A010121 -id:A010121 - cp's OEIS frontend

A010465 Decimal expansion of square root of 7.

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

Offset: 1

N. J. A. Sloane

Continued fraction expansion is 2 followed by {1, 1, 1, 4} repeated. - Harry J. Smith, Jun 01 2009

The convergents to sqrt(7) are given in A041008/A041009. - Wolfdieter Lang, Nov 22 2017

			2.645751311064590590501615753639260425710259183082450180368334459201...

Harry J. Smith, Table of n, a(n) for n = 1..20000
Jason Kimberley, Index of expansions of sqrt(d) in base b.
R. J. Nemiroff & J. Bonnell, The first 1 million digits of the square root of 7.
Simon Plouffe, Plouffe's Inverter, sqrt(7) to 100000 digits.
Index entries for algebraic numbers, degree 2.

Cf. A010121 (continued fraction), A041008/A041009.

SetDefaultRealField(RealField(100)); Sqrt(7); // Vincenzo Librandi, Feb 15 2020

RealDigits[N[Sqrt[7], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)

default(realprecision, 20080); x=sqrt(7); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b010465.txt", n, " ", d));  \\ Harry J. Smith, Jun 01 2009

Equals 8*cos(Pi/14)*sin(2*Pi/14)*cos(3*Pi/14). - Gerry Martens, Mar 13 2025

A233587 Coefficients of the generalized continued fraction expansion sqrt(7) = a(1) +a(1)/(a(2) +a(2)/(a(3) +a(3)/(a(4) +a(4)/....))).

Original entry on oeis.org

2, 3, 30, 34, 111, 235, 3775, 5052, 7352, 9091, 34991, 35530, 53424, 57290, 66023, 1409179, 1519111, 1725990, 1812396, 4370835, 4507156, 4655396, 44257080, 234755198, 261519946, 264374278, 273487975

Offset: 1

Stanislav Sykora, Jan 06 2014

nonn

For more details on Blazys' expansions, see A233582.

Sqrt(7) is the first square root of a natural number with an a-periodic Blazys' expansion (see A233592 and A233593).

Stanislav Sykora, Table of n, a(n) for n = 1..1000

Cf. Blazys' expansions: A233582 (Pi), A233583 (e), A233584 (sqrt(e)), A233585 (1/gamma), A233585 (2*gamma) and Blazys' continued fractions: A233588, A233589, A233590, A233591.

Cf. A010465, A010121.

BlazysExpansion[n_, mx_] := Block[{k = 1, x = n, lmt = mx + 1, s, lst = {}}, While[k < lmt, s = Floor[x]; x = 1/(x/s - 1); AppendTo[lst, s]; k++]; lst]; BlazysExpansion[Sqrt@7, 32] (* Robert G. Wilson v, May 22 2014 *)

bx(x, nmax)={local(c, v, k); \\ Blazys expansion function
v = vector(nmax); c = x; for(k=1, nmax, v[k] = floor(c); c = v[k]/(c-v[k]); ); return (v); }
bx(sqrt(7), 1000) \\ Execution; use very high real precision

sqrt(7) = 2+2/(3+3/(30+30/(34+34/(111+...)))).

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

A171621 Numerator of 1/4 - 1/n^2, each fourth term multiplied by 4.

Original entry on oeis.org

0, 5, 3, 21, 8, 45, 15, 77, 24, 117, 35, 165, 48, 221, 63, 285, 80, 357, 99, 437, 120, 525, 143, 621, 168, 725, 195, 837, 224, 957, 255, 1085, 288, 1221, 323, 1365, 360, 1517, 399, 1677, 440, 1845, 483, 2021, 528

Offset: 2

Paul Curtz, Dec 13 2009

These are the square roots of the fifth column of the array of denominators mentioned in A171522.

Colin Barker, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A005563, A010121, A061037, A078371, A171522.

[-(-5+3*(-1)^n)*(-4+n^2)/8: n in [0..100]]; // G. C. Greubel, Sep 19 2018

A061037 := proc(n) 1/4-1/n^2 ; numer(%) ; end proc:
A171621 := proc(n) if n mod 4 = 2 then 4*A061037(n) ; else A061037(n) ; end if; end proc:
seq(A171621(n),n=2..90) ; # R. J. Mathar, Apr 02 2011

Table[-(-5+3*(-1)^n)*(-4+n^2)/8, {n,0,100}] (* G. C. Greubel, Sep 19 2018 *)
LinearRecurrence[{0,3,0,-3,0,1},{0,5,3,21,8,45},50] (* Harvey P. Dale, Nov 01 2019 *)

concat(0, Vec(x^3*(-5-3*x-6*x^2+x^3+3*x^4)/((x-1)^3*(1+x)^3) + O(x^100))) \\ Colin Barker, Nov 03 2014

a(n) = A061037(n) * A010121(n+2).
a(2n+2) = A005563(n). a(2n+3) = A078371(n).
G.f.: x^3*(-5-3*x-6*x^2+x^3+3*x^4) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Apr 02 2011
a(n) = -(-5+3*(-1)^n)*(-4+n^2)/8. - Colin Barker, Nov 03 2014
Sum_{n>=3} 1/a(n) = 13/12. - Amiram Eldar, Aug 11 2022

A283971 a(n) = n except a(4n + 2) = 2n + 1.

Original entry on oeis.org

0, 1, 1, 3, 4, 5, 3, 7, 8, 9, 5, 11, 12, 13, 7, 15, 16, 17, 9, 19, 20, 21, 11, 23, 24, 25, 13, 27, 28, 29, 15, 31, 32, 33, 17, 35, 36, 37, 19, 39, 40, 41, 21, 43, 44, 45, 23, 47, 48, 49, 25, 51, 52, 53, 27, 55, 56, 57, 29, 59, 60, 61, 31, 63, 64, 65, 33, 67

Offset: 0

Paul Curtz, Mar 18 2017

From Federico Provvedi, Nov 13 2018: (Start)
For n > 1, a(n) is also the cycle length generated by the cycle lengths of the digital roots, in base n, of the powers of k, with k > 0.
Example for n=10 (decimal base): for every h >= 0, the digital roots of 2^h generate a periodic cycle {1,2,4,8,7,5} with period 6; 3^h generates {1,3,9,9,9,9,...} so the periodic cycle {9} has period 1; 4^h generates the periodic cycle {1,4,7} with period 3; etc. So, for n=10 (decimal base representation) the sequence generated by the periods of the digital roots of powers of k (with k > 0) is also periodic {1,6,1,3,6,1,3,2,1} with period 9, hence a(10) = 9. (End)

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A005408, A010121, A022998, A051176, A060819, A186646.

a:=[0,1,1,3,4,5,3,7];; for n in [9..85] do a[n]:=2*a[n-4]-a[n-8]; od; a; # Muniru A Asiru, Jul 20 2018

seq(coeff(series(x*(1+x+3*x^2+4*x^3+3*x^4+x^5+x^6)/((1-x)^2*(1+x)^2*(1+x^2)^2), x,n+1),x,n),n=0..80); # Muniru A Asiru, Jul 20 2018

Table[If[Mod[n, 4] == 2, (n - 2)/2 + 1, n], {n, 67}] (* or *)
CoefficientList[Series[x (1 + x + 3 x^2 + 4 x^3 + 3 x^4 + x^5 + x^6)/((1 - x)^2*(1 + x)^2*(1 + x^2)^2), {x, 0, 67}], x] (* Michael De Vlieger, Mar 19 2017 *)
LinearRecurrence[{0, 0, 0, 2, 0, 0, 0, -1}, {0, 1, 1, 3, 4, 5, 3, 7}, 70] (* Robert G. Wilson v, Jul 23 2018 *)
Table[Length[FindTransientRepeat[(Length[FindTransientRepeat[Mod[#1^Range[b]-1,b-1]+1,2][[2]]]&)/@Range[2, 2*b], 2][[2]]], {b, 2, 100}] (* Federico Provvedi, Nov 13 2018 *)

a(n)=if(n%4==2, n\4*2 + 1, n) \\ Charles R Greathouse IV, Mar 18 2017

concat(0, Vec(x*(1 + x + 3*x^2 + 4*x^3 + 3*x^4 + x^5 + x^6) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2) + O(x^40))) \\ Colin Barker, Mar 19 2017

def A283971(n): return n if (n-2)&3 else n>>1 # Chai Wah Wu, Jan 10 2023

a(2*n) = A022998(n), a(1+2*n) = 1 + 2*n.
a(n) = 2*a(n-4) - a(n-8).
From Colin Barker, Mar 19 2017: (Start)
G.f.: x*(1 + x + 3*x^2 + 4*x^3 + 3*x^4 + x^5 + x^6) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2).
a(n) = -((-1)^n - (-i)^n - i^n - 7)*n/8, where i = sqrt(-1).
(End)
a(n) = A060819(n) * periodic sequence of length 4: repeat [4, 1, 1, 1].
a(n) = a(n-4) + periodic sequence of length 4: repeat [4, 4, 2, 4].
From Werner Schulte, Jul 08 2018: (Start)
For n > 0, a(n) is multiplicative with a(p^e) = p^e for prime p >= 2 and e >= 0 except a(2^1) = 1.
Dirichlet g.f.: (1 - 1/2^s - 1/2^(2*s-1)) * zeta(s-1).
(End)
a(n) = n*(7 + cos(n*Pi/2) - cos(n*Pi) + cos(3*n*Pi/2))/8. - Wesley Ivan Hurt, Oct 04 2018
E.g.f.: (1/4)*x*(4*cosh(x) - sin(x) + 3*sinh(x)). - Franck Maminirina Ramaharo, Nov 13 2018
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

A098457 Farey Bisection Expansion of sqrt(7).

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1

Offset: 1

John W. Layman, Sep 08 2004

We define the Farey Bisection Expansion (FBE) of the nonnegative real number x to be the sequence {a(n)} of 0's and 1's determined as follows. Set na(0)=0, da(0)=1, nb(0)=1 and db(0)=0. For n=1, 2, 3,..., set num=na(n-1)+nb(n-1) and den=da(n-1)+db(n-1); if xA010121.

			G.f. = x + x^2 + x^4 + x^6 + x^7 + x^8 + x^9 + x^11 + x^13 + x^14 + x^15 + ...

N. J. A. Sloane, Stern-Brocot or Farey Tree
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,1).

Cf. A010121, A097853, A098458, A131372.

&cat [[1, 1, 0, 1, 0, 1, 1]^^20]; // Wesley Ivan Hurt, Jul 11 2016

seq(op([1, 1, 0, 1, 0, 1, 1]), n=0..20); # Wesley Ivan Hurt, Jul 11 2016

LinearRecurrence[{0, 0, 0, 0, 0, 0, 1},{1, 1, 0, 1, 0, 1, 1},105] (* Ray Chandler, Aug 26 2015 *)

{a(n) = [1, 1, 0, 1, 0, 1, 1][(n-1)%7+1]}; /* Michael Somos, Dec 26 2016 */

From Wesley Ivan Hurt, Jul 11 2016: (Start)
G.f.: x * (1 + x + x^3 + x^5 + x^6) / (1 - x^7).
a(n) = a(n-7) for n>7.
a(n) = 1 - Sum_{k=1..4} floor((n + k)/7)*(-1)^k. (End)
a(n+1) = (-1)^(mod(mod(n, 7), 3)>0) * A131372(n). - Michael Somos, Dec 26 2016

A109054 Squares and numbers k such that the continued fraction expansion of sqrt(k) is multiplicative.

Original entry on oeis.org

0, 1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 22, 23, 24, 25, 32, 33, 34, 35, 36, 44, 47, 48, 49, 58, 59, 60, 62, 63, 64, 74, 75, 78, 79, 80, 81, 95, 96, 98, 99, 100, 114, 119, 120, 121, 135, 136, 138, 140, 141, 142, 143, 144, 160, 162, 164, 167, 168, 169, 185, 187, 189, 192

Offset: 1

Mitch Harris, Jun 18 2005

If we consider each square k as having a continued fraction expansion c of all zeros after c(0) = sqrt(k)-1, then the continued fraction expansion of sqrt(k) for each square is trivially multiplicative.

For nonsquares, c(1) must be 1 and so k must satisfy m + 1/2 < sqrt(k) <= m+1, for some integer m.

			The continued fraction of sqrt(22) is c = (4; 1, 2, 4, 2, 1, 8, ...) = A010126, which is multiplicative with c(2^e) = 2, c(3^e) = 4, c(p^e) = 1 otherwise.

Union of A000290 and A108575.

Continued fraction expansions: A040001, A010121, A040005, etc.

A187142 Smallest number k such that the continued fraction expansion of sqrt(k) contains n distinct numbers.

Original entry on oeis.org

1, 2, 7, 14, 19, 61, 94, 151, 211, 436, 604, 844, 919, 1324, 1894, 2011, 2731, 3691, 4951, 5086, 6451, 7606, 9619, 10294, 13126, 15814, 17599, 21499, 19231, 21319, 30319, 31606, 34654, 42379, 46006, 53299, 48799, 60811, 76651, 78094, 85999, 90931

Offset: 1

Vladimir Joseph Stephan Orlovsky, Mar 05 2011

nonn

For the first 191 terms, a(n) has the form p*2^i, where p is prime and i >= 0. - T. D. Noe, Mar 07 2011
Looking at just the periodic part of sqrt(k), it is the same sequence without the term a(1). - Robert G. Wilson v, Mar 22 2011
Conjecture: a(n) is of the form p, 2*p or 4*p, where p is prime. For the first 528 terms, a(n) is of the form 4*p only for n = 10, 11, 12, 14, 81 and 277. - Chai Wah Wu, Oct 04 2019

			ContinuedFraction(sqrt(2),x) => 1,2,2,2,...: two distinct terms (1,2);
sqrt(7) => 2,1,1,1,4,1,1,1,...: three distinct terms (1,2,4);
sqrt(14) => four distinct terms (1,2,3,6);
sqrt(19) => five distinct terms (1,2,3,4,8).

Chai Wah Wu, Table of n, a(n) for n = 1..528 (terms 1..191 from Robert G. Wilson v)

Cf. A040000, A010121.

f[n_] := Length@ Union@ Flatten@ ContinuedFraction@ Sqrt@ n; t = Table[ 0, {100}]; Do[a = f@ k; If[ a <= 100 && t[[a]] == 0, t[[a]] = k; Print[{a, k}]], {k, 10^5}]; t

A307453 a(n) is the least prime p for which the continued fraction expansion of sqrt(p) has exactly n consecutive 1's starting at position 2.

Original entry on oeis.org

2, 3, 31, 7, 13, 3797, 5273, 4987, 90371, 79873, 2081, 111301, 1258027, 5325101, 12564317, 9477889, 47370431, 709669249, 1529640443, 2196104969, 392143681, 8216809361, 30739072339, 200758317433, 370949963971, 161356959383, 1788677860531, 7049166342469, 4484287435283, 3690992602753

Offset: 0

Michel Marcus, Apr 09 2019

nonn

			For p = 2,  we have [1; 2, ...]; see A040000.
For p = 3,  we have [1; 1, 2, ...]; see A040001.
For p = 31, we have [5; 1, 1, 3, ...]; see A010129.
For p = 7,  we have [2; 1, 1, 1, 4, ...]; see A010121.

Piotr Miska, Maciej Ulas, On consecutive 1's in continued fractions expansions of square roots of prime numbers, arXiv:1904.03404 [math.NT], 2019. See Table 1 p. 15.
Index entries for continued fractions for constants

Cf. A040000, A040001, A010121, A010129.

isok(p, n) = {my(c=contfrac(sqrt(p)));  for (k=2, n+1, if (c[k] != 1, return (0));); return(c[n+2] !=  1);}
a(n) = {my(p=2); while (! isok(p, n), p = nextprime(p+1)); p;}

Limit_{n->infinity} (sqrt(a(n)) - floor(sqrt(a(n)))) = A094214. - Daniel Suteu, Apr 09 2019

a(21)-a(29) from Daniel Suteu, Apr 09 2019

a(0) added by Chai Wah Wu, Apr 09 2019

A320839 Factorial expansion of sqrt(7) = Sum_{n>=1} a(n)/n!.

Original entry on oeis.org

2, 1, 0, 3, 2, 2, 6, 4, 6, 2, 3, 11, 2, 8, 11, 8, 16, 5, 5, 16, 19, 5, 1, 14, 16, 7, 14, 10, 27, 12, 10, 29, 28, 19, 16, 3, 6, 4, 28, 33, 24, 21, 42, 10, 2, 45, 3, 34, 4, 1, 46, 48, 8, 5, 41, 20, 53, 17, 31, 50, 10, 6, 56, 27, 29, 18, 15, 11, 19, 49, 37, 64, 56, 51, 34, 21, 3, 27, 15, 61

Offset: 1

G. C. Greubel, Dec 10 2018

nonn

			sqrt(7) = 2 + 1/2! + 0/3! + 3/4! + 2/5! + 2/6! + 6/7! + 4/8! + 6/9! + ...

Index entries for factorial base representation

Cf. A010465 (decimal expansion), A010121 (continued fraction).

Cf. A009949 (sqrt(2)), A067881 (sqrt(3)), A068446 (sqrt(5)).

Digits:=200: a:=n->`if`(n=1,floor(sqrt(7)),floor(factorial(n)*sqrt(7))-n*floor(factorial(n-1)*sqrt(7))): seq(a(n),n=1..90); # Muniru A Asiru, Dec 10 2018

cp's OEIS Frontend

A010465 Decimal expansion of square root of 7.

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A233587 Coefficients of the generalized continued fraction expansion sqrt(7) = a(1) +a(1)/(a(2) +a(2)/(a(3) +a(3)/(a(4) +a(4)/....))).

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A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

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A171621 Numerator of 1/4 - 1/n^2, each fourth term multiplied by 4.

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A283971 a(n) = n except a(4*n + 2) = 2*n + 1.

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A098457 Farey Bisection Expansion of sqrt(7).

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Magma

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A283971 a(n) = n except a(4n + 2) = 2n + 1.