A047621 -id:A047621 - cp's OEIS frontend

A066587 Duplicate of A047621.

3, 5, 11, 13, 19, 21, 27, 29, 35, 37, 43, 45, 51, 53, 59, 61, 67, 69, 75, 77, 83, 85, 91, 93

Offset: 1

dead

A014176 Decimal expansion of the silver mean, 1+sqrt(2).

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

Offset: 1

N. J. A. Sloane

From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).
c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c = sqrt(2)-1.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010
Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011
Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013
An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013
n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016
This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020
c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023
c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

			2.414213562373095...

B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Nicholas R. Beaton, Mireille Bousquet-Mélou, Jan de Gier, Hugo Duminil-Copin, and Anthony J. Guttmann, The critical fugacity for surface adsorption of self-avoiding walks on the honeycomb lattice is 1+sqrt(2), arXiv:1109.0358 [math-ph], 2011-2013.
Eric Weisstein's World of Mathematics, Silver Ratio
Wikipedia, Exact trigonometric constants
Wikipedia, Metallic mean
Wikipedia, Silver ratio
Index entries for algebraic numbers, degree 2

Apart from initial digit the same as A002193.
Cf. A000032, A006497, A080039, A179260, A121601.
See A098316 for [3;3,3,...]; A098317 for [4;4,4,...]; A098318 for [5;5,5,...]. - Hieronymus Fischer, Oct 20 2010
Cf. A000129, A049310.

Digits:=100: evalf(1+sqrt(2)); # Wesley Ivan Hurt, Apr 09 2016

RealDigits[1 + Sqrt@ 2, 10, 111] (* Or *)
RealDigits[Exp@ ArcSinh@ 1, 10, 111][[1]] (* Robert G. Wilson v, Aug 17 2011 *)
Circs[n_] := With[{r = Sin[Pi/n]/(1 - Sin[Pi/n])}, Graphics[Append[
  Table[Circle[(r + 1) {Sin[2 Pi k/n], Cos[2 Pi k/n]}, r], {k, n}],   {Blue, Circle[{0, 0}, 1]}]]] Circs[4] (* Charles R Greathouse IV, Jan 14 2013 *)

1+sqrt(2) \\ Charles R Greathouse IV, Jan 14 2013

Conjecture: 1+sqrt(2) = lim_{n->oo} A179807(n+1)/A179807(n).
Equals cot(Pi/8) = tan(Pi*3/8). - Bruno Berselli, Dec 13 2012, and M. F. Hasler, Jul 08 2016
Silver mean = 2 + Sum_{n>=0} (-1)^n/(P(n-1)*P(n)), where P(n) is the n-th Pell number (A000129). - Vladimir Shevelev, Feb 22 2013
Equals exp(arcsinh(1)) which is exp(A091648). - Stanislav Sykora, Nov 01 2013
Limit_{n->oo} exp(asinh(cos(Pi/n))) = sqrt(2) + 1. - Geoffrey Caveney, Apr 23 2014
exp(asinh(cos(Pi/2 - log(sqrt(2)+1)*i))) = exp(asinh(sin(log(sqrt(2)+1)*i))) = i. - Geoffrey Caveney, Apr 23 2014
Equals Product_{k>=1} A047621(k) / A047522(k) = (3/1) * (5/7) * (11/9) * (13/15) * (19/17) * (21/23) * ... . - Dimitris Valianatos, Mar 27 2019
From Wolfdieter Lang, Nov 10 2023:(Start)
Equals lim_{n->oo} A000129(n+1)/A000129(n) (see A000129, Pell).
Equals lim_{n->oo} S(n+1, 2*sqrt(2))/S(n, 2*sqrt(2)), with the Chebyshev S(n,x) polynomial (see A049310). (End)
From  Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 6 for k >= 0.
For example, taking k = 0 and k = 1 yields
sqrt(2) + 1 = 15/(6 + (1*3)/(12 + (5*7)/(12 + (9*11)/(12 + (13*15)/(12 + ... + (4*n + 1)*(4*n + 3)/(12 + ... )))))) and
sqrt(2) + 1 = (715/21) * 1/(14 + (1*3)/(28 + (5*7)/(28 + (9*11)/(28 + (13*15)/(28 + ... + (4*n + 1)*(4*n + 3)/(28 + ... )))))). (End)

A047522 Numbers that are congruent to {1, 7} mod 8.

Original entry on oeis.org

1, 7, 9, 15, 17, 23, 25, 31, 33, 39, 41, 47, 49, 55, 57, 63, 65, 71, 73, 79, 81, 87, 89, 95, 97, 103, 105, 111, 113, 119, 121, 127, 129, 135, 137, 143, 145, 151, 153, 159, 161, 167, 169, 175, 177, 183, 185, 191, 193, 199, 201, 207, 209, 215, 217, 223, 225, 231, 233

Offset: 1

N. J. A. Sloane

Also n such that Kronecker(2,n) = mu(gcd(2,n)). - Jon Perry and T. D. Noe, Jun 13 2003
Also n such that x^2 == 2 (mod n) has a solution. The primes are given in sequence A001132. - T. D. Noe, Jun 13 2003
As indicated in the formula, a(n) is related to the even triangular numbers. - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
A089911(3*a(n)) = 2. - Reinhard Zumkeller, Jul 05 2013
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2)))  = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (For T see A053120.) - Wolfdieter Lang, Jun 04 2023

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 16.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001132, A014494, A056575, A010709, A074378, A047336, A056020, A005408, A047209, A007310, A090771, A175885, A091998, A175886, A175887, A058529, A047621, A179260, A352125.
Cf. A077221 (partial sums).
Cf. A000217, A089911, A113801.
Cf. A010503, A049310, A053120.

a047522 n = a047522_list !! (n-1)
a047522_list = 1 : 7 : map (+ 8) a047522_list
-- Reinhard Zumkeller, Jan 07 2012

Select[Range[1, 191, 2], JacobiSymbol[2, # ]==1&]

a(n)=4*n-2+(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

a(n) = sqrt(8*A014494(n)+1) = sqrt(16*ceiling(n/2)*(2*n+1)+1) = sqrt(8*A056575(n)-8*(2n+1)*(-1)^n+1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
1 - 1/7 + 1/9 - 1/15 + 1/17 - ... = (Pi/8)*(1 + sqrt(2)). [Jolley] - Gary W. Adamson, Dec 16 2006
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = 4n - 2 + (-1)^n = a(n-2) + 8.
G.f.: x(1+6x+x^2)/((1+x)(1-x)^2). (End)
a(n) = 8*n - a(n-1) - 8. - Vincenzo Librandi, Aug 06 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 8*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
E.g.f.: 1 + (4*x - 1)*cosh(x) + (4*x - 3)*sinh(x). - Stefano Spezia, May 13 2021
E.g.f.: 1 + (4*x - 3)*exp(x) + 2*cosh(x). - David Lovler, Jul 16 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2+sqrt(2)) (A179260).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/8)*cosec(Pi/8) (A352125). (End)

A089911 a(n) = Fibonacci(n) mod 12.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1

Offset: 0

Casey Mongoven, Nov 14 2003

From Reinhard Zumkeller, Jul 05 2013: (Start)
Sequence has been applied by several composers to 12-tone equal temperament pitch structure. The complete Fibonacci mod 12 system (a set of 10 periodic sequences) exhausts all possible ordered dyads; that is, every possible combination of two pitches is found in these sets.
a(A008594(n)) = 0;
a(A227144(n)) = 1;
a(3*A047522(n)) = 2;
a(A017569(n)) = a(2*A016933(n)) = a(4*A016777(n)) = 3;
a(2*A017629(n)) = a(3*A017137(n)) = a(6*A004767(n)) = 4;
a(A227146(n)) = 5;
a(nonexistent) = 6;
a(2*A017581(n)) = 7;
a(2*A017557(n)) = a(4*A016813(n)) = 8;
a(A017617(n)) = a(2*A016957(n)) = a(4*A016789(n)) = 9;
a(3*A047621(n)) = 10;
a(2*A017653(n)) = 11. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1199
Ron Knott, Fibonacci Numbers and the Golden Section
M. Renault, The Fibonacci Sequence Modulo M
A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quarterly, Vol.6, No.2 (1968), 139-141.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Index entries for linear recurrences with constant coefficients, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1).

Cf. A000045, A001175, A003893, A010881.

a089911 n = a089911_list !! n
a089911_list = 0 : 1 : zipWith (\u v -> (u + v) `mod` 12)
                       (tail a089911_list) a089911_list
-- Reinhard Zumkeller, Jul 01 2013

[Fibonacci(n) mod 12: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

with(combinat,fibonacci); A089911 := proc(n) fibonacci(n) mod 12; end;

Table[Mod[Fibonacci[n], 12], {n, 0, 100}] (* Vincenzo Librandi, Feb 04 2014 *)

a(n)=fibonacci(n)%12 \\ Charles R Greathouse IV, Feb 03 2014

Has period of 24, restricted period 12 and multiplier 5.

a(n) = (a(n-1) + a(n-2)) mod 12, a(0) = 0, a(1) = 1.

More terms from Ray Chandler, Nov 15 2003

A112070 Square array A(x,y) = y-th odd number 2i+1 (i>=1) for which A112049(2i+1)=x, or 0 if no such i exists; read by descending antidiagonals.

Original entry on oeis.org

3, 5, 7, 11, 9, 23, 13, 15, 25, 49, 19, 17, 47, 71, 121, 21, 31, 73, 119, 311, 169, 27, 33, 95, 191, 551, 479, 289, 29, 39, 97, 239, 671, 1151, 1559, 361, 35, 41, 143, 241, 719, 1319, 2999, 5711, 529, 37, 55, 145, 359, 839, 1679, 3071, 8399, 10559, 841, 43, 57

Offset: 1

Antti Karttunen, Aug 27 2005

This is a permutation of odd numbers greater than unity provided that the sequence A112046 contains only prime values and every prime occurs infinitely many times there. Because the Jacobi symbol is multiplicative with respect to its modulus, it follows that if n occurs on row i and m occurs on row j, then n*m cannot occur before row min(i,j).

			The top left corner of the array:
3,5,11,13,19,21,...
7,9,15,17,31,33,...
23,25,47,73,95,...

A(x, y) = 2*A112060(x, y)+1. Transpose: A112071. Column 1: A112052. Row 1: A047621, Row 2: A112072 Row 3: A112073, Row 4: A112074, Row 5: A112075, Row 6: A112076, Row 7: A112077, Row 8: A112078, Row 9: A112079.

A274651 Triangle read by rows: T(n,k), (1<=k<=n), in which each term is the least positive integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

Original entry on oeis.org

1, 2, 3, 4, 1, 2, 3, 5, 4, 6, 6, 2, 1, 3, 4, 5, 4, 6, 2, 7, 8, 7, 8, 3, 1, 6, 5, 9, 9, 6, 10, 5, 8, 3, 11, 7, 8, 11, 9, 4, 1, 7, 10, 6, 5, 12, 7, 13, 8, 2, 9, 4, 11, 10, 14, 10, 9, 5, 12, 3, 1, 2, 13, 7, 8, 11, 11, 12, 8, 13, 5, 4, 3, 10, 9, 15, 14, 16, 13, 10, 11, 7, 9, 2, 1, 12, 8, 5, 17, 15, 18

Offset: 1

Omar E. Pol, Jul 02 2016

Analog of A269526, but note that this is a right triangle.
The same rule applied to an equilateral triangle gives A269526.
We construct the triangle by reading from left to right in each row, starting with T(1,1) = 1.
Presumably every diagonal and every column is also a permutation of the positive integers, but the proof does not seem so straightforward. Of course, neither the rows nor the antidiagonals are permutations of the positive integers, since they are finite in length.
Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the positive integers is true: see the link in A274650 (duplicated below). - N. J. A. Sloane, Jun 07 2017
It appears that the numbers generally appear for the first time in or near the right border of the triangle.
Theorem 1: the middle diagonal gives A000012 (the all 1's sequence).
Theorem 2: all 1's are in the middle diagonal.
For the proofs of the theorems 1 and 2 see the proofs of the theorems 1 and 2 of A274650 since both sequences are essentially the same.
From Bob Selcoe, Feb 15 2017: (Start)
The columns and diagonal are permutations of the natural numbers. The proofs are essentially the same as the proofs given for the columns and rows (respectively) in A269526.
All coefficients j <= 4 eventually populate in a repeating pattern toward the "middle diagonal" (i.e., relatively near the 1's); this is because we can build the triangle by j in ascending order; that is, we can start by placing all the 1's in the proper cells, then add the 2's, 3's, 4's, 5's, etc. So for i >= 0: since the 1's appear at T(1+2i, 1+i), the 2's appear at T(2+8i, 1+4i), T(3+8i, 3+4i), T(5+8i, 2+4i) and T(6+8i, 4+4i). Accordingly, after the first five 3's appear (at T(2,2), T(4,1), T(5,4), T(7,3) and T(8,6)), the remaining 3's appear at T(11+8i, 5+4i), T(12+8i, 7+4i), T(16+8i, 8+4i) and T(17+8i, 10+4i). Similarly for 4's, after the first 21 appearances, 4's appear at T(44+8i, 21+4i), T(45+8i, 24+4i), T(47+8i, 23+4i) and T(48+8i, 26+4i). So starting at T(41,21), this 16-coefficient pattern repeats at T(41+8i, 21+4i):
n/k  21 22 23 24 25 26
41    1  3
42    2
43    3  1  2
44    4     3
45       2  1  4
46             2
47          4  1
48             3     4
where the next 1 appears at T(49,25), and the pattern repeats at that point from the top left (so T(49,26) = 3, T(50,25) = 2, etc.).
Conjecture: as n gets sufficiently large, all coefficients j>4 will appear in a repeating pattern, populating all rows and diagonals around smaller j's near the "middle diagonal" (while I can offer no formal proof, it appears very likely that this is the case). (End)
From Hartmut F. W. Hoft, Jun 12 2017: (Start)
T(2k+1,k+1) = 1, for all k>=0, and T(n,{n/2,(n+3)/2,(n-1)/2,(n+2)/2}) = 2, for all n>=1 with mod(n,8) = {2,3,5,6} respectively, and no 1's or 2's occur in other positions.
Proof by (recursive) picture:
Positions in the triangle that are empty and those containing the dots of the guiding diagonals contain numbers larger than two.
  n\k 1 2 3 4 5 6 7 8  10  12  14  16  18  20  22  24
   1 |1
   2 |2 .
   3 |  1 2
   4 |      .
   5 |  2 1   .
   6 |      2   .
   7 |      1     .
   8 |____________.
   9 |       |1       .
  10 |       |2 .       .
  11 |       |  1 2       .
  12 |       |      .       .
  13 |       |  2 1   .       .
  14 |       |      2   .       .
  15 |       |      1     .       .
  16 |_____|______________._____.
  17 |       |       |1       .       .
  18 |       |       |2 .       .       .
  19 |       |       |  1 2       .       .
  20 |       |       |      .       .       .
  21 |       |       |  2 1   .       .       .
  22 |       |       |      2   .       .       .
  23 |       |       |      1     .       .       .
  24 |_____|_______|____._______._____._______.
      1 2 3 4 5 6 7 8      12      16      20      24
Consider the center of the triangle. In each octave of rows the columns in the first central quatrain contain a 1 and a 2, and the diagonals in the second central quatrain contain a 1 and a 2. Therefore, no 1's or 2's can occur in the respective downward quatrains of leading columns and trailing diagonals.
The sequence of rows containing 2's is A047447 (n mod 8 = {2,3,5,6}), those containing only 2's is A016825 (n mod 8 = {2,6}), those containing both 1's and 2's is A047621 (n mod 8 = {3,5}), those containing only 1's is A047522 (n mod 8 = {1,7}), and those containing neither 1's nor 2's is A008586 (n mod 8 = {0,4}).
(End)

			Triangle begins:
   1;
   2,  3;
   4,  1,  2;
   3,  5,  4,  6;
   6,  2,  1,  3,  4;
   5,  4,  6,  2,  7,  8;
   7,  8,  3,  1,  6,  5,  9;
   9,  6, 10,  5,  8,  3, 11,  7;
   8, 11,  9,  4,  1,  7, 10,  6,  5;
  12,  7, 13,  8,  2,  9,  4, 11, 10, 14;
  10,  9,  5, 12,  3,  1,  2, 13,  7,  8, 11;
  11, 12,  8, 13,  5,  4,  3, 10,  9, 15, 14, 16;
  13, 10, 11,  7,  9,  2,  1, 12,  8,  5, 17, 15, 18;
  ...
From _Omar E. Pol_, Jun 07 2017: (Start)
The triangle may be reformatted as an isosceles triangle so that the all 1's sequence (A000012) appears in the central column (but note that this is NOT the way the triangle is constructed!):
.
.                  1;
.                2,  3;
.              4,  1,  2;
.            3,  5,  4,  6;
.          6,  2,  1,  3,  4;
.        5,  4,  6,  2,  7,  8;
.      7,  8,  3,  1,  6,  5,  9;
.    9,  6, 10,  5,  8,  3, 11,  7;
.  8, 11,  9,  4,  1,  7, 10,  6,  5;
...
(End)

Michel Marcus, Table of n, a(n) for n = 1..20100
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
N. J. A. Sloane, Notes on A274650 and Proof by Non-Attacking Queens

Cf. A001844 (indices of the 1's).
Cf. A000012 (middle diagonal).
Every diagonal and every column of the right triangle is a permutation of A000027.
Cf. A274650 is the same triangle but with every entry minus 1.
Other sequences of the same family are A269526, A274528, A274820, A274821, A286297, A288530, A288531.
Sequences mentioned in N. J. A. Sloane's proof are A000170, A274616 and A287864.
Cf. A008586, A016825, A047447, A047522, A047621. - Hartmut F. W. Hoft, Jun 12 2017

f[1,1] = 1; (* for 1 < n and 1 <= k <= n *)
f[n_,k_] := f[n,k] = Module[{vals=Sort[Join[Map[f[n, #]&, Range[1, k-1]], Map[f[#, k]&, Range[k, n-1]], Map[f[n-k+#, #]&, Range[1, k-1]], Map[f[n-#, k+#]&, Range[1, Floor[(n-k)/2]]]]], c}, c=Complement[Range[1, Last[vals]], vals]; If[c=={}, Last[vals]+1, First[c]]]
(* computation of rows 1 ... n of triangle *)
a274651[n_] := Prepend[Table[f[i, j], {i, 2, n}, {j, 1, i}], {1}]
Flatten[a274651[13]] (* data *)
TableForm[a274651[13]] (* triangle *)
(* Hartmut F. W. Hoft, Jun 12 2017 *)

T(n,k) = A274650(n-1,k-1) + 1.

A214263 Expansion of f(x^1, x^7) in powers of x where f() is Ramanujan's general theta function.

Original entry on oeis.org

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

Offset: 0

Michael Somos and Omar E. Pol, Jul 09 2012

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Characteristic function of A074377: a(n) = 1 if and only if n is in A074377.

			G.f. = 1 + x + x^7 + x^10 + x^22 + x^27 + x^45 + x^52 + x^76 + x^85 + x^115 + ...
G.f. = q^9 + q^25 + q^121 + q^169 + q^361 + q^441 + q^729 + q^841 + q^1225 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.
Index entries for characteristic functions.

Cf. A047621, A074377, A115359.
A000122, A080995, A010054, A133100, A089801 have g.f. of f(x,x^k) for k=1..5.
Cf. A000122, A000700, A121373.

f[x_, y_] := QPochhammer[-x, x*y]*QPochhammer[-y, x*y]*QPochhammer[x*y, x*y]; Table[SeriesCoefficient[f[q, q^7], {q, 0, n}], {n, 0, 50}] (* G. C. Greubel, Oct 05 2017 *)

PARI
```
{a(n) = issquare(16*n + 9)};
```

Euler transform of period 16 sequence [ 1, -1, 0, 0, 0, 0, 1, -1, 1, 0, 0, 0, 0, -1, 1, -1, ...].
G.f.: f(x, x^7) = sum_{k in Z} x^(4*k^2 - 3*k).
a(n) = A010054(2*n + 1) = A115359(2*n).
Sum_{k=1..n} a(k) ~ sqrt(n). - Amiram Eldar, Jan 13 2024

A101464 Decimal expansion of sqrt(2-sqrt(2)), edge length of a regular octagon with circumradius 1.

Original entry on oeis.org

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

Offset: 0

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Jan 20 2005

			0.765366864730179543456919968060797733522689124971254082867601271255092067920...

G. C. Greubel, Table of n, a(n) for n = 0..5000
Eric Weisstein's World of Mathematics, Octagon.
Index entries for algebraic numbers, degree 4.

Cf. A047621, A101465, A179260 (sqrt(2+sqrt(2))), A182168, A285871, A329246.

RealDigits[Sqrt[2-Sqrt[2]],10,120][[1]] (* Harvey P. Dale, Jun 22 2011 *)

2*sin(Pi/8) \\ Charles R Greathouse IV, Feb 04 2025

polrootsreal(x^4-4*x^2+2)[3] \\ Charles R Greathouse IV, Feb 04 2025

Equals i^(3/4) + i^(-3/4). - Gary W. Adamson, Jul 07 2022
Equals 2*sin(Pi/8) = 2*A182168. - Amiram Eldar, Apr 06 2023
Equals Product_ {k >= 0} ((8*k - 2)*(8*k + 10))/((8*k - 5)*(8*k + 13)). - Antonio Graciá Llorente, Mar 11 2024
Equals Product_{k>=1} (1 + (-1)^k/A047621(k)). - Amiram Eldar, Nov 22 2024
Equals sqrt(A101465) = 1/A285871 = exp(-A329246). - Hugo Pfoertner, Nov 22 2024

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

Original entry on oeis.org

2, 7, 18, 31, 50, 71, 98, 127, 162, 199, 242, 287, 338, 391, 450, 511, 578, 647, 722, 799, 882, 967, 1058, 1151, 1250, 1351, 1458, 1567, 1682, 1799, 1922, 2047, 2178, 2311, 2450, 2591, 2738, 2887, 3042, 3199, 3362, 3527, 3698, 3871, 4050, 4231, 4418, 4607, 4802

Offset: 0

Bruno Berselli, Sep 21 2011 - based on remarks and sequences by Omar E. Pol

Sequence found by reading the numbers in increasing order on the vertical line containing 2 of the square spiral whose vertices are the triangular numbers (A000217) - see Pol's comments in other sequences visible in this numerical spiral.

Also A077591 (without first term) and A157914 interleaved.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A195605.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000217, A077591, A157914.
Cf. A047621 (contains first differences), A016754 (contains the sum of any two consecutive terms).
Cf. A033585, A069129, A077221, A102083, A139098, A139271-A139277, A139592, A139593, A188135, A194268, A194431, A195241 [incomplete list].

[(4*n*(n+2)+(-1)^n+3)/2: n in [0..48]];

CoefficientList[Series[(2 + 3 x + 4 x^2 - x^3) / ((1 + x) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{2,0,-2,1},{2,7,18,31},50] (* Harvey P. Dale, Jan 21 2017 *)

for(n=0, 48, print1((4*n*(n+2)+(-1)^n+3)/2", "));

G.f.: (2+3*x+4*x^2-x^3)/((1+x)*(1-x)^3).
a(n) = a(-n-2) = 2*a(n-1)-2*a(n-3)+a(n-4).
a(n) = A047524(A000982(n+1)).
Sum_{n>=0} 1/a(n) = 1/2 + Pi^2/16 - cot(Pi/(2*sqrt(2)))*Pi/(4*sqrt(2)). - Amiram Eldar, Mar 06 2023

A047425 Numbers that are congruent to {3, 4, 5, 6} mod 8.

Original entry on oeis.org

3, 4, 5, 6, 11, 12, 13, 14, 19, 20, 21, 22, 27, 28, 29, 30, 35, 36, 37, 38, 43, 44, 45, 46, 51, 52, 53, 54, 59, 60, 61, 62, 67, 68, 69, 70, 75, 76, 77, 78, 83, 84, 85, 86, 91, 92, 93, 94, 99, 100, 101, 102, 107, 108, 109, 110, 115, 116, 117, 118, 123, 124

Offset: 1

N. J. A. Sloane

Complement of numbers congruent to {0, 1, 2, 7} mod 8. - Jaroslav Krizek, Dec 19 2009

In general, sequences congruent to {a, a + i, a + 2i, ..., a + pi} mod k and a + p*i < k have a general form of (k - i*p)*floor(n/p) + i*n + a, from offset 0. - Gary Detlefs, Oct 20 2013

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047406, A047621.

[n : n in [0..150] | n mod 8 in [3, 4, 5, 6]]; // Wesley Ivan Hurt, May 31 2016

A047425:=n->8*floor((n-1)/4)+((n-1) mod 4)+3: seq(A047425(n), n=1..100); # Wesley Ivan Hurt, May 31 2016

Flatten[# + {3, 4, 5, 6} &/@(8*Range[0, 15])] (* Harvey P. Dale, Jun 26 2011 *)

G.f.: x*(3+x+x^2+x^3+2*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
a(n) = 8*floor((n-1)/4) + ((n-1) mod 4) + 3.
a(n) = OR(n-1, 1) + OR(n-1, 2). - Gary Detlefs, Oct 20 2013
From Wesley Ivan Hurt, May 31 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (4*n-1-i^(2*n)-(1-i)*i^(-n)-(1+i)*i^n)/2 where i=sqrt(-1).
a(2k) = A047406(k), a(2k-1) = A047621(k). (End)
E.g.f.: 2 + sin(x) - cos(x) + 2*x*sinh(x) + (2*x - 1)*cosh(x). - Ilya Gutkovskiy, May 31 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/16 + (3-sqrt(2))*log(2)/8 + sqrt(2)*log(2-sqrt(2))/4. - Amiram Eldar, Dec 26 2021

cp's OEIS Frontend

A066587 Duplicate of A047621.

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A014176 Decimal expansion of the silver mean, 1+sqrt(2).

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A047522 Numbers that are congruent to {1, 7} mod 8.

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A089911 a(n) = Fibonacci(n) mod 12.

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A112070 Square array A(x,y) = y-th odd number 2i+1 (i>=1) for which A112049(2i+1)=x, or 0 if no such i exists; read by descending antidiagonals.

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A274651 Triangle read by rows: T(n,k), (1<=k<=n), in which each term is the least positive integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

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A214263 Expansion of f(x^1, x^7) in powers of x where f() is Ramanujan's general theta function.

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A101464 Decimal expansion of sqrt(2-sqrt(2)), edge length of a regular octagon with circumradius 1.

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A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.