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A013928 Number of (positive) squarefree numbers < n.

0, 1, 2, 3, 3, 4, 5, 6, 6, 6, 7, 8, 8, 9, 10, 11, 11, 12, 12, 13, 13, 14, 15, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 26, 26, 27, 28, 29, 29, 29, 30, 31, 31, 31, 31, 32, 32, 33, 33, 34, 34, 35, 36, 37, 37, 38, 39, 39, 39, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 47, 47, 48, 49, 50, 50, 50, 51

Offset: 1

Henri Lifchitz

For n >= 1 define an n X n (0, 1) matrix A by A[i, j] = 1 if gcd(i, j) = 1, A[i, j] = 0 if gcd(i, j) > 1 for 1 <= i,j <= n . The rank of A is a(n + 1). Asymptotic expression for a(n) is a(n) ~ n * 6 / Pi^2. - Sharon Sela (sharonsela(AT)hotmail.com), May 06 2002
a(n) = Sum_{k=1..n-1} A008966(k). - Reinhard Zumkeller, Jul 05 2010
For all n >= 1, a(n)/n >= a(176)/176 = 53/88, and the equality occurs only for n=176 (see K. Rogers link). - Michel Marcus, Dec 16 2012 [Thus the Schnirelmann density of the squarefree numbers is 53/88. - Charles R Greathouse IV, Feb 02 2016]
Cohen, Dress, & El Marraki prove that |a(n) - 6n/Pi^2| < 0.02767*sqrt(n) for n >= 438653. - Charles R Greathouse IV, Feb 02 2016

			a(10) = 6 because there are 6 squarefree numbers up to 10: 1, 2, 3, 5, 6, 7.
a(11) = 7 because there are 7 squarefree numbers up to 11: the numbers listed above for 10, plus 10 itself.
a(13) = 8 because the 12 X 12 matrix described in the first comment by Sharon Sela has rank 8. Rows 2,4,8 (the powers of two) are identical, rows 3,9 (the powers of three) are identical, and rows 6 and 12 (same prime factors) are identical. - _Geoffrey Critzer_, Dec 07 2014
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 0, 1, 0, 1, 0, 1, 0, 1, 0  1, 0, ...
1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, ...
1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, ...
1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, ...
1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, ...
1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...
1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, ...
1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, ...
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, ...
1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, ...
.                                   .
.                                    .
.                                     .

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth edition (1979), Clarendon Press, pp. 269-270.
E. Landau, Über den Zusammenhang einiger neuer Sätze der analytischen Zahlentheorie, Wiener Sitzungberichte, Math. Klasse 115 (1906), pp. 589-632. Cited in Sándor, Mitrinović, & Crstici.
József Sándor, Dragoslav S. Mitrinovic, and Borislav Crstici, Handbook of Number Theory I. Springer, 2005. Section VI.18.

Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 1000 terms from T. D. Noe)
Thomas Bloom, Problem 121, Erdős Problems.
Henri Cohen, François Dress, and Mahomed El Marraki, Explicit estimates for summatory functions linked to the Möbius μ-function, Funct. Approx. Comment. Math. 37:1 (2007), pp. 51-63.
G. H. Hardy and S. Ramanujan, The normal number of prime factors of a number n, Q. J. Math., 48 (1917), pp. 76-92.
L. Moser and R. A. MacLeod, The error term for the squarefree integers, Canad. Math. Bull. vol. 9, no. 3, (1966).
Kenneth Rogers, The Schnirelmann density of the squarefree integers, Proc. Amer. Math. Soc. 15 (1964), pp. 515-516.
Terence Tao, Erdős problem database, see no. 121.
A. M. Vaidya, On the order of the error function of the square free numbers, Proc. Nat. Inst. Sci. India Part A 32 (1966), pp. 196-201.
Eric Weisstein's World of Mathematics, Squarefree.

One less than A107079.
Cf. A005117, A002321, A057627, A179211, A000720, A081239, A066779, A179215, A284584.
Cf. A158819 Number of squarefree numbers <= n minus round(n/zeta(2)).

a013928 n = a013928_list !! (n-1)
a013928_list = scanl (+) 0 $ map a008966 [1..]
-- Reinhard Zumkeller, Aug 03 2012

ListTools:-PartialSums([0,seq(numtheory:-mobius(i)^2,i=1..100)]); # Robert Israel, Dec 11 2014

Accumulate[Table[Abs[MoebiusMu[n]], {n, 0, 79}]] (* Alonso del Arte, Oct 07 2012 *)
Accumulate[Table[If[SquareFreeQ[n],1,0],{n,0,80}]] (* Harvey P. Dale, Mar 06 2019 *)

a(n)=sum(i=1,n-1,if(issquarefree(i),1,0)) \\ Lifchitz

a(n)=n--;sum(k=1,sqrtint(n),moebius(k)*(n\k^2)) \\ Benoit Cloitre, Oct 25 2009

a(n)=n--; my(s); forfactored(k=1,sqrtint(n), s += n\k[1]^2*moebius(k)); s \\ Charles R Greathouse IV, Nov 05 2017

a(n)=n--; my(s); forsquarefree(k=1, sqrtint(n), s += n\k[1]^2*moebius(k)); s \\ Charles R Greathouse IV, Jan 08 2018

from sympy.ntheory.factor_  import core
def a(n): return sum ([1 for i in range(1, n) if core(i) == i]) # Indranil Ghosh, Apr 16 2017

from math import isqrt
from sympy import mobius
def A013928(n): return sum(mobius(k)*((n-1)//k**2) for k in range(1,isqrt(n-1)+1)) # Chai Wah Wu, Jan 03 2024

a(n) = Sum_{k = 1..n-1} mu(k)^2. - Vladeta Jovovic, May 18 2001
a(n) = Sum_{d = 1..floor(sqrt(n - 1))} mu(d)*floor((n - 1)/d^2) where mu(d) is the Moebius function (A008683). - Vladeta Jovovic, Apr 06 2001
Asymptotic formula (with error term): a(n) = Sum_{k = 1..n-1} mu(k)^2 = Sum_{k = 1..n-1} |mu(k)| = 6*n/Pi^2 + O(sqrt(n)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jul 20 2002
a(n) = Sum_{k = 0..n} if(k <= n-1, mu(n - k) mod 2, else 0; a(n + 1) = Sum_{k = 0..n} mu(n - k + 1) mod 2. - Paul Barry, May 10 2005
a(n + 1) = Sum_{k = 0..n} abs(mu(n - k + 1)). - Paul Barry, Jul 20 2005
a(n) = Sum_{k = 1..floor(sqrt(n))} mu(k)*floor(n/k^2). - Benoit Cloitre, Oct 25 2009
Landau proved that a(n) = 6*n/Pi^2 + o(sqrt(n)). - Charles R Greathouse IV, Feb 02 2016
Vaidya proved that a(n) = 6*n/Pi^2 + O(n^k) for any k > 2/5 on the Riemann hypothesis. - Charles R Greathouse IV, Feb 02 2016
a(n) = A107079(n)-1. - Antti Karttunen, Oct 07 2016
G.f.: Sum_{k>=1} mu(k)^2*x^(k+1)/(1 - x). - Ilya Gutkovskiy, Feb 06 2017
a(n+1) = n - A057627(n) - Antti Karttunen, Apr 17 2017

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647

Offset: 0

N. J. A. Sloane

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
From Gary W. Adamson, Jul 25 2016: (Start)
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
  1, 1, 0, 0, 0, 0, ...
  2, 0, 3, 0, 0, 0, ...
  2, 0, 0, 3, 0, 0, ...
  2, 0, 0, 0, 3, 0, ...
  2, 0, 0, 0, 0, 3, ...
  ...
The triangle generated from M is:
   1,
   1,  1,
   3,  1, 3,
  11,  3, 3, 9,
  41, 11, 9, 9, 27,
   ...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). (End)
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Indranil Ghosh, Table of n, a(n) for n = 0..1745 (terms 0..200 from T. D. Noe)
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Krassimir T. Atanassov and Anthony G. Shannon, On intercalated Fibonacci sequences, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 218-223.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Pell, Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 43, 56.
Chris Caldwell, Primality Proving, Arndt's theorem.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Tanya Khovanova, Recursive Sequences
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Eugene McDonnell, Heron's Rule and Integer-Area Triangles, At Play With J, 2010.
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Mathematics Stack Exchange.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Bisections are A011943 and A094347.

Cf. A001353, A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

a001075 n = a001075_list !! n
a001075_list =
   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001075 := proc(n)
    orthopoly[T](n,2) ;
end proc:
seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)

{a(n) = subst(poltchebi(abs(n)), x, 2)};

{a(n) = real((2 + quadgen(12))^abs(n))};

{a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016

my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009

def a(n):
    Q = QuadraticField(3, 't')
    u = Q.units()[0]
    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

More terms from James Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A071904 Odd composite numbers.

Original entry on oeis.org

9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99, 105, 111, 115, 117, 119, 121, 123, 125, 129, 133, 135, 141, 143, 145, 147, 153, 155, 159, 161, 165, 169, 171, 175, 177, 183, 185, 187, 189, 195, 201, 203, 205

Offset: 1

Shyam Sunder Gupta, Jun 12 2002

Same as A014076 except for the initial term A014076(1)=1 (which is not a composite number).
Values of quadratic form (2x + 3)*(2y + 3) = 4xy + 6x + 6y + 9 for x, y >= 0. - Anton Joha, Jan 21 2001
Intersection of A002808 and A005408. - Reinhard Zumkeller, Oct 10 2011
Composite numbers n such that (n-1)^(n-1) == 1 (mod n). - Michel Lagneau, Feb 18 2012
There is a rectangular array of n dots (with both sides > 1) with a unique center point if and only if n is in this sequence. - Peter Woodward, Apr 21 2015
First differences <= 6. Cf. A164510. - Zak Seidov, Sep 22 2016
Let r(n) = (a(n)-1)/(a(n)+1) if a(n) mod 4 = 1, (a(n)+1)/(a(n)-1) otherwise; then Product_{n>=1} r(n) = (4/5) * (8/7) * (10/11) * (12/13) * (14/13) * ... = Pi/4. - Dimitris Valianatos, May 24 2017

			45 is in the sequence because it is odd and composite (45 = 3 * 3 * 5).
195 is in the sequence because it is odd and composite (195 = 3 * 5 * 13).

K. D. Bajpai, Table of n, a(n) for n = 1..10000 (first 1000 terms from Zak Seidov).
Joel E. Cohen and Dexter Senft, Gaps of size 2, 4, and (conditionally) 6 between successive odd composite numbers occur infinitely often, Notes on Number Theory and Discrete Mathematics, Volume 31, Number 3, 494-503 (2025). See p. 495.

Cf. A002808, A014076, A005408, A000035, A010051, A020639, A162022.

Cf. A164510.

a071904 n = a071904_list !! (n-1)
a071904_list = filter odd a002808_list
-- Reinhard Zumkeller, Oct 10 2011

remove(isprime, [seq(2*i+1, i = 1 .. 1000)]); # Robert Israel, Apr 22 2015
# alternative
A071904 := proc(n) local a;
    if n = 1 then
        9;
    else
        for a from procname(n-1)+2 by 2 do
            if not isprime(a) then
                return a;
            end if;
        end do:
    end if;
end proc: # R. J. Mathar, Sep 09 2015

Select[Table[n, {n, 9, 300, 2}], !PrimeQ[#] &] (* Vladimir Joseph Stephan Orlovsky, Apr 16 2011 *)
With[{upto = 200}, Complement[Range[9, upto, 2], Prime[Range[ PrimePi[ upto]]]]] (* Harvey P. Dale, Jan 24 2013 *)
With[{upto = 200},oddsequence=Table[2n+1,{n,1,upto}];oddcomposites=Union[Flatten[Range[oddsequence^2,upto,2*oddsequence]]]] (* Ben Engelen, Feb 24 2016 *)

is(n)=n%2 && !isprime(n) && n > 1 \\ Charles R Greathouse IV, Nov 24 2012

lista(nn) = forcomposite(n=1, nn, if (n%2, print1(n, ", "))); \\ Michel Marcus, Sep 24 2016

from sympy import isprime
def ok(n): return n > 3 and n%2 == 1 and not isprime(n)
print(list(filter(ok, range(206)))) # Michael S. Branicky, Sep 15 2021

from sympy import primepi
def A071904(n):
    if n == 1: return 9
    m, k = n, primepi(n) + n + (n>>1)
    while m != k:
        m, k = k, primepi(k) + n + (k>>1)
    return m # Chai Wah Wu, Jul 31 2024

A000035(a(n))*(1-A010051(a(n))) = 1; A020639(a(n)) = A162022(n). - Reinhard Zumkeller, Oct 10 2011
a(n) ~ 2n. - Charles R Greathouse IV, Jul 02 2013
More precisely, a(n) = 2n(1 + 2(1+o(1))/log(n)). - Vladimir Shevelev, Jan 07 2015

A010527 Decimal expansion of sqrt(3)/2.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

This is the ratio of the height of an equilateral triangle to its base.
Essentially the same sequence arises from decimal expansion of square root of 75, which is 8.6602540378443864676372317...
Also the real part of i^(1/3), the cubic root of i. - Stanislav Sykora, Apr 25 2012
Gilbert & Pollak conjectured that this is the Steiner ratio rho_2, the least upper bound of the ratio of the length of the Steiner minimal tree to the length of the minimal tree in dimension 2. (See Ivanov & Tuzhilin for the status of this conjecture as of 2012.) - Charles R Greathouse IV, Dec 11 2012
Surface area of a regular icosahedron with unit edge is 5*sqrt(3), i.e., 10 times this constant. - Stanislav Sykora, Nov 29 2013
Circumscribed sphere radius for a cube with unit edges. - Stanislav Sykora, Feb 10 2014
Also the ratio between the height and the pitch, used in the Unified Thread Standard (UTS). - Enrique Pérez Herrero, Nov 13 2014
Area of a 30-60-90 triangle with shortest side equal to 1. - Wesley Ivan Hurt, Apr 09 2016
If a, b, c are the sides of a triangle ABC and h_a, h_b, h_c the corresponding altitudes, then (h_a+h_b+h_c) / (a+b+c) <= sqrt(3)/2; equality is obtained only when the triangle is equilateral (see Mitrinovic reference). - Bernard Schott, Sep 26 2022

			0.86602540378443864676372317...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 8.2, 8.3 and 8.6, pp. 484, 489, and 504.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §12.4 Theorems and Formulas (Solid Geometry), pp. 450-451.
D. S. Mitrinovic, E. S. Barnes, D. C. B. Marsh, and J. R. M. Radok, Elementary Inequalities, Tutorial Text 1 (1964), P. Noordhoff LTD, Groningen, problem 6.8, page 114.

Harry J. Smith, Table of n, a(n) for n = 0..20000
E. N. Gilbert and H. O. Pollak, Steiner minimal trees, SIAM J. Appl. Math. 16, (1968), pp. 1-29.
A. O. Ivanov and A. A. Tuzhilin, The Steiner ratio Gilbert-Pollak conjecture is still open, Algorithmica 62:1-2 (2012), pp. 630-632.
Matt Parker, The mystery of 0.866025403784438646763723170752936183471402626905190314027903489, Stand-up Maths, YouTube video, Feb 14 2024.
Simon Plouffe, Plouffe's Inverter, sqrt(3)/2 to 10000 digits.
Simon Plouffe, Sqrt(3)/2 to 5000 digits.
Eric Weisstein's World of Mathematics, Lebesgue Minimal Problem.
Wikipedia, Icosahedron.
Wikipedia, Platonic solid.
Wikipedia, Unified Thread Standard.
Index entries for algebraic numbers, degree 2.

Cf. A010153.
Cf. Platonic solids surfaces: A002194 (tetrahedron), A010469 (octahedron), A131595 (dodecahedron).
Cf. Platonic solids circumradii: A010503 (octahedron), A019881 (icosahedron), A179296 (dodecahedron), A187110 (tetrahedron).
Cf. A126664 (continued fraction), A144535/A144536 (convergents).
Cf. A002194, A010502, A020821, A104956, A152623 (other geometric inequalities).

SetDefaultRealField(RealField(100)); Sqrt(3)/2; // G. C. Greubel, Nov 02 2018

Digits:=100: evalf(sqrt(3)/2); # Wesley Ivan Hurt, Apr 09 2016

RealDigits[Sqrt[3]/2, 10, 200][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)

default(realprecision, 20080); x=10*(sqrt(3)/2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b010527.txt", n, " ", d));  \\ Harry J. Smith, Jun 02 2009

sqrt(3)/2 \\ Michel Marcus, Apr 10 2016

Equals cos(30 degrees). - Kausthub Gudipati, Aug 15 2011
Equals A002194/2. - Stanislav Sykora, Nov 30 2013
From Amiram Eldar, Jun 29 2020: (Start)
Equals sin(Pi/3) = cos(Pi/6).
Equals Integral_{x=0..Pi/3} cos(x) dx. (End)
Equals 1/(10*A020832). - Bernard Schott, Sep 29 2022
Equals x^(x^(x^...)) where x = (3/4)^(1/sqrt(3)) (infinite power tower). - Michal Paulovic, Jun 25 2023
Equals 2F1(-1/4,1/4 ; 1/2 ; 3/4) . - R. J. Mathar, Aug 31 2025

Last term corrected and more terms added by Harry J. Smith, Jun 02 2009

A001110 Square triangular numbers: numbers that are both triangular and square.

Original entry on oeis.org

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100

Offset: 0

N. J. A. Sloane

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004
For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007
The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012
Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012
For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013
For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014
The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014
The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

			a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 38, 204.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-259.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.

Kade Robertson, Table of n, a(n) for n = 0..600 [This replaces an earlier b-file computed by T. D. Noe]
Nikola Adžaga, Andrej Dujella, Dijana Kreso, and Petra Tadić, On Diophantine m-tuples and D(n)-sets, 2018.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Tom Beldon and Tony Gardiner, Triangular numbers and perfect squares, The Mathematical Gazette, 2002, pp. 423-431, esp pp. 424-426.
K. S. Brown, Square Triangular Numbers.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
Yong-Gao Chen and Jin-Hui Fang, Triangular numbers in geometric progression, INTEGERS 7 (2007), #A19.
F. Javier de Vega, On the parabolic partitions of a number, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
Colin Dickson et al., ratio of integers = sqrt(2), thread in newsgroup alt.math.recreational, March 7, 2004.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
M. Gardner, Letter to N. J. A. Sloane, circa Aug 11 1980, concerning A001110, A027568, A039596, etc.
Bill Gosper, The Triangular Squares, 2014.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020. Mentions this sequence.
Gillian Hatch, Pythagorean Triples and Triangular Square Numbers, Mathematical Gazette 79 (1995), 51-55.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Roger B. Nelsen, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
A. Nowicki, The numbers a^2+b^2-dc^2, J. Int. Seq. 18 (2015) # 15.2.3.
Matthew Parker, The first 5000 square triangular numbers (7-Zip compressed file).
J. L. Pietenpol, A. V. Sylwester, E. Just, and R. M. Warten, Problem E1473, Amer. Math. Monthly, Vol. 69, No. 2 (Feb. 1962), pp. 168-169. (From the editorial note on p. 169 of this source, we learn that the question about the existence of perfect squares in the sequence of triangular numbers cropped up in the Euler-Goldbach Briefwechsel of 1730; the translation into English of the relevant letters can be found at Correspondence of Leonhard Euler with Christian Goldbach (part II), pp. 614-615.) - _José Hernández_, May 24 2022
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
D. A. Q., Triangular square numbers: a postscript, Math. Gaz., 56 (1972), 311-314.
K. Ramsey, Re_Generalized_Proof_re_Square_Triangular_Numbers. [Broken link]
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Fabio Roman, On certain ratios regarding integer numbers which are both triangulars and squares, arXiv:1703.06701 [math.NT], 2017.
Jon E. Schoenfield, Prime factorization of a(n) for n = 1..300.
Jaap Spies, A Bit of Math, The Art of Problem Solving, Jaap Spies Publishers (2019).
Ralf Stephan, Boring proof of a nonlinearity.
UWC, Problem A, Nieuw Archief voor Wiskunde, Dec 2004; Jaap Spies, Solution.
Michel Waldschmidt, Continued fractions, École de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Square triangular number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001108, A001109.
Other S_r type sequences are S_4=A000290, S_5=A004146, S_7=A054493, S_8=A001108, S_9=A049684, S_20=A049683, S_36=this sequence, S_49=A049682, S_144=A004191^2.
Cf. A001014; intersection of A000217 and A000290; A010052(a(n))*A010054(a(n)) = 1.
Cf. A005214, A054686, A232847 and also A233267 (reveals an interesting divisibility pattern for this sequence).
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
See A229131, A182334, A299921 for near-misses.

a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
   zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
-- Reinhard Zumkeller, Oct 12 2011

[n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015

a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq(A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
A001110:=-(1+z)/((z-1)*(z**2-34*z+1)); # Simon Plouffe in his 1992 dissertation

f[n_]:=n*(n+1)/2; lst={}; Do[If[IntegerQ[Sqrt[f[n]]],AppendTo[lst,f[n]]],{n,0,10!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
Transpose[NestList[Flatten[{Rest[#],34Last[#]-First[#]+2}]&, {0,1},20]][[1]]  (* Harvey P. Dale, Mar 25 2011 *)
LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *)
LinearRecurrence[{6,-1},{0,1},20]^2 (* Harvey P. Dale, Oct 22 2012 *)
(* Square = Triangular = Triangular = A001110 *)
ChebyshevU[#-1,3]^2==Binomial[ChebyshevT[#/2,3]^2,2]==Binomial[(1+ChebyshevT[#,3])/2,2]=={1,36,1225,41616,1413721}[[#]]&@Range[5]
True (* Bill Gosper, Jul 20 2015 *)
L=0;r={};Do[AppendTo[r,L];L=1+17*L+6*Sqrt[L+8*L^2],{i,1,19}];r (* Kebbaj Mohamed Reda, Aug 02 2023 *)

a=vector(100);a[1]=1;a[2]=36;for(n=3,#a,a[n]=34*a[n-1]-a[n-2]+2);a \\ Charles R Greathouse IV, Jul 25 2011

;; With memoizing definec-macro from Antti Karttunen's IntSeq-library.
(definec (A001110 n) (if (< n 2) n (+ 2 (- (* 34 (A001110 (- n 1))) (A001110 (- n 2))))))
;; Antti Karttunen, Dec 06 2013

;; For testing whether n is in this sequence:
(define (inA001110? n) (and (zero? (A068527 n)) (inA001109? (floor->exact (sqrt n)))))
(define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
;; Antti Karttunen, Dec 06 2013

a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001
a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006
a(n) = A001109(n)^2 = A001108(n)*(A001108(n)+1)/2 = (A000129(n)*A001333(n))^2 = (A000129(n)*(A000129(n) + A000129(n-1)))^2. - Henry Bottomley, Apr 19 2000
a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004
From Antonio Alberto Olivares, Nov 07 2003: (Start)
a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
a(n+1) = (17*A029547(n) - A091761(n) - 1)/16. - R. J. Mathar, Nov 16 2007
a(n) = A001333^2 * A000129^2 = A000129(2*n)^2/4 = binomial(A001108,2). - Bill Gosper, Jul 28 2008
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008
a(n) = (1/8)*(sinh(2*n*arcsinh(1)))^2. - Artur Jasinski, Feb 10 2010
a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011
a(n) = (A011900(n) + A001652(n))^2; see the link about the generalized proof of square triangular numbers. - Kenneth J Ramsey, Oct 10 2011
a(2*n+1) = A002315(n)^2*(A002315(n)^2 + 1)/2. - Ivan N. Ianakiev, Oct 10 2012
a(2*n+1) = ((sqrt(t^2 + (t+1)^2))*(2*t+1))^2, where t = (A002315(n) - 1)/2. - Ivan N. Ianakiev, Nov 01 2012
a(2*n) = A001333(2*n)^2 * (A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * (A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013
a(n) = (A001542(n)/2)^2.
From Peter Bala, Apr 03 2014: (Start)
a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = A096979(2*n-1) for n > 0. - Ivan N. Ianakiev, Jun 21 2014
a(n) = (6*sqrt(a(n-1)) - sqrt(a(n-2)))^2. - Arkadiusz Wesolowski, Apr 06 2015
From Daniel Poveda Parrilla, Jul 16 2016 and Sep 21 2016: (Start)
a(n) = A000290(A002965(2*n)*A002965(2*n + 1)) (after Hugh Darwen).
a(n) = A000217(2*(A000129(n))^2 - (A000129(n) mod 2)).
a(n) = A000129(n)^4 + Sum_{k=0..(A000129(n)^2 - (A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers).
a(n) = A002965(2*n)^4 + Sum_{k=A002965(2*n)^2..A002965(2*n)*A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End)
E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016

A090822 Gijswijt's sequence: a(1) = 1; for n>1, a(n) = largest integer k such that the word a(1)a(2)...a(n-1) is of the form xy^k for words x and y (where y has positive length), i.e., the maximal number of repeating blocks at the end of the sequence so far.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 1

Offset: 1

Dion Gijswijt, Feb 27 2004

Here xy^k means the concatenation of the words x and k copies of y.
The name "Gijswijt's sequence" is due to N. J. A. Sloane, not the author!
Fix n and suppose a(n) = k. Let len_y(n) = length of shortest y for this k and let len_x = n-1 - k*len_y(n) = corresponding length of x. A091407 and A091408 give len_y and len_x. For the subsequence when len_x = 0 see A091410 and A091411.
The first 4 occurs at a(220) (see A091409).
The first 5 appears around term 10^(10^23).
We believe that for all N >= 6, the first time N appears is at about position 2^(2^(3^(4^(5^...^(N-1))))). - N. J. A. Sloane and Allan Wilks, Mar 14 2004
For a similar formula, see p. 6 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023
In the first 100000 terms the fraction of [1's, 2's, 3's, 4's] seems to converge, to about [.287, .530, .179, .005] respectively. - Allan Wilks, Mar 04 2004
When k=12 is reached, say, it is treated as the number 12, not as 1,2. This is not a base-dependent sequence.
Does this sequence have a finite average? Does anyone know the exact value? - Franklin T. Adams-Watters, Jan 23 2008
Answer: Given that "...the fraction of [1's, 2's, 3's, 4's] seems to converge, to about [.287, .530, .179, .005]..." that average should be the dot product of these vectors, i.e., about 1.904. - M. F. Hasler, Jan 24 2008
Second answer: The asymptotic densities of the numbers exist, and the average is the dot product. See pp. 56-59 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023
Which is the first step with two consecutive 4's? Or the shortest run found so far between two 4's? - Sergio Pimentel, Oct 10 2016
Answer: The first x such that the x-th and (x+1)-th element are 4, is 255895648634818208370064452304769558261700170817472823... ...398081655524438021806620809813295008281436789493636144. See p. 55 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023

N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

N. J. A. Sloane, Table of n, a(n) for n = 1..24000
Andreas Abel and Andres Loeh, Haskell program
Fokko J. van de Bult, Dion C. Gijswijt, John P. Linderman, N. J. A. Sloane and Allan R. Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
Benjamin Chaffin, John P. Linderman, N. J. A. Sloane, and Allan R. Wilks, On Curling Numbers of Integer Sequences, arXiv:1212.6102 [math.CO], Dec 25 2012.
Benjamin Chaffin, John P. Linderman, N. J. A. Sloane, and Allan R. Wilks, On Curling Numbers of Integer Sequences, Journal of Integer Sequences, Vol. 16 (2013), Article 13.4.3.
Benjamin Chaffin and N. J. A. Sloane, The Curling Number Conjecture, preprint.
Dion C. Gijswijt, Krulgetallen, Pythagoras, 55ste Jaargang, Nummer 3, Jan 2016, (Article in Dutch about this sequence, see pages 10-13, cover and back cover).
Levi van de Pol, The first occurrence of a number in Gijswijt's sequence, arXiv:2209.04657 [math.CO], 2022.
Levi van de Pol, The Growth Rate of Gijswijt's Sequence, J. Int. Seq. (2025) Vol. 28, Art. No. 25.4.6. See p. 2.
N. J. A. Sloane, Seven Staggering Sequences.
N. J. A. Sloane, Confessions of a Sequence Addict (AofA2017), slides of invited talk given at AofA 2017, Jun 19 2017, Princeton. Mentions this sequence.
Index entries for sequences related to Gijswijt's sequence
Index entries for sequences related to curling numbers

Cf. A091407, A091408, A091409 (records), A091410, A091411, A091579, A091586, A091970, A093955-A093958.
A091412 gives lengths of runs. A091413 gives partial sums.
Cf. also A094004, A160766, A217206, A217207.
Generalizations: A094781, A091975, A091976, A092331-A092335.

Haskell
```
-- See link.
```

K:= proc(L)
local n,m,k,i,b;
m:= 0;
n:= nops(L);
for k from 1 do
  if k*(m+1) > n then return(m) fi;
  b:= L[-k..-1];
  for i from 1 while i*k <= n and L[-i*k .. -(i-1)*k-1] = b do od:
  m:= max(m, i-1);
od:
end proc:
A[1]:= 1:
for i from 2 to 220 do
  A[i]:= K([seq(A[j],j=1..i-1)])
od:
seq(A[i],i=1..220); # Robert Israel, Jul 02 2015

ClearAll[a]; reversed = {a[2]=1, a[1]=1}; blocs[len_] := Module[{bloc1, par, pos}, bloc1 = Take[reversed, len]; par = Partition[ reversed, len]; pos = Position[par, bloc_ /; bloc != bloc1, 1, 1]; If[pos == {}, Length[par], pos[[1, 1]] - 1]]; a[n_] := a[n] = Module[{an}, an = Table[{blocs[len], len}, {len, 1, Quotient[n-1, 2]}] // Sort // Last // First; PrependTo[ reversed, an]; an]; A090822 = Table[a[n], {n, 1, 99}] (* Jean-François Alcover, Aug 13 2012 *)

A090822(n,A=[])={while(#Ak||break; k=m);A=concat(A,k));A} \\ M. F. Hasler, Aug 08 2018

def k(s):
    maxk = 1
    for m in range(1, len(s)+1):
        i, y, kk = 1, s[-m:], len(s)//m
        if kk <= maxk: return maxk
        while s[-(i+1)*m:-i*m] == y: i += 1
        maxk = max(maxk, i)
def aupton(terms):
    alst = [1]
    for n in range(2, terms+1):
        alst.append(k(alst))
    return alst
print(aupton(99)) # Michael S. Branicky, Mar 28 2022

A064113 Indices k such that (1/3)*(prime(k)+prime(k+1)+prime(k+2)) is a prime.

Original entry on oeis.org

2, 15, 36, 39, 46, 54, 55, 73, 102, 107, 110, 118, 129, 160, 164, 184, 187, 194, 199, 218, 239, 271, 272, 291, 339, 358, 387, 419, 426, 464, 465, 508, 520, 553, 599, 605, 621, 629, 633, 667, 682, 683, 702, 709, 710, 733, 761, 791, 813, 821, 822, 829, 830

Offset: 1

Jason Earls, Sep 08 2001

n such that d(n) = d(n+1), where d(n) = prime(n+1) - prime(n) = A001223(n).
Of interest because when I generalize it to d(n) = d(n+2), d(n) = d(n+3), etc. I am unable to find any positive number k such that d(n) = d(n+k) has no solution.
From Lei Zhou, Dec 06 2005: (Start)
When (1/3)*(prime(k) + prime(k+1) + prime(k+2)) is prime, then it is equal to prime(k+1).
Also, indices k such that (prime(k)+prime(k+2))/2 = prime(k+1).
The Mathematica program is based on the alternative definition. (End)
Inflection and undulation points of the primes, i.e., positions of zeros in A036263, the second differences of the primes. - Gus Wiseman, Mar 24 2020

			a(2) = 15 because (p(15)+p(16)+p(17)) = 1/3(47 + 53 + 59) = 53 (prime average of three successive primes).
Splitting the prime gaps into anti-runs gives: (1,2), (2,4,2,4,2,4,6,2,6,4,2,4,6), (6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,10,2,6), (6,4,6), ... Then a(n) is the n-th partial sum of the lengths of these anti-runs. - _Gus Wiseman_, Mar 24 2020

Harry J. Smith, Table of n, a(n) for n=1..1000
Thomas Bloom, Let d_n = p_(n+1)-p_n. The set of n such that d_(n+1) >= d_n has density 1/2, and similarly for d_(n+1) <= d_n. Furthermore, there are infinitely many n such that d_(n+1) = d_n, Erdős Problems.
Terence Tao, Erdős problem database, see no. 218.
Wikipedia, Inflection point

Indices of zeros in A036263 (second differences of primes).
Indices (A000720 = primepi) of balanced primes A006562, minus 1.
Cf. A001223, A024675, A075540, A075541.
Cf. A262138.
Complement of A333214.
First differences are A333216.
The version for strict ascents is A258025.
The version for strict descents is A258026.
The version for weak ascents is A333230.
The version for weak descents is A333231.
A triangle for anti-runs of compositions is A106356.
Lengths of maximal runs of prime gaps are A333254.
Anti-runs of compositions in standard order are A333381.
Cf. A000040, A084758, A124762, A124767, A238279.

import Data.List (elemIndices)
a064113 n = a064113_list !! (n-1)
a064113_list = map (+ 1) $ elemIndices 0 a036263_list
-- Reinhard Zumkeller, Jan 20 2012

ct = 0; Do[If[(Prime[k] + Prime[k + 2] - 2*Prime[k + 1]) == 0, ct++; n[ct] = k], {k, 1, 2000}]; Table[n[k], {k, 1, ct}] (* Lei Zhou, Dec 06 2005 *)
Join@@Position[Differences[Array[Prime,100],2],0] (* Gus Wiseman, Mar 24 2020 *)

d(n) = prime(n+1)-prime(n); j=[]; for(n=1,1500, if(d(n)==d(n+1), j=concat(j,n))); j

{ n=0; for (m=1, 10^9, if (d(m)==d(m+1), write("b064113.txt", n++, " ", m); if (n==1000, break)) ) } \\ Using d(n) above. - Harry J. Smith, Sep 07 2009

[n | n<-[1..888], !A036263(n)] \\ M. F. Hasler, Oct 15 2024

\\ More efficient for larges range of n:
A064113_upto(N, n=1, L=List(), q=prime(n+1), d=q-prime(n))={forprime(p=1+q,, if(d==d=p-q, listput(L,n); #LM. F. Hasler, Oct 15 2024

from itertools import count, islice
from sympy import prime, nextprime
def A064113_gen(startvalue=1): # generator of terms >= startvalue
    c = max(startvalue,1)
    p = prime(c)
    q = nextprime(p)
    r = nextprime(q)
    for k in count(c):
        if p+r==(q<<1):
            yield k
        p, q, r = q, r, nextprime(r)
A064113_list = list(islice(A064113_gen(),20)) # Chai Wah Wu, Feb 27 2024

A036263(a(n)) = 0; A122535(n) = A000040(a(n)); A006562(n) = A000040(a(n) + 1); A181424(n) = A000040(a(n) + 2). - Reinhard Zumkeller, Jan 20 2012
A262138(2*a(n)) = 0. - Reinhard Zumkeller, Sep 12 2015
a(n) = A000720(A006562(n)) - 1, where A000720 = (prime)pi, A006562 = balanced primes. - M. F. Hasler, Oct 15 2024

A000048 Number of n-bead necklaces with beads of 2 colors and primitive period n, when turning over is not allowed but the two colors can be interchanged.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 5, 9, 16, 28, 51, 93, 170, 315, 585, 1091, 2048, 3855, 7280, 13797, 26214, 49929, 95325, 182361, 349520, 671088, 1290555, 2485504, 4793490, 9256395, 17895679, 34636833, 67108864, 130150493, 252645135, 490853403, 954437120, 1857283155

Offset: 0

N. J. A. Sloane

Also, for any m which is a multiple of n, the number of 2m-bead balanced binary necklaces of fundamental period 2n that are equivalent to their complements. [Clarified by Aaron Meyerowitz, Jun 01 2024]
Also binary Lyndon words of length n with an odd number of 1's (for n>=1).
Also number of binary irreducible polynomials of degree n having trace 1.
Also number of binary irreducible polynomials of degree n having linear coefficient 1 (this is the same as the trace-1 condition, as the reciprocal of an irreducible polynomial is again irreducible).
Also number of binary irreducible self-reciprocal polynomials of degree 2*n; there is no such polynomial for odd degree except for x+1.
Also number of binary vectors (x_1,...x_n) satisfying Sum_{i=1..n} i*x_i = 1 (mod n+1) = size of Varshamov-Tenengolts code VT_1(n).
Also the number of dynamical cycles of period 2n of a threshold Boolean automata network which is a quasi-minimal negative circuit of size nq where q is odd and which is updated in parallel. - Mathilde Noual (mathilde.noual(AT)ens-lyon.fr), Mar 03 2009
Also the number of 3-elements orbits of the symmetric group S3 action on irreducible polynomials of degree 2n, n>1, over GF(2). - Jean Francis Michon, Philippe Ravache (philippe.ravache(AT)univ-rouen.fr), Oct 04 2009
Conjecture: Also the number of caliber-n cycles of Zagier-reduced indefinite binary quadratic forms with sum invariant equal to s, where (s-1)/n is an odd integer. - Barry R. Smith, Dec 14 2014
The Metropolis, Stein, Stein (1973) reference on page 31 Table II lists a(k) for k = 2 to 15 and is actually for sequence A056303 since there a(k) = 0 for k<2. - Michael Somos, Dec 20 2014

			a(5) = 3 corresponding to the necklaces 00001, 00111, 01011.
a(6) = 5 from 000001, 000011, 000101, 000111, 001011.

B. D. Ginsburg, On a number theory function applicable in coding theory, Problemy Kibernetiki, No. 19 (1967), pp. 249-252.
H. Kawakami, Table of rotation sequences of x_{n+1} = x_n^2 - lambda, pp. 73-92 of G. Ikegami, Editor, Dynamical Systems and Nonlinear Oscillations, Vol. 1, World Scientific, 1986.
Robert M. May, "Simple mathematical models with very complicated dynamics." Nature, Vol. 261, June 10, 1976, pp. 459-467; reprinted in The Theory of Chaotic Attractors, pp. 85-93. Springer, New York, NY, 2004. The sequences listed in Table 2 are A000079, A027375, A000031, A001037, A000048, A051841. - N. J. A. Sloane, Mar 17 2019
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..3320 (terms 0..200 from T. D. Noe)
Joerg Arndt, Matters Computational (The Fxtbook), p.408 and p.848.
L. Carlitz, A theorem of Dickson on irreducible polynomials, Proc. Amer. Math. Soc. 3, (1952). 693-700.
CombOS - Combinatorial Object Server, Generate necklaces, Lyndon words, chord diagrams, and relatives
J. Demongeot, M. Noual and S. Sene, On the number of attractors of positive and negative threshold Boolean automata circuits, hal-00647877 preprint (2009). [From Mathilde Noual (mathilde.noual(AT)ens-lyon.fr), Mar 03 2009]
N. J. Fine, Classes of periodic sequences, Illinois J. Math., 2 (1958), 285-302.
H. L. Fisher, Letter to N. J. A. Sloane, Mar 16 1989
E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.
R. W. Hall and P. Klingsberg, Asymmetric rhythms and tiling canons, Amer. Math. Monthly, 113 (2006), 887-896.
J. E. Iglesias, A formula for the number of closest packings of equal spheres having a given repeat period, Z. Krist. 155 (1981) 121-127, Table 2.
J. E. Iglesias, Enumeration of polytypes MX and MX_2 through the use of the symmetry of the Zhadanov symbol, Acta Cryst. A 62 (3) (2006) 178-194, Table 1.
Veronika Irvine, Lace Tessellations: A mathematical model for bobbin lace and an exhaustive combinatorial search for patterns, PhD Dissertation, University of Victoria, 2016.
A. A. Kulkarni, N. Kiyavash and R. Sreenivas, On the Varshamov-Tenengolts Construction on Binary Strings, 2013.
R. P. Loh, A. G. Shannon, A. F. Horadam, Divisibility Criteria and Sequence Generators Associated with Fermat Coefficients, Preprint, 1980.
R. M. May, Simple mathematical models with very complicated dynamics, Nature, 261 (Jun 10, 1976), 459-467.
N. Metropolis, M. L. Stein and P. R. Stein, On finite limit sets for transformations on the unit interval, J. Combin. Theory, A 15 (1973), 25-44; reprinted in P. Cvitanovic, ed., Universality in Chaos, Hilger, Bristol, 1986, pp. 187-206.
H. Meyn and W. Götz, Self-reciprocal polynomials over finite fields, Séminaire Lotharingien de Combinatoire, B21d (1989), 8 pp.
Simon Michalowsky, Bahman Gharesifard, Christian Ebenbauer, A Lie bracket approximation approach to distributed optimization over directed graphs, arXiv:1711.05486 [math.OC], 2017.
Tilman Piesk, Lists of the three sets of necklaces for n=1..12
R. Pries and C. Weir, The Ekedahl-Oort type of Jacobians of Hermitian curves, arXiv preprint arXiv:1302.6261 [math.NT], 2013.
Frank Ruskey, Number of q-ary Lyndon words with given trace mod q
Frank Ruskey, Number of Lyndon words over GF(q) with given trace
N. J. A. Sloane, On single-deletion-correcting codes, arXiv:math/0207197 [math.CO], 2002; in Codes and Designs (Columbus, OH, 2000), 273-291, Ohio State Univ. Math. Res. Inst. Publ., 10, de Gruyter, Berlin, 2002.
N. J. A. Sloane, On single-deletion-correcting codes
B. R. Smith, Reducing quadratic forms by kneading sequences, Journal of Integer Sequences, 17 (2014), article 14.11.8.
P. R. Stein, Letter to N. J. A. Sloane, Jun 02 1971
J.-Y. Thibon, The cycle enumerator of unimodal permutations, arXiv:math/0102051 [math.CO], 2001.
Index entries for "core" sequences
Index entries for sequences related to Lyndon words
Index entries for sequences related to subset sums modulo m

Like A000013, but primitive necklaces. Half of A064355.
Equals A042981 + A042982.
Cf. A002823, A000016, A053633, A051841, A001037, A002075, A002076, A008683.
Cf. also A001037, A056303.
Very close to A006788 [Fisher, 1989].
bisection (odd terms) is A131203

with(numtheory); A000048 := proc(n) local d,t1; if n = 0 then RETURN(1) else t1 := 0; for d from 1 to n do if n mod d = 0 and d mod 2 = 1 then t1 := t1+mobius(d)*2^(n/d)/(2*n); fi; od; RETURN(t1); fi; end;

a[n_] := Total[ MoebiusMu[#]*2^(n/#)& /@ Select[ Divisors[n], OddQ]]/(2n); a[0] = 1; Table[a[n], {n,0,35}] (* Jean-François Alcover, Jul 21 2011 *)
a[ n_] := If[ n < 1, Boole[n == 0], DivisorSum[ n, MoebiusMu[#] 2^(n/#) &, OddQ] / (2 n)]; (* Michael Somos, Dec 20 2014 *)

A000048(n) = sumdiv(n,d,(d%2)*(moebius(d)*2^(n/d)))/(2*n) \\ Michael B. Porter, Nov 09 2009

L(n, k) = sumdiv(gcd(n,k), d, moebius(d) * binomial(n/d, k/d) );
a(n) = sum(k=0, n, if( (n+k)%2==1, L(n, k), 0 ) ) / n;
vector(55,n,a(n)) \\ Joerg Arndt, Jun 28 2012

from sympy import divisors, mobius
def a(n): return 1 if n<1 else sum(mobius(d)*2**(n//d) for d in divisors(n) if d%2)//(2*n) # Indranil Ghosh, Apr 28 2017

a(n) = (1/(2*n)) * Sum_{odd d divides n} mu(d)*2^(n/d), where mu is the Mobius function A008683.
a(n) = A056303(n) for all integer n>=2. - Michael Somos, Dec 20 2014
Sum_{k dividing m for which m/k is odd} k*a(k) = 2^(m-1). (This explains the observation that the sequence is very close to A006788. Unless m has some nontrivial odd divisors that are small relative to m, the term m*a(m) will dominate the sum. Thus, we see for instance that a(n) = A006788(n) when n has one of the forms 2^m or 2^m*p where p is an odd prime with a(2^m) < p.) - Barry R. Smith, Oct 24 2015
A000013(n) = Sum_{d|n} a(d). - Robert A. Russell, Jun 09 2019
G.f.: 1 + Sum_{k>=1} mu(2*k)*log(1 - 2*x^k)/(2*k). - Ilya Gutkovskiy, Nov 11 2019

Additional comments from Frank Ruskey, Dec 13 1999

A143823 Number of subsets {x(1),x(2),...,x(k)} of {1,2,...,n} such that all differences |x(i)-x(j)| are distinct.

Original entry on oeis.org

1, 2, 4, 7, 13, 22, 36, 57, 91, 140, 216, 317, 463, 668, 962, 1359, 1919, 2666, 3694, 5035, 6845, 9188, 12366, 16417, 21787, 28708, 37722, 49083, 63921, 82640, 106722, 136675, 174895, 222558, 283108, 357727, 451575, 567536, 712856, 890405, 1112081, 1382416, 1717540

Offset: 0

John W. Layman, Sep 02 2008

When the set {x(1),x(2),...,x(k)} satisfies the property that all differences |x(i)-x(j)| are distinct (or alternately, all the sums are distinct), then it is called a Sidon set. So this sequence is basically the number of Sidon subsets of {1,2,...,n}. - Sayan Dutta, Feb 15 2024
See A143824 for sizes of the largest subsets of {1,2,...,n} with the desired property.
Also the number of subsets of {1..n} such that every orderless pair of (not necessarily distinct) elements has a different sum. - Gus Wiseman, Jun 07 2019

			{1,2,4} is a subset of {1,2,3,4}, with distinct differences 2-1=1, 4-1=3, 4-2=2 between pairs of elements, so {1,2,4} is counted as one of the 13 subsets of {1,2,3,4} with the desired property.  Only 2^4-13=3 subsets of {1,2,3,4} do not have this property: {1,2,3}, {2,3,4}, {1,2,3,4}.
From _Gus Wiseman_, May 17 2019: (Start)
The a(0) = 1 through a(5) = 22 subsets:
  {}  {}   {}     {}     {}       {}
      {1}  {1}    {1}    {1}      {1}
           {2}    {2}    {2}      {2}
           {1,2}  {3}    {3}      {3}
                  {1,2}  {4}      {4}
                  {1,3}  {1,2}    {5}
                  {2,3}  {1,3}    {1,2}
                         {1,4}    {1,3}
                         {2,3}    {1,4}
                         {2,4}    {1,5}
                         {3,4}    {2,3}
                         {1,2,4}  {2,4}
                         {1,3,4}  {2,5}
                                  {3,4}
                                  {3,5}
                                  {4,5}
                                  {1,2,4}
                                  {1,2,5}
                                  {1,3,4}
                                  {1,4,5}
                                  {2,3,5}
                                  {2,4,5}
(End)

Fausto A. C. Cariboni, Table of n, a(n) for n = 0..100 (terms 0..60 from Alois P. Heinz, 61..81 from Vaclav Kotesovec)
Thomas Bloom, Problem 861, Erdős Problems.
Terence Tao, Erdős problem database, see no. 861.
Wikipedia, Sidon sequence.

First differences are A308251.
Second differences are A169942.
Row sums of A381476.
The maximal case is A325879.
The integer partition case is A325858.
The strict integer partition case is A325876.
Heinz numbers of the counterexamples are given by A325992.
Cf. A143824, A054578, A169947.
Cf. A000079, A108917, A143824, A169942, A308251, A325676, A325677, A325679, A325683, A325860, A325864, A241688, A241689, A241690.

b:= proc(n, s) local sn, m;
      if n<1 then 1
    else sn:= [s[], n];
         m:= nops(sn);
         `if`(m*(m-1)/2 = nops(({seq(seq(sn[i]-sn[j],
           j=i+1..m), i=1..m-1)})), b(n-1, sn), 0) +b(n-1, s)
      fi
    end:
a:= proc(n) option remember;
       b(n-1, [n]) +`if`(n=0, 0, a(n-1))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 14 2011

b[n_, s_] := Module[{ sn, m}, If[n<1, 1, sn = Append[s, n]; m = Length[sn]; If[m*(m-1)/2 == Length[Table[sn[[i]] - sn[[j]], {i, 1, m-1}, {j, i+1, m}] // Flatten // Union], b[n-1, sn], 0] + b[n-1, s]]]; a[n_] := a[n] = b[n - 1, {n}] + If[n == 0, 0, a[n-1]]; Table [a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 08 2015, after Alois P. Heinz *)
Table[Length[Select[Subsets[Range[n]],UnsameQ@@Abs[Subtract@@@Subsets[#,{2}]]&]],{n,0,15}] (* Gus Wiseman, May 17 2019 *)

from itertools import combinations
def is_sidon_set(s):
    allsums = []
    for i in range(len(s)):
        for j in range(i, len(s)):
            allsums.append(s[i] + s[j])
    if len(allsums)==len(set(allsums)):
        return True
    return False
def a(n):
    sidon_count = 0
    for r in range(n + 1):
        subsets = combinations(range(1, n + 1), r)
        for subset in subsets:
            if is_sidon_set(subset):
                sidon_count += 1
    return sidon_count
print([a(n) for n in range(20)]) # Sayan Dutta, Feb 15 2024

from functools import cache
def b(n, s):
    if n < 1: return 1
    sn = s + [n]
    m = len(sn)
    return (b(n-1, sn) if m*(m-1)//2 == len(set(sn[i]-sn[j] for i in range(m-1) for j in range(i+1, m))) else 0) + b(n-1, s)
@cache
def a(n): return b(n-1, [n]) + (0 if n==0 else a(n-1))
print([a(n) for n in range(31)]) # Michael S. Branicky, Feb 15 2024 after Alois P. Heinz

a(n) = A169947(n-1) + n + 1 for n>=2. - Nathaniel Johnston, Nov 12 2010

a(n) = A054578(n) + 1 for n>0. - Alois P. Heinz, Jan 17 2013

a(21)-a(29) from Nathaniel Johnston, Nov 12 2010

Corrected a(21)-a(29) and more terms from Alois P. Heinz, Sep 14 2011

A001025 Powers of 16: a(n) = 16^n.

Original entry on oeis.org

1, 16, 256, 4096, 65536, 1048576, 16777216, 268435456, 4294967296, 68719476736, 1099511627776, 17592186044416, 281474976710656, 4503599627370496, 72057594037927936, 1152921504606846976, 18446744073709551616, 295147905179352825856, 4722366482869645213696, 75557863725914323419136, 1208925819614629174706176

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 16), L(1, 16), P(1, 16), T(1, 16). Essentially same as Pisot sequences E(16, 256), L(16, 256), P(16, 256), T(16, 256). See A008776 for definitions of Pisot sequences.
Convolution-square (auto-convolution) of A098430. - R. J. Mathar, May 22 2009
Subsequence of A161441: A160700(a(n)) = 1. - Reinhard Zumkeller, Jun 10 2009
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 16-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Muniru A Asiru, Table of n, a(n) for n = 0..820 (terms n = 0..100 from T. D. Noe)
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 280
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Index entries for linear recurrences with constant coefficients, signature (16).

Partial sums give A131865.

Cf. A000079, A000302, A001018, A005843, A008586.

List([0..20],n->16^n); # Muniru A Asiru, Nov 07 2018

a001025 = (16 ^)
a001025_list = iterate (* 16) 1  -- Reinhard Zumkeller, Nov 07 2012

A001025:=-1/(-1+16*z); # Simon Plouffe in his 1992 dissertation

Table[4^(2*n), {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Mar 01 2009 *)

A001025(n):=16^n$
makelist(A001025(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=1<<(4*n) \\ Charles R Greathouse IV, Feb 01 2012

print([16**n for n in range(20)]) # Stefano Spezia, Nov 10 2018

[lucas_number1(n,16,0) for n in range(1, 18)] # Zerinvary Lajos, Apr 29 2009

G.f.: 1/(1-16*x).
E.g.f.: exp(16*x).
From Muniru A Asiru, Nov 07 2018: (Start)
a(n) = 16^n.
a(0) = 1, a(n) = 16*a(n-1). (End)
a(n) = 4^A005843(n) = 2^A008586(n) = A000302(n)^2 = A000079(n)*A001018(n). - Muniru A Asiru, Nov 10 2018
a(n) = ( Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k) ) * ( Sum_{k = 0..n} (-1)^k/(2*k + 1)*binomial(2*n + 1, n - k) ). - Peter Bala, Feb 12 2019
a(n) = Sum_{k = 0..2*n} A000984(k) * A000984(2*n-k). - Peter Bala, Aug 23 2025

cp's OEIS Frontend

A013928 Number of (positive) squarefree numbers < n.

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A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

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A071904 Odd composite numbers.

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A010527 Decimal expansion of sqrt(3)/2.

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A001110 Square triangular numbers: numbers that are both triangular and square.

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