A091998 -id:A091998 - cp's OEIS frontend

A007310 Numbers congruent to 1 or 5 mod 6.

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175

Offset: 1

C. Christofferson (Magpie56(AT)aol.com)

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with 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 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1.  Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...

K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016-2018.
B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers.
L. B. W. Jolley, Summation of Series, Dover, 1961
Cedric A. B. Smith, Prime factors and recurring duodecimals, Math. Gaz. 59 (408) (1975) 106-109.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N).
Eric Weisstein's World of Mathematics, Rough Number.
Eric Weisstein's World of Mathematics, Pi Formulas. [_Jaume Oliver Lafont_, Oct 23 2009]
Index entries for sequences related to smooth numbers [_Michael B. Porter_, Oct 09 2009]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Filtered([1..150],n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018

a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016

seq(seq(6*i+j,j=[1,5]),i=0..100); # Robert Israel, Sep 08 2014

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *)
a[n_] := (6 n + (-1)^n - 3)/2; a[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)
Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)
Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)

isA007310(n) = gcd(n,6)==1 \\ Michael B. Porter, Oct 09 2009

A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014

\\ given an element from the sequence, find the next term in the sequence.
nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016

def A007310(n): return (n+(n>>1)<<1)-1 # Chai Wah Wu, Oct 10 2023

[i for i in range(150) if gcd(6,i) == 1] # Zerinvary Lajos, Apr 21 2009

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)

A047209 Numbers that are congruent to {1, 4} mod 5.

Original entry on oeis.org

1, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 109, 111, 114, 116, 119, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154

Offset: 1

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 72 ).
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 our case, a(n)^2 - 1 == 0 (mod 5). - Bruno Berselli, Nov 17 2010
The sum of the alternating series (-1)^(n+1)/a(n) from n=1 to infinity is (Pi/5)*cot(Pi/5), that is (1/5)*sqrt(1 + 2/sqrt(5))*Pi. - Jean-François Alcover, May 03 2013
These numbers appear in the product of a Rogers-Ramanujan identity. See A003114 also for references. - Wolfdieter Lang, Oct 29 2016
Let m be a product of any number of terms of this sequence. Then m - 1 or m + 1 is divisible by 5. Closed under multiplication. - David A. Corneth, May 11 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
William A. Stein, The modular forms database.
Eric Weisstein's World of Mathematics, Determined by Spectrum.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000566, A036666, A001622, A003114, A203776, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887, A352324.
Cf. A005408 (n=1 or 3 mod 4), A007310 (n=1 or 5 mod 6).
Cf. A045468 (primes), A032527 (partial sums).

a047209 = (flip div 2) . (subtract 2) . (* 5)
a047209_list = 1 : 4 : (map (+ 5) a047209_list)
-- Reinhard Zumkeller, Jul 19 2013, Jan 05 2011

seq(floor(5*k/2)-1, k=1..100); # Wesley Ivan Hurt, Sep 27 2013

Select[Range[0, 200], MemberQ[{1, 4}, Mod[#, 5]] &] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
LinearRecurrence[{1,1,-1},{1,4,6},70] (* Harvey P. Dale, Jul 19 2024 *)

a(n)=(10*n+(-1)^n-5)/4 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: (1+3x+x^2)/((1-x)(1-x^2)).
a(n) = floor((5n-2)/2). [corrected by Reinhard Zumkeller, Jul 19 2013]
a(1) = 1, a(n) = 5(n-1) - a(n-1). - Benoit Cloitre, Apr 12 2003
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = (10*n + (-1)^n - 5)/4.
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 5 for n > 2.
a(n) = 5*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
a(n)^2 = 5*A036666(n) + 1 (cf. also Comments). (End)
a(n) = 5*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
E.g.f.: 1 + ((10*x - 5)*exp(x) + exp(-x))/4. - David Lovler, Aug 23 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = phi (A001622).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/5) * cosec(Pi/5) (A352324). (End)

Edited by Michael Somos, Sep 22 2002

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)

A056020 Numbers that are congruent to +-1 mod 9.

Original entry on oeis.org

1, 8, 10, 17, 19, 26, 28, 35, 37, 44, 46, 53, 55, 62, 64, 71, 73, 80, 82, 89, 91, 98, 100, 107, 109, 116, 118, 125, 127, 134, 136, 143, 145, 152, 154, 161, 163, 170, 172, 179, 181, 188, 190, 197, 199, 206, 208, 215, 217, 224, 226, 233, 235, 242, 244, 251, 253

Offset: 1

Robert G. Wilson v, Jun 08 2000

Or, numbers k such that k^2 == 1 (mod 9).

Or, numbers k such that the iterative cycle j -> sum of digits of j^2 when started at k contains a 1. E.g., 8 -> 6+4 = 10 -> 1+0+0 = 1 and 17 -> 2+8+9 = 19 -> 3+6+1 = 10 -> 1+0+0 = 1. - Asher Auel, May 17 2001

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

Cf. A007953, A047522 (n=1 or 7 mod 8), A090771 (n=1 or 9 mod 10).
Cf. A129805 (primes), A195042 (partial sums).
Cf. A005408, A019676, A019968, A047209, A007310, A047336, A175885, A091998, A175886, A113801, A175887, A332437.
Cf. A381319 (general case mod n^2).

a056020 n = a056020_list !! (n-1)
a05602_list = 1 : 8 : map (+ 9) a056020_list
-- Reinhard Zumkeller, Jan 07 2012

Select[ Range[ 300 ], PowerMod[ #, 2, 3^2 ]==1& ]
(* or *)
LinearRecurrence[{1, 1, -1}, {1, 8, 10}, 67] (* Mike Sheppard, Feb 18 2025 *)

a(n)=9*(n>>1)+if(n%2,1,-1) \\ Charles R Greathouse IV, Jun 29 2011

for(n=1,40,print1(9*n-8,", ",9*n-1,", ")) \\ Charles R Greathouse IV, Jun 29 2011

a(1) = 1; a(n) = 9(n-1) - a(n-1). - Rolf Pleisch, Jan 31 2008 [Offset corrected by Jon E. Schoenfield, Dec 22 2008]
From R. J. Mathar, Feb 10 2008: (Start)
O.g.f.: 1 + 5/(4(x+1)) + 27/(4(-1+x)) + 9/(2(-1+x)^2).
a(n+1) - a(n) = A010697(n). (End)
a(n) = (9*A132355(n) + 1)^(1/2). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = a(n-2) + 9, for n > 2.
a(n) = 9*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), n > 1. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/9)*cot(Pi/9) = A019676 * A019968. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((18*x - 9)*exp(x) + 5*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/9) (A332437).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/9)*cosec(Pi/9). (End)
From Mike Sheppard, Feb 18 2025: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) ~ (3^2/2)*n. (End)

A090771 Numbers that are congruent to {1, 9} mod 10.

Original entry on oeis.org

1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69, 71, 79, 81, 89, 91, 99, 101, 109, 111, 119, 121, 129, 131, 139, 141, 149, 151, 159, 161, 169, 171, 179, 181, 189, 191, 199, 201, 209, 211, 219, 221, 229, 231, 239, 241, 249, 251, 259, 261, 269, 271, 279, 281

Offset: 1

Giovanni Teofilatto, Feb 07 2004

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 10). - Bruno Berselli, Nov 17 2010

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

Cf. A000217, A000796, A090298, A005408, A047209, A007310, A019970, A047336, A047522, A091998, A094888, A113801, A175886, A175887, A188593.
Cf. A056020 (n = 1 or 8 mod 9), A175885 (n = 1 or 10 mod 11).
Cf. A045468 (primes), A195142 (partial sums).

a090771 n = a090771_list !! (n-1)
a090771_list = 1 : 9 : map (+ 10) a090771_list
-- Reinhard Zumkeller, Jan 07 2012

Flatten[Table[10n - {9, 1}, {n, 30}]] (* Alonso del Arte, Sep 02 2014 *)
LinearRecurrence[{1,1,-1},{1,9,11},60] (* Harvey P. Dale, Jul 05 2020 *)

a(n)=n\2*10-(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

a(n) = sqrt(40*A057569(n) + 1). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Sep 16 2010 - Nov 17 2010: (Start)
  G.f.: x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
  a(n) = (10*n + 3*(-1)^n - 5)/2.
  a(n) = -a(-n + 1) = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 10.
  a(n) = 10*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 10*n - a(n-1) - 10 (with a(1) = 1). - Vincenzo Librandi, Nov 16 2010
a(n) = sqrt(10*A132356(n-1) + 1). - Ivan N. Ianakiev, Nov 09 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/10)*cot(Pi/10) = A000796 * A019970 / 10 = sqrt(5 + 2*sqrt(5))*Pi/10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((10*x - 5)*exp(x) + 3*exp(-x))/2. - David Lovler, Sep 03 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(phi+2) (A188593).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi*phi/5 = A094888/10. (End)

Edited and extended by Ray Chandler, Feb 10 2004

A047336 Numbers that are congruent to {1, 6} mod 7.

Original entry on oeis.org

1, 6, 8, 13, 15, 20, 22, 27, 29, 34, 36, 41, 43, 48, 50, 55, 57, 62, 64, 69, 71, 76, 78, 83, 85, 90, 92, 97, 99, 104, 106, 111, 113, 118, 120, 125, 127, 132, 134, 139, 141, 146, 148, 153, 155, 160, 162, 167, 169, 174, 176, 181, 183, 188, 190, 195, 197, 202, 204, 209

Offset: 1

N. J. A. Sloane

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 7). - Bruno Berselli, Nov 17 2010

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

Cf. A007310, A019674, A047522, A045472 (primes), A195041 (partial sums), A005408, A047209, A056020, A090771, A091998, A113801, A160389, A175885, A175886, A175887, A178818, A371858.

a047336 n = a047336_list !! (n-1)
a047336_list = 1 : 6 : map (+ 7) a047336_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..210]| n mod 7 in {1,6}]; // Bruno Berselli, Feb 22 2011

Rest[Flatten[Table[{7i-1,7i+1},{i,0,40}]]] (* Harvey P. Dale, Nov 20 2010 *)

a(n)=n\2*7-(-1)^n \\ Charles R Greathouse IV, May 02 2016

a(1) = 1; a(n) = 7(n-1) - a(n-1). - Rolf Pleisch, Jan 31 2008 (corrected by Jon E. Schoenfield, Dec 22 2008)
a(n) = (7/2)*(n-(1-(-1)^n)/2) - (-1)^n. - Rolf Pleisch, Nov 02 2010
From Bruno Berselli, Nov 17 2010: (Start)
G.f.: x*(1+5*x+x^2)/((1+x)*(1-x)^2).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = a(n-2)+7.
a(n) = 7*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 7*floor(n/2)+(-1)^(n+1). - Gary Detlefs, Dec 29 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/7)*cot(Pi/7) = A019674 * A178818. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((14*x - 7)*exp(x) + 3*exp(-x))/4. - David Lovler, Sep 01 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/7) (A160389).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/7) * cosec(Pi/7) (A371858). (End)

More terms from Jon E. Schoenfield, Jan 18 2009

A113801 Numbers that are congruent to {1, 13} mod 14.

Original entry on oeis.org

1, 13, 15, 27, 29, 41, 43, 55, 57, 69, 71, 83, 85, 97, 99, 111, 113, 125, 127, 139, 141, 153, 155, 167, 169, 181, 183, 195, 197, 209, 211, 223, 225, 237, 239, 251, 253, 265, 267, 279, 281, 293, 295, 307, 309, 321, 323, 335, 337, 349, 351, 363, 365, 377, 379

Offset: 1

Giovanni Teofilatto, Jan 22 2006

If 14k+1 is a perfect square..(0,12,16,52,60,120..) then the square root of 14k+1 = a(n) - Gary Detlefs, Feb 22 2010

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 our case, a(n)^2-1==0 (mod 14). Also a(n)^2-1==0 (mod 28). - Bruno Berselli, Oct 26 2010 - Nov 17 2010

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

Cf. A000217, A113802, A113803, A113804, A113805, A113806, A113807, A008589, A045472 (primes), A195145 (partial sums), A005408, A047209, A007310, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887.

a113801 n = a113801_list !! (n-1)
a113801_list = 1 : 13 : map (+ 14) a113801_list
-- Reinhard Zumkeller, Jan 07 2012

LinearRecurrence[{1,1,-1},{1,13,15},60] (* or *) Select[Range[500], MemberQ[{1,13},Mod[#,14]]&] (* Harvey P. Dale, May 11 2011 *)

a(n)=n\2*14-(-1)^n \\ Charles R Greathouse IV, Sep 15 2015

a(n) = 14*(n-1)-a(n-1), n>1. - R. J. Mathar, Jan 30 2010
From Bruno Berselli, Oct 26 2010: (Start)
a(n) = -a(-n+1) = (14*n+5*(-1)^n-7)/2.
G.f.: x*(1+12*x+x^2)/((1+x)*(1-x)^2).
a(n) = a(n-2)+14 for n>2.
a(n) = 14*A000217(n-1)+1 - 2*sum[i=1..n-1] a(i) for n>1. (End)
a(0)=1, a(1)=13, a(2)=15, a(n)=a(n-1)+a(n-2)-a(n-3). - Harvey P. Dale, May 11 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/14)*cot(Pi/14). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((14*x - 7)*exp(x) + 5*exp(-x))/2. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/14).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/14)*cosec(Pi/14). (End)

Corrected and extended by Giovanni Teofilatto, Nov 14 2008

Replaced the various formulas by a correct one - R. J. Mathar, Jan 30 2010

A195817 Multiples of 10 and odd numbers interleaved.

Original entry on oeis.org

0, 1, 10, 3, 20, 5, 30, 7, 40, 9, 50, 11, 60, 13, 70, 15, 80, 17, 90, 19, 100, 21, 110, 23, 120, 25, 130, 27, 140, 29, 150, 31, 160, 33, 170, 35, 180, 37, 190, 39, 200, 41, 210, 43, 220, 45, 230, 47, 240, 49, 250, 51, 260, 53, 270, 55, 280, 57, 290, 59, 300

Offset: 0

Omar E. Pol, Sep 29 2011

A008592 and A005408 interleaved.
Partial sums give the generalized 14-gonal (or tetradecagonal) numbers A195818.
a(n) is also the length of the n-th line segment of a rectangular spiral on the infinite square grid. The vertices of the spiral are the generalized 14-gonal numbers. - Omar E. Pol, Jul 27 2018

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

Column 10 of A195151.
Sequences whose partial sums give the generalized n-gonal numbers, if n>=5: A026741, A001477, zero together with A080512, A022998, A195140, zero together with A165998, A195159, A195161, A195312, this sequence.
Cf. A091998, A195152.

[(2*(-1)^n+3)*n: n in [0..60]]; // Vincenzo Librandi, Sep 30 2011

With[{nn=30},Riffle[10*Range[0,nn],Range[1,2*nn+1,2]]] (* or *) LinearRecurrence[{0,2,0,-1},{0,1,10,3},70] (* Harvey P. Dale, Nov 24 2013 *)

a(n) = (2*(-1)^n+3)*n; \\ Andrew Howroyd, Jul 23 2018

a(n) = (2*(-1)^n+3)*n. - Vincenzo Librandi, Sep 30 2011
From Bruno Berselli, Sep 30 2011: (Start)
G.f.: x*(1+10*x+x^2)/((1-x)^2*(1+x)^2).
a(n) = -a(-n) = a(n-2)*n/(n-2) = 2*a(n-2)-a(n-4).
a(n) * a(n+1) = a(n(n+1)).
a(n) + a(n+1) = A091998(n+1). (End)
a(0)=0, a(1)=1, a(2)=10, a(3)=3, a(n)=2*a(n-2)-a(n-4). - Harvey P. Dale, Nov 24 2013
Multiplicative with a(2^e) = 5*2^e, a(p^e) = p^e for odd prime p. - Andrew Howroyd, Jul 23 2018
Dirichlet g.f.: zeta(s-1) * (1 + 2^(3-s)). - Amiram Eldar, Oct 25 2023

A097933 Primes p that divide 3^((p-1)/2) - 1.

Original entry on oeis.org

11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167, 179, 181, 191, 193, 227, 229, 239, 241, 251, 263, 277, 311, 313, 337, 347, 349, 359, 373, 383, 397, 409, 419, 421, 431, 433, 443, 457, 467, 479, 491, 503, 541, 563, 577, 587, 599, 601, 613

Offset: 1

Cino Hilliard, Sep 04 2004

nonn

Rational primes that decompose in the field Q[sqrt(3)]. - N. J. A. Sloane, Dec 26 2017
For all primes p > 2 and integers gcd(x, y, p) = 1, x^((p-1)/2) +- y^((p-1)/2) is divisible by p. This is because (x^((p-1)/2) - y^((p-1)/2))(x^((p-1)/2) + y^((p-1)/2)) = x^(p-1) - y^(p-1) is divisible by p according to Fermat's Little Theorem (FLT). This sequence lists p that divides 3^((p-1)/2) - 1^((p-1)/2), and A003630 lists the '+' case.
Apart from initial terms, this and A038874 are the same. - N. J. A. Sloane, May 31 2009
Primes in A091998. - Reinhard Zumkeller, Jan 07 2012
Also, primes congruent to 1 or 11 (mod 12). - Vincenzo Librandi, Mar 23 2013
Conjecture: Let r(n) = (a(n) - 1)/(a(n) + 1) if a(n) mod 4 = 1, (a(n) + 1)/(a(n) - 1) otherwise; then Product_{n>=1} r(n) = (6/5) * (6/7) * (12/11) * (18/19) * ... = 2/sqrt(3). - Dimitris Valianatos, Mar 27 2017
Primes p such that Kronecker(12,p) = +1 (12 is the discriminant of Q[sqrt(3)]), that is, odd primes that have 3 as a quadratic residue. - Jianing Song, Nov 21 2018
Comment from Richard R. Forberg, Feb 07 2023: (Start)
Conjecture: These are the exclusive prime factors of the set of integers d > 1 such that there exist primitive Heronian triangles with sides {b, b+d, b+2d} for one or more integers b.
Also b is always > d. For d=11 the b values begin {15, 17, 65, 75, 267, 305, 1025, ...}. For d=1 (not prime, thus not listed) the b values are given by A016064. (End)

			For p = 5, 3^2 - 1 = 8 <> 3*k for any integer k, so 5 is not in this sequence.
For p = 11, 3^5 - 1 = 242 = 11*22, so 11 is in this sequence.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index to sequences related to decomposition of primes in quadratic fields

Cf. A003630, A010051, A038874, A091998.

a097933 n = a097933_list !! (n-1)
a097933_list = [x | x <- a091998_list, a010051 x == 1]
-- Reinhard Zumkeller, Jan 07 2012

[p: p in PrimesUpTo(1000) | p mod 24 in [1, 11, 13, 23]]; // Vincenzo Librandi, Mar 23 2013

Select[Prime[Range[300]], MemberQ[{1, 11, 13, 23}, Mod[#, 24]]&] (* Vincenzo Librandi, Mar 23 2013 *)
Select[Prime[Range[2,200]],PowerMod[3,(#-1)/2,#]==1&] (* Harvey P. Dale, Jun 02 2020 *)

/* s = +-1, d=diff */ ptopm1d2(n,x,d,s) = { forprime(p=3,n,p2=(p-1)/2; y=x^p2 + s*(x-d)^p2; if(y%p==0,print1(p","))) }

{a(n)= local(m, c); if(n<1, 0, c=0; m=0; while( cMichael Somos, Aug 28 2006 */

cp's OEIS Frontend

A229098 Smallest area A of Heron triangles with sides (a, b, c) in arithmetic progression of the form b - d(n), b, b + d(n), where d(n) = A091998(n) = 12*n +- 1.

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A007310 Numbers congruent to 1 or 5 mod 6.

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A047209 Numbers that are congruent to {1, 4} mod 5.

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

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A056020 Numbers that are congruent to +-1 mod 9.

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A090771 Numbers that are congruent to {1, 9} mod 10.

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