A017605 -id:A017605 - cp's OEIS frontend

A016921 a(n) = 6*n + 1.

1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127, 133, 139, 145, 151, 157, 163, 169, 175, 181, 187, 193, 199, 205, 211, 217, 223, 229, 235, 241, 247, 253, 259, 265, 271, 277, 283, 289, 295, 301, 307, 313, 319, 325, 331

Offset: 0

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 22 ).
Also solutions to 2^x + 3^x == 5 (mod 7). - Cino Hilliard, May 10 2003
Except for 1, exponents n > 1 such that x^n - x^2 - 1 is reducible. - N. J. A. Sloane, Jul 19 2005
Let M(n) be the n X n matrix m(i,j) = min(i,j); then the trace of M(n)^(-2) is a(n-1) = 6*n - 5. - Benoit Cloitre, Feb 09 2006
If Y is a 3-subset of an (2n+1)-set X then, for n >= 3, a(n-1) is the number of 3-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
All composite terms belong to A269345 as shown in there. - Waldemar Puszkarz, Apr 13 2016
First differences of the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 773", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
For b(n) = A103221(n) one has b(a(n)-1) = b(a(n)+1) = b(a(n)+2) = b(a(n)+3) = b(a(n)+4) = n+1 but b(a(n)) = n. So-called "dips" in A103221. See the Avner and Gross remark on p. 178. - Wolfdieter Lang, Sep 16 2016
A (n+1,n) pebbling move involves removing n + 1 pebbles from a vertex in a simple graph and placing n pebbles on an adjacent vertex. A two-player impartial (n+1,n) pebbling game involves two players alternating (n+1,n) pebbling moves. The first player unable to make a move loses. The sequence a(n) is also the minimum number of pebbles such that any assignment of those pebbles on a complete graph with 3 vertices is a next-player winning game in the two player impartial (k+1,k) pebbling game. These games are represented by A347637(3,n). - Joe Miller, Oct 18 2021
Interleaving of A017533 and A017605. - Leo Tavares, Nov 16 2021

			From _Ilya Gutkovskiy_, Apr 15 2016: (Start)
Illustration of initial terms:
                      o
                    o o o
              o     o o o
            o o o   o o o
      o     o o o   o o o
    o o o   o o o   o o o
o   o o o   o o o   o o o
n=0  n=1     n=2     n=3
(End)

Avner Ash and Robert Gross, Summing it up, Princeton University Press, 2016, p. 178.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, and Tony W. H. Wong, Generalized Impartial Two-player Pebbling Games on K_3 and C_4, J. Int. Seq. (2024) Vol. 27, Issue 5, Art. No. 24.5.8. See p. 3.
Tanya Khovanova, Recursive Sequences.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N)).
William A. Stein, The modular forms database.
Leo Tavares, Illustration: Hexagonal Lines.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093563 ((6, 1) Pascal, column m=1).
Cf. A000567 (partial sums), A002476 (primes), A005408, A008588, A016813, A016933, A016945, A016957, A017281, A017533, A128470, A158057, A161700, A161705, A161709, A161714.
a(n) = A007310(2*(n+1)); complement of A016969 with respect to A007310.
Cf. A287326 (second column).
Cf. A000217, A003215.
Cf. A001477, A017605.

List([0..60], n-> 6*n+1); # G. C. Greubel, Sep 18 2019

a016921 = (+ 1) . (* 6)
a016921_list = [1, 7 ..]  -- Reinhard Zumkeller, Jan 15 2013

[6*n+1: n in [0..60]]; // G. C. Greubel, Sep 18 2019

a[1]:=1:for n from 2 to 100 do a[n]:=a[n-1]+6 od: seq(a[n], n=1..56); # Zerinvary Lajos, Mar 16 2008

Range[1, 500, 6] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)

a(n)=6*n+1 \\ Charles R Greathouse IV, Mar 22 2016

for n in range(0,10**5):print(6*n+1) # Soumil Mandal, Apr 14 2016

a(n) = 6*n + 1, n >= 0 (see the name).
G.f.: (1+5*x)/(1-x)^2.
A008615(a(n)) = n. - Reinhard Zumkeller, Feb 27 2008
A157176(a(n)) = A013730(n). - Reinhard Zumkeller, Feb 24 2009
a(n) = 4*(3*n-1) - a(n-1) (with a(0)=1). - Vincenzo Librandi, Nov 20 2010
E.g.f.: (1 + 6*x)*exp(x). - G. C. Greubel, Sep 18 2019
a(n) = A003215(n) - 6*A000217(n-1). See Hexagonal Lines illustration. - Leo Tavares, Sep 10 2021
From Leo Tavares, Oct 27 2021: (Start)
a(n) = 6*A001477(n-1) + 7
a(n) = A016813(n) + 2*A001477(n)
a(n) = A017605(n-1) + A008588(n-1)
a(n) = A016933(n) - 1
a(n) = A008588(n) + 1. (End)
Sum_{n>=0} (-1)^n/a(n) = Pi/6 + sqrt(3)*arccoth(sqrt(3))/3. - Amiram Eldar, Dec 10 2021

A165460 The height at the 1/3 point of Jacobi-bridge, computed for 12n+7. a(n) = Sum_{i=0..(4n+2)} J(i,12n+7), where J(i,m) is the Jacobi symbol.

Original entry on oeis.org

2, 2, 6, 2, 8, 2, 10, 4, 10, 4, 10, 6, 14, 2, 4, 4, 18, 6, 14, 4, 12, 8, 22, 6, 16, 6, 20, 6, 2, 8, 18, 6, 28, 4, 20, 4, 30, 12, 14, 0, 14, 6, 28, 10, 28, 6, 32, 10, 16, 8, 26, 10, 26, 6, 24, 8, 36, 10, 28, 8, 26, 10, 30, 8, 0, 10, 32, 14, 18, 12, 0, 14, 44, 6, 32, 6, 38, 0, 32, 8, 22

Offset: 0

Antti Karttunen, Oct 06 2009

nonn

Conjecture: a(2n) = 2*A165605(2n) and a(2n+1) = (2/3)*A165605(2n+1). - Antti Karttunen, Oct 05 2009. (If true, then implies also the truth of conjecture in A165462.)

A. Karttunen, Table of n, a(n) for n = 0..21845

Cf. A165461, A165462, A165463, A165605.

Table[Sum[JacobiSymbol[i, 12n + 7], {i, 0, 4n + 2}], {n, 0, 100}] (* Indranil Ghosh, May 13 2017 *)

a(n) = sum(i=0, 4*n + 2, kronecker(i, 12*n + 7)); \\ Indranil Ghosh, May 13 2017

from sympy import jacobi_symbol as J
def a(n): return sum([J(i, 12*n + 7) for i in range(4*n + 3)]) # Indranil Ghosh, May 13 2017

A369461 Number of representations of 12n-5 as a sum (pq + pr + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 1, 1, 1, 0, 0, 2, 0, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 3, 0, 0, 0, 2, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 3, 0, 0, 1, 1, 2, 0, 0, 2, 1, 1, 0, 1, 0, 0, 1, 2, 0, 0, 1, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 1, 2, 1, 0, 1, 1, 0, 0, 0, 4, 0, 1, 1, 0, 0, 0, 1, 2, 0, 1, 0, 0

Offset: 1

Antti Karttunen, Jan 23 2024

nonn

The sequence seems to contain an infinite number of zeros. See A369451 for the cumulative sum, and comments there.

Question: Are there any sections of this sequence, with parameters k >= 2, 0 <= i < k, for which a((k*n)-i) = 0 for all n >= 1? - Antti Karttunen, Nov 20 2024

Antti Karttunen, Table of n, a(n) for n = 1..100000

Trisection of A369055.

Cf. A017605, A369054, A369451 (partial sums), A369460, A369462.

A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));
A369461(n) = A369054((12*n)-5);

a(n) = A369054(A017605(n-1)) = A369054((12*n)-5).

a(n) = A369055((3*n)-1).

A056530 Sequence remaining after third round of Flavius Josephus sieve; remove every fourth term of A047241.

Original entry on oeis.org

1, 3, 7, 13, 15, 19, 25, 27, 31, 37, 39, 43, 49, 51, 55, 61, 63, 67, 73, 75, 79, 85, 87, 91, 97, 99, 103, 109, 111, 115, 121, 123, 127, 133, 135, 139, 145, 147, 151, 157, 159, 163, 169, 171, 175, 181, 183, 187, 193, 195, 199, 205, 207, 211, 217, 219, 223, 229, 231

Offset: 1

Henry Bottomley, Jun 19 2000

Numbers {1, 3, 7} mod 12: A017533, A017557, A017605 interleaved.

Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Index entries for sequences related to the Josephus Problem

We have A000027 after 0 rounds of sieving, A005408 after 1 round of sieving, A047241 after 2 rounds, A056530 after 3 rounds, A056531 after 4 rounds, A000960 after all rounds. After n rounds the remaining sequence comprises A002944(n) numbers mod A003418(n+1), i.e. 1/(n+1) of them.

LinearRecurrence[{1,0,1,-1},{1,3,7,13},60] (* Harvey P. Dale, Oct 19 2022 *)

From Chai Wah Wu, Jul 24 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n > 4.
G.f.: x*(5*x^3 + 4*x^2 + 2*x + 1)/(x^4 - x^3 - x + 1). (End)
a(n) = 4*n - (13 + 2*A131713(n))/3. - R. J. Mathar, Jun 22 2020

A051895 Partial sums of second pentagonal numbers with even index (A049453).

Original entry on oeis.org

0, 7, 33, 90, 190, 345, 567, 868, 1260, 1755, 2365, 3102, 3978, 5005, 6195, 7560, 9112, 10863, 12825, 15010, 17430, 20097, 23023, 26220, 29700, 33475, 37557, 41958, 46690, 51765, 57195, 62992, 69168, 75735, 82705, 90090, 97902, 106153, 114855, 124020, 133660

Offset: 0

Barry E. Williams, Dec 17 1999

For A049453(n+1), the corresponding formula would be a(n)=(n+1)*(6*n+7) and its partial sums would be given by a(n)=(n+1)*(n+2)*(4*n+7)/2.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002492, A016061, A017605, A049453.

I:=[0, 7, 33, 90]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Apr 27 2012

Table[(n(4n-1)(n-1))/2,{n,40}]  (* Harvey P. Dale, Mar 11 2011 *)
CoefficientList[Series[x*(7+5*x)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Apr 27 2012 *)

a(n) = n*(n+1)*(4*n+3)/2; \\ Altug Alkan, Apr 20 2018

a(n) = n*(n+1)*(4*n+3)/2.
G.f.: x*(7+5*x)/(1-x)^4. - Colin Barker, Jan 12 2012
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). - Vincenzo Librandi, Apr 27 2012
a(n) = A002492(n) + A016061(n). - J. M. Bergot, Apr 20 2018

A142241 a(n) = 24*n + 14.

Original entry on oeis.org

14, 38, 62, 86, 110, 134, 158, 182, 206, 230, 254, 278, 302, 326, 350, 374, 398, 422, 446, 470, 494, 518, 542, 566, 590, 614, 638, 662, 686, 710, 734, 758, 782, 806, 830, 854, 878, 902, 926, 950, 974, 998, 1022, 1046, 1070, 1094, 1118, 1142, 1166, 1190, 1214, 1238

Offset: 0

Paul Curtz, Sep 17 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A017605.

[24*n + 14: n in [0..50]]; // Vincenzo Librandi, Aug 08 2011

Range[14, 2000, 24] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011 *)

a(n)=24*n+14 \\ Charles R Greathouse IV, Jun 14 2011

From R. J. Mathar and Omar E. Pol, Sep 19 2008: (Start)
a(n) = 2*A017605(n).
G.f.: 2*(7+5*x)/(1-x)^2. (End)
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: 2*exp(x)*(7 + 12*x).
a(n) = 2*a(n-1) - a(n-2). (End)

Edited by N. J. A. Sloane, Sep 19 2008

A166050 a(n) = Sum_{i=0..(2n+1)} J(i,12n+7), where J(i,k) is the Jacobi symbol.

Original entry on oeis.org

1, -1, 3, -1, 4, -1, 5, -2, 5, -2, 5, -3, 7, -1, 2, -2, 9, -3, 7, -2, 6, -4, 11, -3, 8, -3, 10, -3, 1, -4, 9, -3, 14, -2, 10, -2, 15, -6, 7, 0, 7, -3, 14, -5, 14, -3, 16, -5, 8, -4, 13, -5, 13, -3, 12, -4, 18, -5, 14, -4, 13, -5, 15, -4, 0, -5, 16, -7, 9, -6, 0, -7, 22, -3, 16, -3

Offset: 0

Antti Karttunen, Oct 13 2009. Erroneous name corrected Oct 20 2009

sign

The height at the 1/6 point of "Jacobi-bridge/path", computed for each odd integer of the form 12n+7.

A. Karttunen, Table of n, a(n) for n = 0..131071
Wikipedia, Jacobi symbol

Bisections: A166268, A166269 (see conjectures there). Cf. A017605. Scheme-code for jacobi-symbol is given at A165601.

A378704 Array read by ascending antidiagonals: A(n, k) is the total area of n-Fibonacci polyominoes with k columns, where k > 0.

Original entry on oeis.org

2, 3, 7, 4, 11, 16, 5, 15, 31, 35, 6, 19, 43, 73, 70, 7, 23, 55, 111, 168, 136, 8, 27, 67, 143, 261, 370, 256, 9, 31, 79, 175, 351, 602, 790, 473, 10, 35, 91, 207, 431, 816, 1350, 1658, 860, 11, 39, 103, 239, 511, 1023, 1865, 2966, 3425, 1545, 12, 43, 115, 271, 591, 1215, 2346, 4178, 6414, 6989, 2748

Offset: 2

Stefano Spezia, Dec 04 2024

			The array begins as:
  2,  7, 16,  35,  70,  136,  256, ...
  3, 11, 31,  73, 168,  370,  790, ...
  4, 15, 43, 111, 261,  602, 1350, ...
  5, 19, 55, 143, 351,  816, 1865, ...
  6, 23, 67, 175, 431, 1023, 2346, ...
  7, 27, 79, 207, 511, 1215, 2815, ...
  ...

Juan F. Pulido, José L. Ramírez, and Andrés R. Vindas-Meléndez, Generating Trees and Fibonacci Polyominoes, arXiv:2411.17812 [math.CO], 2024. See pages 8-9.

Cf. A004767 (k=2), A017605 (k=3), A356888, A378706, A378707, A378716.

A[n_,k_]:=SeriesCoefficient[y(n^2(1-y)^2y^n+2y(1-y^n)-n(1-y)(2-y^n+y^(n+1)))/(2(-1+y)(1-2y+y^(n+1))^2),{y,0,k}]; Table[A[n-k+1,k],{n,2,12},{k,n-1}]//Flatten

A(n, k) = [y^k] y*(n^2*(1 - y)^2*y^n + 2*y*(1 - y^n) - n(1 - y)*(2- y^n + y^(n+1)))/(2*(-1 + y)*(1 - 2*y + y^(n+1))^2).

A(n, n-1) = A356888(n) - 1.

A272975 Numbers that are congruent to {0,7} mod 12.

Original entry on oeis.org

0, 7, 12, 19, 24, 31, 36, 43, 48, 55, 60, 67, 72, 79, 84, 91, 96, 103, 108, 115, 120, 127, 132, 139, 144, 151, 156, 163, 168, 175, 180, 187, 192, 199, 204, 211, 216, 223, 228, 235, 240, 247, 252, 259, 264, 271, 276, 283, 288, 295, 300, 307, 312, 319, 324

Offset: 1

Wesley Ivan Hurt, May 30 2016

Numbers that are not congruent to {1, 2, 3, 4, 5, 6, 8, 9, 10, 11} mod 12.

Bisection of A083032.

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

Cf. A008594, A017605, A083032, A049453, A049598.

[n : n in [0..400] | n mod 12 in [0, 7]];

A272975:=n->(12*n-11+(-1)^n)/2: seq(A272975(n), n=1..100);

Mathematica
```
Table[(12n - 11 + (-1)^n)/2, {n, 80}]
```

concat(0, Vec(x^2*(7+5*x)/((x-1)^2*(x+1)) + O(x^99))) \\ Altug Alkan, May 31 2016

G.f.: x^2*(7+5*x) / ((x-1)^2*(x+1)).
a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
a(n) = (12*n - 11 + (-1)^n)/2.
a(2k) = A017605(k-1) k>0, a(2k-1) = A008594(k-1) k>0, a(2k)-a(2k-1) = 7.
a(n)-a(-n) = A008594(n) for n>0.
Sum_{i=1..n} a(2*i) = A049453(n) for n>0.
Sum_{i=1..n} a(2*i-1) = A049598(n-1) for n>0.
E.g.f.: 5 + ((12*x - 11)*exp(x) + exp(-x))/2. - David Lovler, Sep 04 2022
Sum_{n>=2} (-1)^n/a(n) = log(2)/4 + log(3)/8 - ((sqrt(3)-1)*Pi + 2*(sqrt(3)+3)*log(sqrt(3)+2))/(24*(sqrt(3)+1)). - Amiram Eldar, Sep 17 2023

A110599 Balanced numbers n such that n mod 12 = 7.

Original entry on oeis.org

24871, 58435, 140335, 1529983, 2086903, 3722875, 3830827, 8697535, 13932919, 16408315, 21578755, 27882595, 76319155, 126245119, 183531439, 192871987, 198394675, 207619555, 229523371, 337800463, 361504507, 416690995, 440127655, 535044055, 693298315, 729802255

Offset: 1

Walter Kehowski, Sep 13 2005

nonn

For the first 26 terms, the quotient (sigma(n)/phi(n)) is 2 or 3.

Amiram Eldar, Table of n, a(n) for n = 1..7901 (terms below 6.5*10^14, calculated using data from Jud McCranie)

Intersection of A017605 and A020492.

Cf. A000010, A000203, A062699.

with(numtheory); BNM7:=[]: for z from 1 to 1 do for m from 1 to 1000000 do n:=12*m+7; if sigma(n) mod phi(n) = 0 then BNM7:=[op(BNM7),n] fi; od; od; BNM7;

Select[Range[10^7], Mod[#, 12] == 7 && Divisible[DivisorSigma[1, #], EulerPhi[#]] &] (* Amiram Eldar, Dec 04 2019 *)

Duplicate terms removed and a(8)-a(26) from Donovan Johnson, Aug 30 2012

cp's OEIS Frontend

A016921 a(n) = 6*n + 1.

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A165460 The height at the 1/3 point of Jacobi-bridge, computed for 12n+7. a(n) = Sum_{i=0..(4n+2)} J(i,12n+7), where J(i,m) is the Jacobi symbol.

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A369461 Number of representations of 12n-5 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

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A056530 Sequence remaining after third round of Flavius Josephus sieve; remove every fourth term of A047241.

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A051895 Partial sums of second pentagonal numbers with even index (A049453).

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A142241 a(n) = 24*n + 14.

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A166050 a(n) = Sum_{i=0..(2n+1)} J(i,12n+7), where J(i,k) is the Jacobi symbol.

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A369461 Number of representations of 12n-5 as a sum (pq + pr + q*r) with three odd primes p <= q <= r.