 01 2000

A083577 Primes in A003154.

Original entry on oeis.org

13, 37, 73, 181, 337, 433, 541, 661, 937, 1093, 2053, 2281, 2521, 3037, 3313, 5581, 5953, 6337, 6733, 7561, 7993, 8893, 10333, 10837, 11353, 12421, 12973, 13537, 15913, 18481, 20533, 21961, 25741, 27337, 32413, 33301, 36037, 36973, 42841
Offset: 1

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Author

Jason Earls, Jun 13 2003

Keywords

Comments

Except for the first 3 terms, a subsequence of A057199 (see comment in A057199). - Chai Wah Wu, Sep 16 2019
All terms are primes of the form 4*k + 1 (A002144), because terms of A003154 are of the form 12*k + 1. - John Elias, Jan 25 2021

Crossrefs

Programs

  • Mathematica
    Select[Array[6*#*(#-1)+1 &, 100], PrimeQ] (* Paolo Xausa, Mar 05 2024 *)
  • PARI
    for(n=1,150,s=6*n*(n-1)+1; if(isprime(s),print1(s",")))

A156712 Star numbers (A003154) that are also triangular numbers (A000217).

Original entry on oeis.org

1, 7, 91, 1261, 17557, 244531, 3405871, 47437657, 660721321, 9202660831, 128176530307, 1785268763461, 24865586158141, 346332937450507, 4823795538148951, 67186804596634801, 935791468814738257, 13033893758809700791, 181538721154521072811, 2528508202404485318557
Offset: 1

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Author

Aaron Meyerowitz, Feb 14 2009

Keywords

Comments

From Colin Barker, Jan 06 2015: (Start)
Also indices of centered square numbers (A001844) which are also centered triangular numbers (A005448).
Also indices of centered octagonal numbers (A016754) which are also centered hexagonal numbers (A003215).
Also positive integers y in the solutions to 3*x^2-4*y^2-3*x+4*y = 0, the corresponding values of x being A001922.
(End)

Crossrefs

Programs

  • Magma
    [(Evaluate(ChebyshevSecond(n+1),7) - 13*Evaluate(ChebyshevU(n), 7) + 1)/2: n in [1..30]]; // G. C. Greubel, Oct 07 2022
    
  • Maple
    f:= gfun[rectoproc]({a(n+3)=15*a(n+2)-15*a(n+1)+a(n),a(1)=1,a(2)=7,a(3)=91},a(n),'remember'):
    seq(f(n), n=1..30); # Robert Israel, Jan 01 2015
  • Mathematica
    f[n_] := (Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)] + 4)/8; Array[f, 17] (* Robert G. Wilson v, Oct 28 2010 *)
  • PARI
    Vec(-x*(x^2-8*x+1)/((x-1)*(x^2-14*x+1)) + O(x^100)) \\ Colin Barker, Jan 01 2015
    
  • SageMath
    def A156712(n): return (1 + chebyshev_U(n, 7) - 13*chebyshev_U(n-1, 7))/2
    [A156712(n) for n in range(1,31)] # G. C. Greubel, Oct 07 2022

Formula

a(n+3) = 15*a(n+2) - 15*a(n+1) + a(n).
If x^2 - 3*y^2 = 1 with x even then a(y) = (y+2)/4 evidently related to A001570 by: add 1 and halve.
G.f.: x*(1 - 8*x + x^2)/((1-x)*(1 - 14*x + x^2)). - Alexander R. Povolotsky, Feb 15 2009
a(n) = (4 + (2 + sqrt(3))*(7 - 4*sqrt(3))^n + (2 - sqrt(3))*(7 + 4*sqrt(3))^n)/8. - Colin Barker, Mar 05 2016
a(n) = (1/2)*( 1 + ChebyshevU(n, 7) - 13*ChebyshevU(n-1, 7) ). - G. C. Greubel, Oct 07 2022

Extensions

a(11) onwards from Robert G. Wilson v, Oct 28 2010

A218172 Centered 12-gonal numbers which are semiprimes, intersection of A003154 and A001358.

Original entry on oeis.org

121, 253, 793, 1261, 1441, 1633, 1837, 2773, 3601, 3901, 4213, 4537, 4873, 5221, 7141, 9841, 11881, 14113, 14701, 16537, 17173, 17821, 19153, 19837, 21241, 22693, 23437, 24193, 24961, 28153, 28981, 29821, 30673, 34201, 37921, 38881, 39853, 40837, 41833, 43861, 45937, 48061, 49141, 50233, 53581, 55873
Offset: 1

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Author

Zak Seidov, Oct 22 2012

Keywords

Comments

Might also be called 'semiprime star numbers'.
A083749 and A006061 are subsequences.

Examples

			a(1) = 121 = 11^2 = A001358(40) = A003154(5) = A083749(1) = A006061(1) = A078972(11).
a(2) = 253 = 11*23 = A001358(81) = A003154(7) = A083749(2) = A078972(18).
		

Crossrefs

Programs

  • Mathematica
    Select[Table[6n(n-1)+1,{n,100}],PrimeOmega[#]==2&] (* Harvey P. Dale, Sep 01 2014 *)
  • PARI
    lista(nn) = {for (n = 1, nn, if (bigomega(v = 6*n*(n-1) + 1) == 2, print1(v, ", ")););} \\ Michel Marcus, Nov 09 2013

A306980 Decimal expansion of the sum of the reciprocals of the star numbers (A003154).

Original entry on oeis.org

1, 1, 5, 9, 1, 7, 3, 3, 1, 9, 6, 3, 2, 1, 7, 4, 6, 9, 5, 0, 7, 8, 5, 6, 6, 6, 2, 4, 0, 6, 7, 4, 3, 7, 8, 7, 6, 8, 4, 4, 6, 3, 5, 5, 1, 1, 9, 5, 5, 9, 3, 9, 5, 5, 3, 9, 0, 2, 4, 4, 9, 7, 4, 6, 9, 9, 6, 9, 5, 3, 5, 6, 5, 1, 4, 1, 1, 6, 8, 4, 3, 4, 5, 0, 0, 4, 2
Offset: 1

Views

Author

Amiram Eldar, Mar 18 2019

Keywords

Examples

			1.15917331963217469507856662406743787684463551195593...
		

Crossrefs

Cf. A003154, A093766 (Pi/(2*sqrt(3))).

Programs

  • Mathematica
    s[n_]:=6*n*(n-1) + 1; RealDigits[Sum[1/s[n], {n,1,Infinity}], 10, 100][[1]]

Formula

Equals Sum_{n>=1} 1/(6*n*(n-1) + 1) = x * tan(x) with x = Pi/(2*sqrt(3)).

A083748 Star numbers (A003154) whose digital sum and product of nonzero digits are both triangular numbers.

Original entry on oeis.org

1, 37, 73, 433, 15301, 39853, 66781, 78661, 110161, 146953, 156493, 202033, 220033, 305101, 355753, 416593, 509833, 593461, 805933, 1011061, 1503001, 1563661, 2952613, 4000033, 4019653, 4609513, 5262193, 6211837, 6359221, 6608701
Offset: 1

Views

Author

Jason Earls, Jun 17 2003

Keywords

Examples

			a(2)=37 because
........()........
.......()().......
..()()()()()()()..
...()()()()()()...
....()()()()()....
...()()()()()()...
..()()()()()()()..
.......()().......
........()........
and 3+7 is
....()....
...()()...
..()()()..
.()()()().
and 3*7 is
......()......
.....()().....
....()()()....
...()()()()...
..()()()()()..
.()()()()()().
		

A338795 Each term of A003215 (centered hexagonal numbers) is multiplied by the corresponding term of A003154 (centered dodecagonal numbers).

Original entry on oeis.org

1, 91, 703, 2701, 7381, 16471, 32131, 56953, 93961, 146611, 218791, 314821, 439453, 597871, 795691, 1038961, 1334161, 1688203, 2108431, 2602621, 3178981, 3846151, 4613203, 5489641, 6485401, 7610851, 8876791, 10294453, 11875501, 13632031, 15576571, 17722081, 20081953
Offset: 1

Views

Author

David Z Crookes, Nov 09 2020

Keywords

Comments

The digital root (A010888) of each term is 1.

Examples

			The centered hexagonal number of 4 is 37, and the centered dodecagonal number of 4 is 73, so the fourth term of the series is 37*73 = 2701.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{1,91,703,2701,7381},40] (* Harvey P. Dale, May 13 2022 *)

Formula

a(n) = A003215(n)*A003154(n).
a(n) = 18*n^4 - 36*n^3 + 27*n^2 - 9*n + 1.
From Elmo R. Oliveira, Sep 01 2025: (Start)
G.f.: -x*(1 + 86*x + 258*x^2 + 86*x^3 + x^4)/(x - 1)^5.
E.g.f.: -1 + exp(x)*(1 + 45*x^2 + 72*x^3 + 18*x^4).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)

A016754 Odd squares: a(n) = (2n+1)^2. Also centered octagonal numbers.

Original entry on oeis.org

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025
Offset: 0

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Author

Keywords

Comments

The brown rat (rattus norwegicus) breeds very quickly. It can give birth to other rats 7 times a year, starting at the age of three months. The average number of pups is 8. The present sequence gives the total number of rats, when the intervals are 12/7 of a year and a young rat starts having offspring at 24/7 of a year. - Hans Isdahl, Jan 26 2008
Numbers n such that tau(n) is odd where tau(x) denotes the Ramanujan tau function (A000594). - Benoit Cloitre, May 01 2003
If Y is a fixed 2-subset of a (2n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Binomial transform of [1, 8, 8, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
All terms of this sequence are of the form 8k+1. For numbers 8k+1 which aren't squares see A138393. Numbers 8k+1 are squares iff k is a triangular number from A000217. And squares have form 4n(n+1)+1. - Artur Jasinski, Mar 27 2008
Sequence arises from reading the line from 1, in the direction 1, 25, ... and the line from 9, in the direction 9, 49, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Equals the triangular numbers convolved with [1, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
First differences: A008590(n) = a(n) - a(n-1) for n>0. - Reinhard Zumkeller, Nov 08 2009
Central terms of the triangle in A176271; cf. A000466, A053755. - Reinhard Zumkeller, Apr 13 2010
Odd numbers with odd abundance. Odd numbers with even abundance are in A088828. Even numbers with odd abundance are in A088827. Even numbers with even abundance are in A088829. - Jaroslav Krizek, May 07 2011
Appear as numerators in the non-simple continued fraction expansion of Pi-3: Pi-3 = K_{k>=1} (1-2*k)^2/6 = 1/(6+9/(6+25/(6+49/(6+...)))), see also the comment in A007509. - Alexander R. Povolotsky, Oct 12 2011
Ulam's spiral (SE spoke). - Robert G. Wilson v, Oct 31 2011
All terms end in 1, 5 or 9. Modulo 100, all terms are among { 1, 9, 21, 25, 29, 41, 49, 61, 69, 81, 89 }. - M. F. Hasler, Mar 19 2012
Right edge of both triangles A214604 and A214661: a(n) = A214604(n+1,n+1) = A214661(n+1,n+1). - Reinhard Zumkeller, Jul 25 2012
Also: Odd numbers which have an odd sum of divisors (= sigma = A000203). - M. F. Hasler, Feb 23 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective even leg b (A231100); sequence gives values c-b, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
For n>1 a(n) is twice the area of the irregular quadrilateral created by the points ((n-2)*(n-1),(n-1)*n/2), ((n-1)*n/2,n*(n+1)/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+2)*(n+3)/2,(n+1)*(n+2)/2). - J. M. Bergot, May 27 2014
Number of pairs (x, y) of Z^2, such that max(abs(x), abs(y)) <= n. - Michel Marcus, Nov 28 2014
Except for a(1)=4, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
a(n) is the sum of 2n+1 consecutive numbers, the first of which is n+1. - Ivan N. Ianakiev, Dec 21 2016
a(n) is the number of 2 X 2 matrices with all elements in {0..n} with determinant = 2*permanent. - Indranil Ghosh, Dec 25 2016
Engel expansion of Pi*StruveL_0(1)/2 where StruveL_0(1) is A197037. - Benedict W. J. Irwin, Jun 21 2018
Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; the segments on the hypotenuse {p = a(n)/A001844(n), q = A060300(n)/A001844(n) = A001844(n) - p} and their ratio p/q = a(n)/A060300(n) are irreducible fractions in Q\Z. X values are A005408, Y values are A046092, Z values are A001844. - Ralf Steiner, Feb 25 2020
a(n) is the number of large or small squares that are used to tile primitive squares of type 2 (A344332). - Bernard Schott, Jun 03 2021
Also, positive odd integers with an odd number of odd divisors (for similar sequence with 'even', see A348005). - Bernard Schott, Nov 21 2021
a(n) is the least odd number k = x + y, with 0 < x < y, such that there are n distinct pairs (x,y) for which x*y/k is an integer; for example, a(2) = 25 and the two corresponding pairs are (5,20) and (10,15). The similar sequence with 'even' is A016742 (see Comment of Jan 26 2018). - Bernard Schott, Feb 24 2023
From Peter Bala, Jan 03 2024: (Start)
The sequence terms are the exponents of q in the series expansions of the following infinite products:
1) q*Product_{n >= 1} (1 - q^(16*n))*(1 + q^(8*n)) = q + q^9 + q^25 + q^49 + q^81 + q^121 + q^169 + ....
2) q*Product_{n >= 1} (1 + q^(16*n))*(1 - q^(8*n)) = q - q^9 - q^25 + q^49 + q^81 - q^121 - q^169 + + - - ....
3) q*Product_{n >= 1} (1 - q^(8*n))^3 = q - 3*q^9 + 5*q^25 - 7*q^49 + 9*q^81 - 11*q^121 + 13*q^169 - + ....
4) q*Product_{n >= 1} ( (1 + q^(8*n))*(1 - q^(16*n))/(1 + q^(16*n)) )^3 = q + 3*q^9 - 5*q^25 - 7*q^49 + 9*q^81 + 11*q^121 - 13*q^169 - 15*q^225 + + - - .... (End)

References

  • L. Lorentzen and H. Waadeland, Continued Fractions with Applications, North-Holland 1992, p. 586.

Crossrefs

Cf. A000447 (partial sums).
Cf. A348005, A379481 [= a(A048673(n)-1)].
Partial sums of A022144.
Positions of odd terms in A341528.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = 1 + Sum_{i=1..n} 8*i = 1 + 8*A000217(n). - Xavier Acloque, Jan 21 2003; Zak Seidov, May 07 2006; Robert G. Wilson v, Dec 29 2010
O.g.f.: (1+6*x+x^2)/(1-x)^3. - R. J. Mathar, Jan 11 2008
a(n) = 4*n*(n + 1) + 1 = 4*n^2 + 4*n + 1. - Artur Jasinski, Mar 27 2008
a(n) = A061038(2+4n). - Paul Curtz, Oct 26 2008
Sum_{n>=0} 1/a(n) = Pi^2/8 = A111003. - Jaume Oliver Lafont, Mar 07 2009
a(n) = A000290(A005408(n)). - Reinhard Zumkeller, Nov 08 2009
a(n) = a(n-1) + 8*n with n>0, a(0)=1. - Vincenzo Librandi, Aug 01 2010
a(n) = A033951(n) + n. - Reinhard Zumkeller, May 17 2009
a(n) = A033996(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = (A005408(n))^2. - Zak Seidov, Nov 29 2011
From George F. Johnson, Sep 05 2012: (Start)
a(n+1) = a(n) + 4 + 4*sqrt(a(n)).
a(n-1) = a(n) + 4 - 4*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 8.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2).
(a(n+1) - a(n-1))/8 = sqrt(a(n)).
a(n+1)*a(n-1) = (a(n)-4)^2.
a(n) = 2*A046092(n) + 1 = 2*A001844(n) - 1 = A046092(n) + A001844(n).
Limit_{n -> oo} a(n)/a(n-1) = 1. (End)
a(n) = binomial(2*n+2,2) + binomial(2*n+1,2). - John Molokach, Jul 12 2013
E.g.f.: (1 + 8*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 23 2016
a(n) = A101321(8,n). - R. J. Mathar, Jul 28 2016
Product_{n>=1} A033996(n)/a(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A014105(n) + A000384(n+1). - Bruce J. Nicholson, Nov 11 2017
a(n) = A003215(n) + A002378(n). - Klaus Purath, Jun 09 2020
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 13*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/e. (End)
Sum_{n>=0} (-1)^n/a(n) = A006752. - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi/4 (A003881). (End)
From Leo Tavares, Nov 24 2021: (Start)
a(n) = A014634(n) - A002943(n). See Diamond Triangles illustration.
a(n) = A003154(n+1) - A046092(n). See Diamond Stars illustration. (End)
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 ))))).
3/2 - 2*log(2) = Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 - ... ))))).
Row 2 of A142992. (End)
From Peter Bala, Mar 26 2024: (Start)
8*a(n) = (2*n + 1)*(a(n+1) - a(n-1)).
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 1/2 - Pi/8 = 1/(9 + (1*3)/(8 + (3*5)/(8 + ... + (4*n^2 - 1)/(8 + ... )))). For the continued fraction use Lorentzen and Waadeland, p. 586, equation 4.7.9 with n = 1. Cf. A057813. (End)

Extensions

Additional description from Terrel Trotter, Jr., Apr 06 2002

A003462 a(n) = (3^n - 1)/2.

Original entry on oeis.org

0, 1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573, 265720, 797161, 2391484, 7174453, 21523360, 64570081, 193710244, 581130733, 1743392200, 5230176601, 15690529804, 47071589413, 141214768240, 423644304721, 1270932914164
Offset: 0

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Comments

Partial sums of A000244. Values of base 3 strings of 1's.
a(n) = (3^n-1)/2 is also the number of different nonparallel lines determined by pair of vertices in the n dimensional hypercube. Example: when n = 2 the square has 4 vertices and then the relevant lines are: x = 0, y = 0, x = 1, y = 1, y = x, y = 1-x and when we identify parallel lines only 4 remain: x = 0, y = 0, y = x, y = 1 - x so a(2) = 4. - Noam Katz (noamkj(AT)hotmail.com), Feb 11 2001
Also number of 3-block bicoverings of an n-set (if offset is 1, cf. A059443). - Vladeta Jovovic, Feb 14 2001
3^a(n) is the highest power of 3 dividing (3^n)!. - Benoit Cloitre, Feb 04 2002
Apart from the a(0) and a(1) terms, maximum number of coins among which a lighter or heavier counterfeit coin can be identified (but not necessarily labeled as heavier or lighter) by n weighings. - Tom Verhoeff, Jun 22 2002, updated Mar 23 2017
n such that A001764(n) is not divisible by 3. - Benoit Cloitre, Jan 14 2003
Consider the mapping f(a/b) = (a + 2b)/(2a + b). Taking a = 1, b = 2 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the sequence 1/2, 4/5, 13/14, 40/41, ... converging to 1. Sequence contains the numerators = (3^n-1)/2. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a+b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Binomial transform of A000079 (with leading zero). - Paul Barry, Apr 11 2003
With leading zero, inverse binomial transform of A006095. - Paul Barry, Aug 19 2003
Number of walks of length 2*n + 2 in the path graph P_5 from one end to the other one. Example: a(2) = 4 because in the path ABCDE we have ABABCDE, ABCBCDE, ABCDCDE and ABCDEDE. - Emeric Deutsch, Apr 02 2004
The number of triangles of all sizes (not counting holes) in Sierpiński's triangle after n inscriptions. - Lee Reeves (leereeves(AT)fastmail.fm), May 10 2004
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n + 1, s(0) = 1, s(2n+1) = 4. - Herbert Kociemba, Jun 10 2004
Number of non-degenerate right-angled incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1. - Alex Fink and R. K. Guy, Aug 18 2005
Also numerator of the sum of the reciprocals of the first n powers of 3, with A000244 being the sequence of denominators. With the exception of n < 2, the base 10 digital root of a(n) is always 4. In base 3 the digital root of a(n) is the same as the digital root of n. - Alonso del Arte, Jan 24 2006
The sequence 3*a(n), n >= 1, gives the number of edges of the Hanoi graph H_3^{n} with 3 pegs and n >= 1 discs. - Daniele Parisse, Jul 28 2006
Numbers n such that a(n) is prime are listed in A028491 = {3, 7, 13, 71, 103, 541, 1091, ...}. 2^(m+1) divides a(2^m*k) for m > 0. 5 divides a(4k). 5^2 divides a(20k). 7 divides a(6k). 7^2 divides a(42k). 11^2 divides a(5k). 13 divides a(3k). 17 divides a(16k). 19 divides a(18k). 1093 divides a(7k). 41 divides a(8k). p divides a((p-1)/5) for prime p = {41, 431, 491, 661, 761, 1021, 1051, 1091, 1171, ...}. p divides a((p-1)/4) for prime p = {13, 109, 181, 193, 229, 277, 313, 421, 433, 541, ...}. p divides a((p-1)/3) for prime p = {61, 67, 73, 103, 151, 193, 271, 307, 367, ...} = A014753, 3 and -3 are both cubes (one implies other) mod these primes p = 1 mod 6. p divides a((p-1)/2) for prime p = {11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, ...} = A097933(n). p divides a(p-1) for prime p > 7. p^2 divides a(p*(p-1)k) for all prime p except p = 3. p^3 divides a(p*(p-1)*(p-2)k) for prime p = 11. - Alexander Adamchuk, Jan 22 2007
Let P(A) be the power set of an n-element set A. Then a(n) = the number of [unordered] pairs of elements {x,y} of P(A) for which x and y are disjoint [and both nonempty]. Wieder calls these "disjoint usual 2-combinations". - Ross La Haye, Jan 10 2008 [This is because each of the elements of {1, 2, ..., n} can be in the first subset, in the second or in neither. Because there are three options for each, the total number of options is 3^n. However, since the sets being empty is not an option we subtract 1 and since the subsets are unordered we then divide by 2! (The number of ways two objects can be arranged.) Thus we obtain (3^n-1)/2 = a(n). - Chayim Lowen, Mar 03 2015]
Also, still with P(A) being the power set of a n-element set A, a(n) is the number of 2-element subsets {x,y} of P(A) such that the union of x and y is equal to A. Cf. A341590. - Fabio Visonà, Feb 20 2021
Starting with offset 1 = binomial transform of A003945: (1, 3, 6, 12, 24, ...) and double bt of (1, 2, 1, 2, 1, 2, ...); equals polcoeff inverse of (1, -4, 3, 0, 0, 0, ...). - Gary W. Adamson, May 28 2009
Also the constant of the polynomials C(x) = 3x + 1 that form a sequence by performing this operation repeatedly and taking the result at each step as the input at the next. - Nishant Shukla (n.shukla722(AT)gmail.com), Jul 11 2009
It appears that this is A120444(3^n-1) = A004125(3^n) - A004125(3^n-1), where A004125 is the sum of remainders of n mod k for k = 1, 2, 3, ..., n. - John W. Layman, Jul 29 2009
Subsequence of A134025; A171960(a(n)) = a(n). - Reinhard Zumkeller, Jan 20 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j] = 1, A[i, i] := 3, (i > 1), A[i, i-1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = det(A). - Milan Janjic, Jan 27 2010
This is the sequence A(0, 1; 2, 3; 2) = A(0, 1; 4, -3; 0) of the family of sequences [a, b:c, d:k] considered by Gary Detlefs, and treated as A(a, b; c, d; k) in the Wolfdieter Lang link given below. - Wolfdieter Lang, Oct 18 2010
It appears that if s(n) is a first order rational sequence of the form s(0) = 0, s(n) = (2*s(n-1)+1)/(s(n-1)+2), n > 0, then s(n)= a(n)/(a(n)+1). - Gary Detlefs, Nov 16 2010
This sequence also describes the total number of moves to solve the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Towers of Hanoi puzzle (cf. A183111 - A183125).
From Adi Dani, Jun 08 2011: (Start)
a(n) is number of compositions of odd numbers into n parts less than 3. For example, a(3) = 13 and there are 13 compositions odd numbers into 3 parts < 3:
1: (0, 0, 1), (0, 1, 0), (1, 0, 0);
3: (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0), (1, 1, 1);
5: (1, 2, 2), (2, 1, 2), (2, 2, 1).
(End)
Pisano period lengths: 1, 2, 1, 2, 4, 2, 6, 4, 1, 4, 5, 2, 3, 6, 4, 8, 16, 2, 18, 4, ... . - R. J. Mathar, Aug 10 2012
a(n) is the total number of holes (triangles removed) after the n-th step of a Sierpiński triangle production. - Ivan N. Ianakiev, Oct 29 2013
a(n) solves Sum_{j = a(n) + 1 .. a(n+1)} j = k^2 for some integer k, given a(0) = 0 and requiring smallest a(n+1) > a(n). Corresponding k = 3^n. - Richard R. Forberg, Mar 11 2015
a(n+1) equals the number of words of length n over {0, 1, 2, 3} avoiding 01, 02 and 03. - Milan Janjic, Dec 17 2015
For n >= 1, a(n) is also the total number of words of length n, over an alphabet of three letters, such that one of the letters appears an odd number of times (See A006516 for 4 letter words, and the Balakrishnan reference there). - Wolfdieter Lang, Jul 16 2017
Also, the number of maximal cliques, maximum cliques, and cliques of size 4 in the n-Apollonian network. - Andrew Howroyd, Sep 02 2017
For n > 1, the number of triangles (cliques of size 3) in the (n-1)-Apollonian network. - Andrew Howroyd, Sep 02 2017
a(n) is the largest number that can be represented with n trits in balanced ternary. Correspondingly, -a(n) is the smallest number that can be represented with n trits in balanced ternary. - Thomas König, Apr 26 2020
These form Sierpinski nesting-stars, which alternate pattern on 3^n+1/2 star numbers A003154, based on the square configurations of 9^n. The partial sums of 3^n are delineated according to the geometry of a hexagram, see illustrations in links. (3*a(n-1) + 1) create Sierpinski-anti-triangles, representing the number of holes in a (n+1) Sierpinski triangle (see illustrations). - John Elias, Oct 18 2021
For n > 1, a(n) is the number of iterations necessary to calculate the hyperbolic functions with CORDIC. - Mathias Zechmeister, Jul 26 2022
a(n) is the least number k such that A065363(k) = n. - Amiram Eldar, Sep 03 2022
For all n >= 0, Sum_{k=a(n)+1..a(n+1)} 1/k < Sum_{j=a(n+1)+1..a(n+2)} 1/j. These are the minimal points which partition the infinite harmonic series into a monotonically increasing sequence. Each partition approximates log(3) from below as n tends to infinity. - Joseph Wheat, Apr 15 2023
a(n) is also the number of 3-cycles in the n-Dorogovtsev-Goltsev-Mendes graph (using the convention the 0-Dorogovtsev-Goltsev-Mendes graph is P_2). - Eric W. Weisstein, Dec 06 2023

Examples

			There are 4 3-block bicoverings of a 3-set: {{1, 2, 3}, {1, 2}, {3}}, {{1, 2, 3}, {1, 3}, {2}}, {{1, 2, 3}, {1}, {2, 3}} and {{1, 2}, {1, 3}, {2, 3}}.
Ternary........Decimal
0.................0
1.................1
11................4
111..............13
1111.............40 etc. - _Zerinvary Lajos_, Jan 14 2007
There are altogether a(3) = 13 three letter words over {A,B,C} with say, A, appearing an odd number of times: AAA; ABC, ACB, ABB, ACC; BAC, CAB, BAB, CAC; BCA, CBA, BBA, CCA. - _Wolfdieter Lang_, Jul 16 2017
		

References

  • J. G. Mauldon, Strong solutions for the counterfeit coin problem, IBM Research Report RC 7476 (#31437) 9/15/78, IBM Thomas J. Watson Research Center, P. O. Box 218, Yorktown Heights, N. Y. 10598.
  • Paulo Ribenboim, The Book of Prime Number Records, Springer-Verlag, NY, 2nd ed., 1989, p. 60.
  • Paulo Ribenboim, The Little Book of Big Primes, Springer-Verlag, NY, 1991, p. 53.
  • Amir Sapir, The Tower of Hanoi with Forbidden Moves, The Computer J. 47 (1) (2004) 20, case three-in-a row, sequence a(n).
  • Robert Sedgewick, Algorithms, 1992, pp. 109.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sequences used for Shell sort: A033622, A003462, A036562, A036564, A036569, A055875.
Cf. A179526 (repeats), A113047 (characteristic function).
Cf. A000225, A000392, A004125, A014753, A028491 (indices of primes), A059443 (column k = 3), A065363, A097933, A120444, A321872 (sum reciprocals).
Cf. A064099 (minimal number of weightings to detect lighter or heavier coin among n coins).
Cf. A039755 (column k = 1).
Cf. A006516 (binomial transform, and special 4 letter words).
Cf. A341590.
Cf. A003462(n) (3-cycles), A367967(n) (5-cycles), A367968(n) (6-cycles).

Programs

Formula

G.f.: x/((1-x)*(1-3*x)).
a(n) = 4*a(n-1) - 3*a(n-2), n > 1. a(0) = 0, a(1) = 1.
a(n) = 3*a(n-1) + 1, a(0) = 0.
E.g.f.: (exp(3*x) - exp(x))/2. - Paul Barry, Apr 11 2003
a(n+1) = Sum_{k = 0..n} binomial(n+1, k+1)*2^k. - Paul Barry, Aug 20 2004
a(n) = Sum_{i = 0..n-1} 3^i, for n > 0; a(0) = 0.
a(n) = A125118(n, 2) for n > 1. - Reinhard Zumkeller, Nov 21 2006
a(n) = StirlingS2(n+1, 3) + StirlingS2(n+1, 2). - Ross La Haye, Jan 10 2008
a(n) = Sum_{k = 0..n} A106566(n, k)*A106233(k). - Philippe Deléham, Oct 30 2008
a(n) = 2*a(n-1) + 3*a(n-2) + 2, n > 1. - Gary Detlefs, Jun 21 2010
a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3), a(0) = 0, a(1) = 1, a(2) = 4. Observation by G. Detlefs. See the W. Lang comment and link. - Wolfdieter Lang, Oct 18 2010
A008344(a(n)) = 0, for n > 1. - Reinhard Zumkeller, May 09 2012
A085059(a(n)) = 1 for n > 0. - Reinhard Zumkeller, Jan 31 2013
G.f.: Q(0)/2 where Q(k) = 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 - 1/(3*9^k - 27*x*81^k/(9*x*9^k - 1/Q(k+1)))))); (continued fraction ). - Sergei N. Gladkovskii, Apr 12 2013
a(n) = A001065(3^n) where A001065(m) is the sum of the proper divisors of m for positive integer m. - Chayim Lowen, Mar 03 2015
a(n) = A000244(n) - A007051(n) = A007051(n)-1. - Yuchun Ji, Oct 23 2018
Sum_{n>=1} 1/a(n) = A321872. - Amiram Eldar, Nov 18 2020

Extensions

More terms from Michael Somos
Corrected my comment of Jan 10 2008. - Ross La Haye, Oct 29 2008
Removed comment that duplicated a formula. - Joerg Arndt, Mar 11 2010
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