A047522 -id:A047522 - cp's OEIS frontend

A175887 Numbers that are congruent to {1, 14} mod 15.

1, 14, 16, 29, 31, 44, 46, 59, 61, 74, 76, 89, 91, 104, 106, 119, 121, 134, 136, 149, 151, 164, 166, 179, 181, 194, 196, 209, 211, 224, 226, 239, 241, 254, 256, 269, 271, 284, 286, 299, 301, 314, 316, 329, 331, 344, 346, 359, 361, 374, 376, 389, 391, 404

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1==0 (mod h); in this case, a(n)^2-1==0 (mod 15).

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

Cf. A000217, A019693, A019976, A113801, A175886, A091998, A175885, A090771, A056020, A047522, A047336, A007310, A047209, A005408, A001651.

Cf. A132240 (primes).

a175887 n = a175887_list !! (n-1)
a175887_list = 1 : 14 : map (+ 15) a175887_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..450] | n mod 15 in [1,14]];

[(30*n+11*(-1)^n-15)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013

Select[Range[1, 450], MemberQ[{1,14}, Mod[#, 15]]&]
CoefficientList[Series[(1 + 13 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)

a(n)=(30*n+11*(-1)^n-15)/4 \\ Charles R Greathouse IV, Sep 28 2015

G.f.: x*(1+13*x+x^2)/((1+x)*(1-x)^2).
a(n) = (30*n+11*(-1)^n-15)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = 15*A000217(n-1) -2*sum(a(i), i=1..n-1) +1 for n>1.
a(n) = A047209(A047225(n+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/15)*cot(Pi/15) = A019693 * A019976 / 10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((30*x - 15)*exp(x) + 11*exp(-x))/4. - David Lovler, Sep 05 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/15)*cosec(Pi/15).
Product_{n>=2} (1 + (-1)^n/a(n)) = 2*cos(Pi/15). (End)

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.

A214264 Expansion of f(x^3, x^5) in powers of x where f() is Ramanujan's two-variable theta function.

Original entry on oeis.org

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

Offset: 0

Michael Somos, Jul 09 2012

nonn

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

			1 + x^3 + x^5 + x^14 + x^18 + x^33 + x^39 + x^60 + x^68 + x^95 + x^105 +
q + q^49 + q^81 + q^225 + q^289 + q^529 + q^625 + q^961 + q^1089 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A010054, A047522, A074378.

Cf. A000122, A000700, A121373.

f[x_, y_]:= QPochhammer[-x, x*y]*QPochhammer[-y, x*y]*QPochhammer[x*y, x*y]; A214264[n_] := SeriesCoefficient[f[x^3, x^5], {x, 0, n}]; Table[A214264[n], {n, 0, 50}] (* G. C. Greubel, Dec 03 2017 *)

PARI
```
{a(n) = issquare( 16*n + 1)}
```

Euler transform of period 16 sequence [ 0, 0, 1, 0, 1, -1, 0, -1, 0, -1, 1, 0, 1, 0, 0, -1, ...].
G.f.: Sum_{k} x^(((8*k + 1)^2 - 1) / 16).
Characteristic function of A074378. a(n) = 1 if and only if n is in A074378.
a(n) = A010054(2*n).
Sum_{k=1..n} a(k) ~ sqrt(n). - Amiram Eldar, Jan 13 2024

A352125 Decimal expansion of Pisqrt(2)sqrt(2 + sqrt(2))/8.

Original entry on oeis.org

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

Offset: 1

Stefano Spezia, Mar 05 2022

			1.02617215297703088887146778087283197497962...

Jean-François Pabion, Éléments d'Analyse Complexe, licence de Mathématiques, page 111, Ellipses, 1995.

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, pp. 235-236.
Index entries for transcendental numbers.

Cf. A000796, A047522, A121601.

Integral_{x=0..oo} 1/(1+x^m) dx: A019669 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), A019670 (m=6), this sequence (m=8), A094888 (m=10).

First[RealDigits[N[Pi*Sqrt[2]Sqrt[2+Sqrt[2]]/8,95]]]

Pi*sqrt(4 + 2*sqrt(2))/8 \\ Michel Marcus, Mar 07 2022

Equals Integral_{x=0..oo} 1/(1 + x^8) dx.
Equals Pi*csc(Pi/8)/8.
Equals 1/Product_{k>=1} (1 - 1/(8*k)^2). - Amiram Eldar, Mar 12 2022
Equals Product_{k>=2} (1 + (-1)^k/A047522(k)). - Amiram Eldar, Nov 22 2024

A225822 Lesser of adjacent odd numbers with different parity of binary weight and both isolated from odd numbers of same parity of binary weight.

Original entry on oeis.org

7, 23, 31, 39, 55, 71, 87, 95, 103, 119, 127, 135, 151, 159, 167, 183, 199, 215, 223, 231, 247, 263, 279, 287, 295, 311, 327, 343, 351, 359, 375, 383, 391, 407, 415, 423, 439, 455, 471, 479, 487, 503, 511, 519, 535, 543, 551, 567, 583, 599, 607, 615, 631

Offset: 1

Brad Clardy, Jul 30 2013

Write the sequence of odious odd numbers above the sequence of evil odd numbers connecting all that are 2 apart:
1 7 11-13 19-21 25 31 35-37 41 47-49 55 59-61 67-69 73 79-81 87 91-93 97
3-5  9 15-17 23 27-29 33 39 43-45 51-53 57 63-65 71 75-77 83-85 89 95 99-
Remove the connected numbers:
1 7             25 31       41       55             73       87       97
     9       23       33 39             57       71             89  95
Define these as "isolated".
The sequence is the smaller of the remaining pairs that are 2 apart.
The 1 is not a member since there is no change in parity between 1 and 7.
All of the differences between adjacent numbers in both the evil and odious sequences are either 2, 4 or 6, with 4 being the indicator that a transition in parity occurs. The program provided utilizes that fact to produce the sequence.
The sequence that includes all numbers along this path is A047522 (numbers congruent to {1,7} mod 8). This is also the same as the odd terms of A199398 (XOR of the first n odd numbers).
This sequence is similar to A044449 (numbers n such that string 1,3 occurs in the base 4 representation of n but not of n+1), but it contains additional terms. An example is 119. Its base 4 representation is 1313 while the base 4 representation of 120 is 1320. It may be that another workable definition of the sequence is -- numbers n in base 4 representation such that string 1,3 occurs one less time in n+1 than n, but I have not been able to check this.
The difference between the numbers in the sequence is always either 8 or 16, however there appears to be no recurring repetitions in it. Writing the 8 as a 0 and the 16 as a 1 (or dividing the difference pattern by 2 and subtracting a 1) produces a difference pattern of: 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1... which is an infinite word.
A similar process writing Even Odious over Even Evils produces 6, 22, 30, 38, 54, 70... which is twice A131323 (Odd numbers n such that the binary expansion ends in an even number of 1's), with all numbers along the path given by A047451 (numbers congruent to  {0,6} mod 8) and yields the same difference pattern which produces the same infinite word.

Brad Clardy, Table of n, a(n) for n = 1..1000

Cf. A001969 (evil numbers), A129771 (odd evil numbers).
Cf. A000069 (odious numbers), A092246 (odd odious numbers).
Cf. A047522 (numbers congruent to {1,7} mod 8).
Cf. A199398 (XOR of first n odd numbers).
Cf. A044449 (a subset of this sequence).
Cf. A131323 (odd numbers n such that the binary expansion ends in an even number of 1's).
Cf. A047451 (numbers congruent to {0,6} mod 8).
Cf. A000120 (binary weight of n).
Cf. A079523.

//Function Bweight calculates the binary weight of an integer
Bweight := function(m)
Bweight:=0;
adigs := Intseq(m,2);
for n:= 1 to Ilog2(m)+1 do
  Bweight:=Bweight+adigs[n];
end for;
return Bweight;
end function;
prevodi:=0;
currodi:=0;
m:=0;
count:=0;
for n:= 1 to 20000 by 2 do
    m:=m+1;
    if (Bweight(n) mod 2 eq 1) then odious:=true; else odious:=false; end if;
    if (odious) then currodi:=n; end if;
       if (currodi - prevodi eq 4) then
          if (m mod 2 eq 1) then count:=count+1; count,n-2;
             else count:=count+1;count,n-4;
       end if;
    end if;
    if(odious) then prevodi:=currodi; end if;
end for;

2*Select[Range[1, 320, 2], EvenQ[IntegerExponent[# + 1, 2]] &] + 1 (* Amiram Eldar, Jul 24 2023 *)

is(n)=n%4==3 && valuation(n\4+1, 2)%2 \\ Charles R Greathouse IV, Aug 20 2013

from itertools import count, islice
def A225822_gen(startvalue=1): # generator of terms >= startvalue
    return map(lambda m:(m<<1)+1,filter(lambda n:n&1 and not (~(n+1)&n).bit_length()&1,count(max(startvalue,1))))
A225822__list = list(islice(A225822_gen(),30)) # Chai Wah Wu, Jul 09 2022

a(n) = 2*A131323(n) + 1.
a(n) = 4*A079523(n) + 3. - Charles R Greathouse IV, Aug 20 2013
a(n) ~ 12n. (In particular, a(n) = 12n + O(log n).) - Charles R Greathouse IV, Aug 20 2013

A047530 Numbers that are congruent to {0, 1, 3, 7} mod 8.

Original entry on oeis.org

0, 1, 3, 7, 8, 9, 11, 15, 16, 17, 19, 23, 24, 25, 27, 31, 32, 33, 35, 39, 40, 41, 43, 47, 48, 49, 51, 55, 56, 57, 59, 63, 64, 65, 67, 71, 72, 73, 75, 79, 80, 81, 83, 87, 88, 89, 91, 95, 96, 97, 99, 103, 104, 105, 107, 111, 112, 113, 115, 119, 120, 121, 123

Offset: 1

N. J. A. Sloane

Numbers n such that the n-th homotopy group of the topological group O(oo) does not vanish [see Baez]. Cf. A195679.

The a(n+1) determine the maximal number of linearly independent smooth nowhere zero vector fields on a (2n+1)-sphere, see A053381. - Johannes W. Meijer, Jun 07 2011

John C. Baez, The Octonions, Bull. Amer. Math. Soc., Vol. 39, No. 2 (2002), pp. 145-205; Errata, ibid., Vol. 42, No. 2 (2005), p. 213; alternative link.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A008621. - Johannes W. Meijer, Jun 07 2011

Cf. A047470, A047522, A053381, A195679.

A047530 := proc(n): ceil(n/4) + 2*ceil((n-1)/4) + 4*ceil((n-2)/4) + ceil((n-3)/4) end: seq(A047530(n), n=0..47); # Johannes W. Meijer, Jun 07 2011
A047530:=n->(8*n-9+I^(2*n)+(2+I)*I^(-n)+(2-I)*I^n)/4: seq(A047530(n), n=1..100); # Wesley Ivan Hurt, May 21 2016

Table[(8n-9+I^(2n)+(2+I)*I^(-n)+(2-I)*I^n)/4, {n, 80}] (* Wesley Ivan Hurt, May 21 2016 *)

a(n)=n>>2<<3+[-1,0,1,3][n%4+1] \\ Charles R Greathouse IV, Jun 09 2011

From Johannes W. Meijer, Jun 07 2011: (Start)
a(n) = ceiling(n/4) + 2*ceiling((n-1)/4) + 4*ceiling((n-2)/4) + ceiling((n-3)/4).
a(n+1) = A053381(2^p). (End)
G.f.: x^2*(1+2*x+4*x^2+x^3) / ((1+x)*(x^2+1)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, May 21 2016: (Start)
a(n) = a(n-1)+a(n-4)-a(n-5) for n>5.
a(n) = (8n-9+i^(2n)+(2+i)*i^(-n)+(2-i)*i^n)/4, where i=sqrt(-1).
a(2n) = A047522(n), a(2n-1) = A047470(n). (End)
E.g.f.: (2 + sin(x) + 2*cos(x) + (4*x - 5)*sinh(x) + 4*(x - 1)*cosh(x))/2. - Ilya Gutkovskiy, May 21 2016
Sum_{n>=2} (-1)^n/a(n) = (8-3*sqrt(2))*log(2)/16 + 3*sqrt(2)*log(2+sqrt(2))/8 - (sqrt(2)-1)*Pi/16. - Amiram Eldar, Dec 20 2021

More terms from Wesley Ivan Hurt, May 21 2016

A047527 Numbers that are congruent to {0, 1, 2, 7} mod 8.

Original entry on oeis.org

0, 1, 2, 7, 8, 9, 10, 15, 16, 17, 18, 23, 24, 25, 26, 31, 32, 33, 34, 39, 40, 41, 42, 47, 48, 49, 50, 55, 56, 57, 58, 63, 64, 65, 66, 71, 72, 73, 74, 79, 80, 81, 82, 87, 88, 89, 90, 95, 96, 97, 98, 103, 104, 105, 106, 111, 112, 113, 114, 119, 120

Offset: 1

N. J. A. Sloane

Complement of numbers that are congruent to {3, 4, 5, 6} mod 8 (A047425). - Jaroslav Krizek, Dec 19 2009

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

Cf. A103127, A047425, A047467, A047522.

[n : n in [0..100] | n mod 8 in [0, 1, 2, 7]]; // Wesley Ivan Hurt, May 21 2016

seq(3*n-4*floor((n-2)/4)-6+(-1)^n, n=1..61); # Gary Detlefs, Mar 27 2010

Select[Range[0,200], MemberQ[{0,1,2,7}, Mod[#,8]]&] (* or *) LinearRecurrence[{1,0,0,1,-1}, {0,1,2,7,8}, 200] (* Harvey P. Dale, Sep 05 2014 *)

a(n) = 3*n-4*floor((n-2)/4)-6+(-1)^n. - Gary Detlefs, Mar 27 2010
G.f.: x^2*(1+x+5*x^2+x^3) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - Harvey P. Dale, Sep 05 2014
From Wesley Ivan Hurt, May 21 2016: (Start)
a(n) = (4n-5+i^(2n)+(1+i)*i^(-n)+(1-i)*i^n)/2 where i = sqrt(-1).
a(2n) = A047522(n), a(2n-1) = A047467(n). (End)
Sum_{n>=2} (-1)^n/a(n) = (5-sqrt(2))*log(2)/8 + sqrt(2)*log(2+sqrt(2))/4 - Pi/16. - Amiram Eldar, Dec 20 2021

A047551 Numbers that are congruent to {0, 1, 6, 7} mod 8.

Original entry on oeis.org

0, 1, 6, 7, 8, 9, 14, 15, 16, 17, 22, 23, 24, 25, 30, 31, 32, 33, 38, 39, 40, 41, 46, 47, 48, 49, 54, 55, 56, 57, 62, 63, 64, 65, 70, 71, 72, 73, 78, 79, 80, 81, 86, 87, 88, 89, 94, 95, 96, 97, 102, 103, 104, 105, 110, 111, 112, 113, 118, 119, 120, 121, 126

Offset: 1

N. J. A. Sloane

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

Cf. A010873, A030308, A047451, A047522.

[n : n in [0..150] | n mod 8 in [0, 1, 6, 7]]; // Wesley Ivan Hurt, May 29 2016

A047551:=n->(4*n-3-I^(2*n)+(1-I)*I^(-n)+(1+I)*I^n)/2: seq(A047551(n), n=1..100); # Wesley Ivan Hurt, May 29 2016

Table[(4n-3-I^(2n)+(1-I)*I^(-n)+(1+I)*I^n)/2, {n, 80}] (* Wesley Ivan Hurt, May 29 2016 *)

a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=6 and b(k)=2^(k+1) for k>1. - Philippe Deléham, Oct 19 2011
a(n) = 2n - A010873(n+1). - Wesley Ivan Hurt, Jul 07 2013
G.f.: x^2*(1+5*x+x^2+x^3) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Jul 14 2013
From Wesley Ivan Hurt, May 29 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (4*n-3-i^(2*n)+(1-i)*i^(-n)+(1+i)*i^n)/2 where i=sqrt(-1).
a(2k) = A047522(k), a(2k-1) = A047451(k). (End)
E.g.f.: 1 - sin(x) + cos(x) + (2*x - 1)*sinh(x) + 2*(x - 1)*cosh(x). - Ilya Gutkovskiy, May 29 2016
Sum_{n>=2} (-1)^n/a(n) = Pi/16 + (5-sqrt(2))*log(2)/8 + sqrt(2)*log(2+sqrt(2))/4. - Amiram Eldar, Dec 20 2021

A125725 Numbers whose base-7 representation is 222....2.

Original entry on oeis.org

0, 2, 16, 114, 800, 5602, 39216, 274514, 1921600, 13451202, 94158416, 659108914, 4613762400, 32296336802, 226074357616, 1582520503314, 11077643523200, 77543504662402, 542804532636816, 3799631728457714, 26597422099204000

Offset: 1

Zerinvary Lajos, Feb 02 2007

			base 7.......decimal
0..................0
2..................2
22................16
222..............114
2222.............800
22222...........5602
222222.........39216
2222222.......274514
22222222.....1921600
222222222...13451202
etc...........etc.

Davis Smith, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8,-7).

Cf. A007310, A023000, A047522, A165759.

Cf. also A002276, A005610, A020988, A024023, A125831, A125835, A125857 for related or similarly constructed sequences.

List([1..25], n-> (7^(n-1) -1)/3); # G. C. Greubel, May 23 2019

[0] cat [n:n in [1..15000000]| Set(Intseq(n,7)) subset [2]]; // Marius A. Burtea, May 06 2019

[(7^(n-1)-1)/3: n in [1..25]]; // Marius A. Burtea, May 06 2019

Maple
```
seq(2*(7^n-1)/6, n=0..25);
```

FromDigits[#,7]&/@Table[PadLeft[{2},n,2],{n,0,25}]  (* Harvey P. Dale, Apr 13 2011 *)
(7^(Range[25]-1) - 1)/3 (* G. C. Greubel, May 23 2019 *)

vector(25, n, (7^(n-1)-1)/3) \\ Davis Smith, Apr 04 2019

[(7^(n-1) -1)/3 for n in (1..25)] # G. C. Greubel, May 23 2019

a(n) = (7^(n-1) - 1)/3 = 2*A023000(n-1).
a(n) = 7*a(n-1) + 2, with a(1)=0. - Vincenzo Librandi, Sep 30 2010
G.f.: 2*x^2 / ( (1-x)*(1-7*x) ). - R. J. Mathar, Sep 30 2013
From Davis Smith, Apr 04 2019: (Start)
A007310(a(n) + 1) = 7^(n - 1).
A047522(a(n + 1)) = -1*A165759(n). (End)
E.g.f.: (exp(7*x) - 7*exp(x) + 6)/21. - Stefano Spezia, Jan 12 2025

Offset corrected by N. J. A. Sloane, Oct 02 2010

A047479 Numbers that are congruent to {0, 1, 5, 7} mod 8.

Original entry on oeis.org

0, 1, 5, 7, 8, 9, 13, 15, 16, 17, 21, 23, 24, 25, 29, 31, 32, 33, 37, 39, 40, 41, 45, 47, 48, 49, 53, 55, 56, 57, 61, 63, 64, 65, 69, 71, 72, 73, 77, 79, 80, 81, 85, 87, 88, 89, 93, 95, 96, 97, 101, 103, 104, 105, 109, 111, 112, 113, 117, 119, 120, 121, 125

Offset: 1

N. J. A. Sloane

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

Cf. A047522, A047615.

I:=[0, 1, 5, 7, 8]; [n le 5 select I[n] else Self(n-1)+Self(n-4)-Self(n-5): n in [1..70]]; // Vincenzo Librandi, May 16 2012

A047479:=n->(-7-I^(2*n)+(2-I)*(-I)^n+(2+I)*I^n+8*n)/4: seq(A047479(n), n=1..100); # Wesley Ivan Hurt, Jun 01 2016

Select[Range[0,300], MemberQ[{0,1,5,7}, Mod[#,8]]&] (* Vincenzo Librandi, May 16 2012 *)

my(x='x+O('x^100)); concat(0, Vec(x^2*(1+4*x+2*x^2+x^3)/((1-x)^2*(1+x)*(1+x^2)))) \\ Altug Alkan, Dec 24 2015

From Colin Barker, May 14 2012: (Start)
a(n) = (-7-(-1)^n+(2-i)*(-i)^n+(2+i)*i^n+8*n)/4 where i=sqrt(-1).
G.f.: x^2*(1+4*x+2*x^2+x^3)/((1-x)^2*(1+x)*(1+x^2)). (End)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - Vincenzo Librandi, May 16 2012
a(2k) = A047522(k), a(2k-1) = A047615(k). - Wesley Ivan Hurt, Jun 01 2016
E.g.f.: (2 - sin(x) + 2*cos(x) + (4*x - 3)*sinh(x) + 4*(x - 1)*cosh(x))/2. - Ilya Gutkovskiy, Jun 02 2016
Sum_{n>=2} (-1)^n/a(n) = (sqrt(2)-1)*Pi/16 + (8-3*sqrt(2))*log(2)/16 + 3*sqrt(2)*log(2+sqrt(2))/8. - Amiram Eldar, Dec 20 2021

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