A011944 -id:A011944 - cp's OEIS frontend

A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367

Offset: 1

E. K. Lloyd

If k is in the sequence, then it has successor 7*k + 4*sqrt(3*(k^2 - 1)). - Lekraj Beedassy, Jun 28 2002
Chebyshev's polynomials T(n,x) evaluated at x=7.
a(n+1) give all (nontrivial) solutions of Pell equation a(n+1)^2 - 48*b(n+1)^2 = +1 with b(n+1)=A007655(n+2), n >= 0.
Also all solutions for x in Pell equation x^2 - 12*y^2 = 1. A011944 gives corresponding values for y. - Herbert Kociemba, Jun 05 2022
Also numbers x of the form 3j+1 such that x^2 = 3m^2+1. Also solutions of x in x^2 - 3*y^2 = 1 in A001075 if x = 3j+1, j=1,2,... - Cino Hilliard, Mar 05 2005
In addition to having integral standard deviation, these k consecutive integers also have integral mean. This question was posed by Jim Delany of Cal Poly in 1989. The solution appeared in the American Mathematical Monthly Vol. 97, No. 5, (May, 1990), pp. 432 as problem E3302. - Ronald S. Tiberio, Jun 23 2008
Lebl and Lichtblau give the formula a(d) = ((7+4*sqrt(3))^d + (7-4*sqrt(3))^d)/2 in Theorem 1.2(iii), p. 4. - Jonathan Vos Post, Aug 05 2008
In a (square pyramidal) pile of cannonballs, paint the ball at the top and the balls on 2 opposite sides of the base. The sequence, after its 1st term, gives the numbers of painted balls, k, in piles where the total number of balls is twice a square multiple of k. - Peter Munn, Feb 06 2025

P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Robert Israel, Table of n, a(n) for n = 1..788
Jim Delany, Roger Douglass, Mike Breen and Roger B. Eggleton, Problem E 3302: Averaging to Integers, The American Mathematical Monthly, Vol. 97, No. 5 (May, 1990), p. 432.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Tanya Khovanova, Recursive Sequences
Jiri Lebl and Daniel Lichtblau, Uniqueness of certain polynomials constant on a hyperplane, arXiv:0808.0284 [math.CV], 2008-2010.
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Ronald S. Tiberio, Solution to Problem E 3302 [Broken link]
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = A001075(2n).
Row 2 of array A188644
Cf. A000330, A007654, A007655, A011944, A102344.

I:=[1,7]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 19 2015

seq(orthopoly[T](n,7), n = 0..50); # Robert Israel, Jun 02 2015
a := n -> (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4):
seq(simplify(a(n)), n=1..20); # Peter Luschny, Jul 26 2020

LinearRecurrence[{14,-1},{1,7},30] (* Harvey P. Dale, Dec 16 2013 *)
a[n_]:=1/2((7-4 Sqrt[3])^n+(7+4 Sqrt[3])^n); Table[a[n] // Simplify,{n,0,20}] (* Gerry Martens, May 30 2015 *)

PARI
```
a(n)=if(n<0,0,subst(poltchebi(n),x,7))
```

g(n) = forstep(x=1,n,3,y=(x^2-1)/3;if(issquare(y),print1(x","))) \\ Cino Hilliard, Mar 05 2005

a(n) = 14*a(n-1) - a(n-2).
a(n) = sqrt(12*A011944(n)^2 + 1).
a(n) ~ (1/2)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = T(n, 7) = (S(n, 14)-S(n-2, 14))/2 = T(2*n, 2) with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 14)=A007655(n+2).
a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.
a(n) = sqrt(48*A007655(n+1)^2 + 1).
G.f.: (1-7*x)/(1-14*x+x^2).
a(n) = cosh(2n*arcsinh(sqrt(3))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - Peter Luschny, Jul 26 2020
E.g.f.: exp(7*x)*cosh(4*sqrt(3)*x). - Stefano Spezia, Dec 12 2022

Better description from Lekraj Beedassy, Jun 27 2002
Chebyshev comments from Wolfdieter Lang, Nov 08 2002
More terms from Vincenzo Librandi, Apr 19 2015

A041041 Denominators of continued fraction convergents to sqrt(26).

Original entry on oeis.org

1, 10, 101, 1020, 10301, 104030, 1050601, 10610040, 107151001, 1082120050, 10928351501, 110365635060, 1114584702101, 11256212656070, 113676711262801, 1148023325284080, 11593909964103601, 117087122966320090, 1182465139627304501, 11941738519239365100

Offset: 0

N. J. A. Sloane

Generalized Fibonacci sequence.
Sqrt(26) = 10/2 + 10/101 + 10/(101*10301) + 10/(10301*1050601) + ... - Gary W. Adamson, Jun 13 2008
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 10's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0, 1, ..., 10} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Bruno Berselli, May 03 2018: (Start)
Numbers k for which m*k^2 + (-1)^k is a perfect square:
m =  2: 0, 1, 2, 5, 12, 29, 70, 169, ...   (A000129);
m =  3: 0, 4, 56, 780, 10864, 151316, ...  (4*A007655);
m =  5: 0, 1, 4, 17, 72, 305, 1292, ...    (A001076);
m =  6: 0, 2, 20, 198, 1960, 19402, ...    (A001078);
m =  7: 0, 48, 12192, 3096720, ...         (2*A175672);
m =  8: 0, 6, 204, 6930, 235416, ...       (A082405);
m = 10: 0, 1, 6, 37, 228, 1405, 8658, ...  (A005668);
m = 11: 0, 60, 23880, 9504180, ...         [°];
m = 12: 0, 2, 28, 390, 5432, 75658, ...    (A011944);
m = 13: 0, 5, 180, 6485, 233640, ...       (5*A041613);
m = 14: 0, 4, 120, 3596, 107760, ...       (A068204);
m = 15: 0, 8, 496, 30744, 1905632, ...     [°];
m = 17: 0, 1, 8, 65, 528, 4289, 34840, ... (A041025);
m = 18: 0, 4, 136, 4620, 156944, ...       (A202299);
m = 19: 0, 13260, 1532829480, ...          [°];
m = 20: 0, 2, 36, 646, 11592, 208010, ...  (A207832);
m = 21: 0, 12, 1320, 145188, ...           (A174745);
m = 22: 0, 42, 16548, 6519870, ...         (A174766);
m = 23: 0, 240, 552480, 1271808720, ...    [°];
m = 24: 0, 10, 980, 96030, 9409960, ...    (A168520);
m = 26: 0, 1, 10, 101, 1020, 10301, ...    (this sequence);
m = 27: 0, 260, 702520, 1898208780, ...    [°];
m = 28: 0, 24, 6096, 1548360, ...          (A175672);
m = 29: 0, 13, 1820, 254813, 35675640, ... [°];
m = 30: 0, 2, 44, 966, 21208, 465610, ...  (2*A077421), etc.
[°] apparently without related sequences in the OEIS.
(End)
From Michael A. Allen, Mar 12 2023: (Start)
Also called the 10-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 10 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
Sergio Falcón and Ángel Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals (2007).
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,1).

Cf. A099374 (squares).
Cf. A041040.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A243399.
Row n=10 of A073133, A172236 and A352361.

I:=[1,10]; [n le 2 select I[n] else 10*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

seq(combinat:-fibonacci(n+1, 10), n=0..19); # Peter Luschny, May 04 2018

Denominator[Convergents[Sqrt[26], 30]] (* Vincenzo Librandi, Dec 10 2013 *)
LinearRecurrence[{10,1}, {1,10}, 30] (* G. C. Greubel, Jan 24 2018 *)

x='x+O('x^30); Vec(1/(1-10*x-x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,10,-1) for n in range(1, 19)] # Zerinvary Lajos, Apr 26 2009

G.f.: 1/(1 - 10*x - x^2).
a(n) = 10*a(n-1) + a(n-2), n>=1; a(-1):=0, a(0)=1.
a(n) = S(n, 10*i)*(-i)^n where i^2:=-1 and S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind. See A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = 5+sqrt(26), am = -1/ap = 5-sqrt(26).
a(n) = F(n+1, 10), the (n+1)-th Fibonacci polynomial evaluated at x=10. - T. D. Noe, Jan 19 2006
a(n) = Sum_{i=0..floor(n/2)} binomial(n-i,i)*10^(n-2*i). - Sergio Falcon, Sep 24 2007

Extended by T. D. Noe, May 23 2011

A007654 Numbers k such that the standard deviation of 1,...,k is an integer.

Original entry on oeis.org

0, 3, 48, 675, 9408, 131043, 1825200, 25421763, 354079488, 4931691075, 68689595568, 956722646883, 13325427460800, 185599261804323, 2585064237799728, 36005300067391875, 501489136705686528, 6984842613812219523, 97286307456665386800, 1355023461779503195683

Offset: 1

N. J. A. Sloane

Gives solutions k to the Diophantine equation m^2 = k*(k+1)/3. - Anton Lorenz Vrba (anton(AT)a-l-v.net), Jun 28 2005
If x=a(n), y=a(n+1), z=a(n+2) are three consecutive terms, then x^2 - 16*y*x + 14*x*z + 16*y^2 - 16*z*y + z^2 = 144. The formula is symmetric in x and z, so it is also valid for x=a(n+2), y=a(n+1), z=a(n). - Alexander Samokrutov, Jul 02 2015
From Bernard Schott, Apr 09 2021 (Start):
Corresponding solutions m (of first comment) are in A011944.
Equivalently, numbers k such that k/3 and k+1 are both perfect squares. (End)

Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 13, Questions proposées aux élèves de Terminale S, Exercice 1, p. 220, p. 223.
D. A. Benaron, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..100
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A007655, A011944, A055793, A098301.

I:=[0,3]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+6: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016

RecurrenceTable[{a[m] == 14 a[m - 1] - a[m - 2] + 6, a[1] == 0, a[2] == 3}, a, {m, 1, 17}] (* Michael De Vlieger, Jul 02 2015 *)
CoefficientList[Series[-3 x^2*(1 + x)/(-1 + x)/(1 - 14 x + x^2), {x, 0, 17}], x] (* Michael De Vlieger, Feb 02 2016 *)

concat(0,3*Vec((1+x)/(1-x)/(1-14*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, May 14 2013

a(n) = 3*A098301(n-1).
a(m) = 14*a(m-1) - a(m-2) + 6.
G.f.: -3*x^2*(1+x)/(-1+x)/(1-14*x+x^2) = -3 + (1/2)/(-1+x) + (1/2)*(-97*x+7)/(1-14*x+x^2). - R. J. Mathar, Nov 20 2007
a(n) = (-2 + (7-4*sqrt(3))^n*(7+4*sqrt(3)) + (7-4*sqrt(3))*(7+4*sqrt(3))^n)/4. - Colin Barker, Mar 05 2016
From Bernard Schott, Apr 09 2021: (Start)
a(n) = 3 * A001353(n-1)^2.
a(n) = A055793(n+1) - 1 = A001075(n-1)^2 - 1. (End)
2*a(n) = A011943(n)-1. - R. J. Mathar, Mar 16 2023

Corrected by Keith Lloyd, Mar 15 1996

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

Original entry on oeis.org

0, 2, 36, 646, 11592, 208010, 3732588, 66978574, 1201881744, 21566892818, 387002188980, 6944472508822, 124613502969816, 2236098580947866, 40125160954091772, 720016798592704030

Offset: 0

Gary Detlefs, Feb 20 2012

Denote as {a,b,c,d} the second-order linear recurrence a(n) = c*a(n-1) + d*a(n-2) with initial terms a, b. The following sequences and recurrence formulas are related to integer solutions of k*x^2 + 1 = y^2.
.
   k             x                        y
   -  -----------------------  -----------------------
A001542 {0,2,6,-1}       A001541 {1,3,6,-1}
A001353 {0,1,4,-1}       A001075 {1,2,4,-1}
A060645 {0,4,18,-1}      A023039 {1,9,18,-1}
A001078 {0,2,10,-1}      A001079 {1,5,10,-1}
A001080 {0,3,16,-1}      A001081 {1,8,16,-1}
A001109 {0,1,6,-1}       A001541 {1,3,6,-1}
A084070 {0,1,38,-1}      A078986 {1,19,38,-1}
A001084 {0,3,20,-1}      A001085 {1,10,20,-1}
A011944 {0,2,14,-1}      A011943 {1,7,14,-1}
A075871 {0,180,1298,-1}  A114047 {1,649,1298,-1}
A068204 {0,4,30,-1}      A069203 {1,15,30,-1}
A001090 {0,1,8,-1}       A001091 {1,4,8,-1}
A121740 {0,8,66,-1}      A099370 {1,33,66,-1}
A202299 {0,4,34,-1}      A056771 {1,17,34,-1}
A174765 {0,39,340,-1}    A114048 {1,179,340,-1}
a(n)    {0,2,18,-1}      A023039 {1,9,18,-1}
A174745 {0,12,110,-1}    A114049 {1,55,110,-1}
A174766 {0,42,394,-1}    A114050 {1,197,394,-1}
A174767 {0,5,48,-1}      A114051 {1,24,48,-1}
A004189 {0,1,10,-1}      A001079 {1,5,10,-1}
A174768 {0,10,102,-1}    A099397 {1,51,102,-1}
The sequence of the c parameter is listed in A180495.

Bruno Berselli, Table of n, a(n) for n = 0..500
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039, A049660, A079962.

m:=16; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2*x/(1-18*x+x^2))); // Bruno Berselli, Jun 19 2019

readlib(issqr):for x from 1 to 720016798592704030 do if issqr(20*x^2+1) then print(x) fi od;

LinearRecurrence[{18, -1}, {0, 2}, 16] (* Bruno Berselli, Feb 21 2012 *)
Table[2 ChebyshevU[-1 + n, 9], {n, 0, 16}]  (* Herbert Kociemba, Jun 05 2022 *)

makelist(expand(((2+sqrt(5))^(2*n)-(2-sqrt(5))^(2*n))/(4*sqrt(5))), n, 0, 15); /* Bruno Berselli, Jun 19 2019 */

a(n) = 18*a(n-1) - a(n-2).
From Bruno Berselli, Feb 21 2012: (Start)
G.f.: 2*x/(1-18*x+x^2).
a(n) = -a(-n) = 2*A049660(n) = ((2 + sqrt(5))^(2*n)-(2 - sqrt(5))^(2*n))/(4*sqrt(5)). (End)
a(n) = Fibonacci(6*n)/4. - Bruno Berselli, Jun 19 2019
For n>=1, a(n) = A079962(6n-3). - Christopher Hohl, Aug 22 2021

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A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

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A041041 Denominators of continued fraction convergents to sqrt(26).

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A007654 Numbers k such that the standard deviation of 1,...,k is an integer.

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A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

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