cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A007310 Numbers congruent to 1 or 5 mod 6.

Original entry on oeis.org

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

Views

Author

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

Keywords

Comments

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

Examples

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

References

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

Links

Crossrefs

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).

Programs

Formula

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)

A164577 Integer averages of the first perfect cubes up to some n^3.

Original entry on oeis.org

1, 12, 25, 45, 112, 162, 225, 396, 507, 637, 960, 1156, 1377, 1900, 2205, 2541, 3312, 3750, 4225, 5292, 5887, 6525, 7936, 8712, 9537, 11340, 12321, 13357, 15600, 16810, 18081, 20812, 22275, 23805, 27072, 28812, 30625, 34476, 36517, 38637, 43120
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Keywords

Comments

Integers of the form A000537(k)/k, created by the k>0 listed in A042965. - R. J. Mathar, Aug 20 2009
Integers of the form (1/4)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009

Examples

			The average of the first cube is 1^3/1=1=a(1).
The average of the first two cubes is (1^3+2^3)/2=9/2, not integer, and does not contribute to the sequence.
The average of the first three cubes is (1^3+2^3+3^3)/3=12, integer, and defines a(2).

Links

Crossrefs

Cf. A000537, A050248, A051456, A078617, A078618, A154293, A164576

Programs

  • Mathematica
    Timing[s=0;lst={};Do[a=(s+=n^3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n, 5!}];lst]
With[{nn=80},Select[#[[1]]/#[[2]]&/@Thread[{Accumulate[Range[ nn]^3],Range[ nn]}],IntegerQ]] (* or *) LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{1,12,25,45,112,162,225,396,507,637},50] (* Harvey P. Dale, Mar 14 2020 *)

Formula

G.f.: ( x*(1+11*x+13*x^2+17*x^3+34*x^4+11*x^5+6*x^6+3*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011

Extensions

Changed comments to examples - R. J. Mathar, Aug 20 2009

A164578 Integers of the form (k+1)*(2k+1)/12.

Original entry on oeis.org

10, 23, 65, 94, 168, 213, 319, 380, 518, 595, 765, 858, 1060, 1169, 1403, 1528, 1794, 1935, 2233, 2390, 2720, 2893, 3255, 3444, 3838, 4043, 4469, 4690, 5148, 5385, 5875, 6128, 6650, 6919, 7473, 7758, 8344, 8645, 9263, 9580, 10230, 10563, 11245, 11594
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Keywords

Comments

This can also be defined as integer averages of the first k halved squares, 1^2/2, 2^2/2, 3^2/2,... , 3^k/2, because sum_{j=1..k} j^2/2 = k*(k+1)*(2k+1)/12. The generating k are in A168489.

Links

Crossrefs

Cf. A078617, A078618, A154293, A164576, A164577.

Programs

  • Mathematica
    s=0;lst={};Do[a=(s+=(n^2)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,2*6!}];lst
Select[Table[((n+1)(2n+1))/12,{n,300}],IntegerQ] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{10,23,65,94,168},60] (* Harvey P. Dale, Jun 14 2017 *)
  • PARI
    Vec(x*(10+13*x+22*x^2+3*x^3)/((1-x)^3*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 26 2016

Formula

a(n) = +a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5). G.f. x*(-10-13*x-22*x^2-3*x^3) / ((1+x)^2*(x-1)^3). - R. J. Mathar, Jan 25 2011
From Colin Barker, Jan 26 2016: (Start)
a(n) = (24*n^2+6*n-(-1)^n*(8*n+1)+1)/4.
a(n) = (12*n^2-n)/2 for n even.
a(n) = (12*n^2+7*n+1)/2 for n odd.
(End)

A164579 Integer averages of halves of first cubes of natural numbers (n^3)/2 for some n.

Original entry on oeis.org

6, 56, 81, 198, 480, 578, 950, 1656, 1875, 2646, 3968, 4356, 5670, 7800, 8405, 10406, 13536, 14406, 17238, 21560, 22743, 26550, 32256, 33800, 38726, 46008, 47961, 54150, 63200, 65610, 73206, 84216, 87131, 96278, 109440, 112908, 123750, 139256
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Keywords

Comments

Also, integers of the form (1/8)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009

Examples

			1/2, 9/4, 6, 25/2, 45/2, 147/4, 56, 81, ...

Links

Crossrefs

Cf. A050248, A051456, A078617, A078618, A136116, A154293, A164576, A164577, A164578

Programs

  • Mathematica
    s=0;lst={};Do[a=(s+=(n^3)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,3*5!}];lst
LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{6,56,81,198,480,578,950,1656,1875,2646},40] (* Harvey P. Dale, Jul 26 2017 *)
Module[{nn=200,ac},ac=Accumulate[Range[nn]^3/2];Select[#[[1]]/#[[2]]&/@ Thread[{ac,Range[nn]}],IntegerQ]] (* Harvey P. Dale, Jan 28 2020 *)
  • PARI
    forstep(n=3, 150, [4,1,3], print1(n*(n+1)^2>>3, ", ")); \\ Charles R Greathouse IV, Nov 02 2009

Formula

G.f.: ( x*(6+50*x+25*x^2+99*x^3+132*x^4+23*x^5+39*x^6+10*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011

A215574 Minimal sum s of n distinct squares such that s is divisible by n.

Original entry on oeis.org

1, 10, 21, 84, 55, 156, 140, 240, 342, 470, 506, 864, 819, 1176, 1395, 1616, 1785, 2214, 2470, 3260, 3570, 4092, 4324, 5184, 5525, 6578, 7101, 8456, 8555, 9750, 10416, 11712, 13134, 14314, 14910, 16776, 17575, 19570, 21099, 22760, 23821, 26166, 27434, 30096
Offset: 1

Views

Author

Zak Seidov, Aug 16 2012

Keywords

Comments

At n = 6k +/- 1, there is a simple formula a(n) = n*(n+1)*(2n+1)/6 and set of squares is the first n squares {1..n}^2.
Companion sequence of minimal integral averages of n distinct squares is a(n)/n: 1, 5, 7, 21, 11, 26, 20, 30, 38, 47, 46, 72, 63, 84, 93, 101, 105, 123, 130, 163, 170, 186, 188, 216, 221, ... .

Examples

			a(1) =  1 = 1^2 = 1*1.
a(2) = 10 = 1^2 + 3^2 = 2*5.
a(3) = 21 = 1^2 + 2^2 + 4^2 = 3*7.
a(4) = 84 = 1^2 + 3^2 + 5^2 + 7^2 = 4*21.
a(5) = 55 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 5*11.

Links

Crossrefs

Cf. A000330, A164576.

Programs

  • Maple
    b:= proc(n, i, t) option remember; local h; h:= t*(t+1)*(2*t+1)/6;
      i>=t and n>=h and (n=h or t>0 and
      (b(n, i-1, t) or i<=n and b(n-i^2, i-1, t-1)))
    end:
a:= proc(n) local s;
      for s from n*ceil((n+1)*(2*n+1)/6) by n
        while not b(s, iroot(s, 2)+1, n)
      do od; s
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Aug 16 2012
  • Mathematica
    $RecursionLimit = 2000;
b[n_, i_, t_] := b[n, i, t] = Module[{h = t(t+1)(2t+1)/6}, i >= t && n >= h && (n==h || t>0 && (b[n, i-1, t] || i <= n && b[n-i^2, i-1, t-1]))];
a[n_] := Module[{s}, For[s = n Ceiling[(n+1)(2n+1)/6], !b[s, Floor @ Sqrt[s] + 1, n], s += n]; s];
Array[a, 50] (* Jean-François Alcover, Nov 22 2020, after Alois P. Heinz *)

Extensions

More terms from Alois P. Heinz, Aug 16 2012

A164619 Integers of the form A164577(k)/3.

Original entry on oeis.org

4, 15, 54, 75, 132, 169, 320, 459, 735, 847, 1104, 1250, 1764, 2175, 2904, 3179, 3780, 4107, 5200, 6027, 7425, 7935, 9024, 9604, 11492, 12879, 15162, 15979, 17700, 18605, 21504, 23595, 26979, 28175, 30672, 31974, 36100, 39039, 43740, 45387, 48804
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Aug 17 2009

Keywords

Comments

The sequence members are the third of the average of a set of smallest cubes, if integer.

Examples

			A third of the average of the first cube, A164577(1)/3=1/3, is not an integer and does not contribute to the sequence.
A third of the average of the first two cubes, A164577(2)/3=4, is an integer and defines a(1)=4 of the sequence.

Links

Crossrefs

Cf. A050248, A051456, A078617, A078618, A136116, A154293, A164286, A164576, A164577, A164578, A164579.

Programs

  • Mathematica
    s=0;lst={};Do[a=(s+=(n^3)/3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,2*5!}]; lst
LinearRecurrence[{2,-1,-1,2,-1,2,-4,2,2,-4,2,-1,2,-1,-1,2,-1},{4,15,54,75,132,169,320,459,735,847,1104,1250,1764,2175,2904,3179,3780},50] (* Harvey P. Dale, Apr 06 2016 *)
  • PARI
    Vec(x*(x^14 +x^13 +16*x^12 +10*x^11 +47*x^10 -22*x^9 +61*x^8 +10*x^7 +88*x^6 +8*x^5 +43*x^4 -14*x^3 +28*x^2 +7*x +4) / ((x -1)^4*(x +1)^3*(x^2 -x +1)^3*(x^2 +x +1)^2) + O(x^100)) \\ Colin Barker, Oct 27 2014

Formula

a(n) = +2*a(n-1) -a(n-2) -a(n-3) +2*a(n-4) -a(n-5) +2*a(n-6) -4*a(n-7) +2*a(n-8) +2*a(n-9) -4*a(n-10) +2*a(n-11) -a(n-12) +2*a(n-13) -a(n-14) -a(n-15) +2*a(n-16) -a(n-17). - R. J. Mathar, Jan 25 2011
G.f.: x*(x^14 +x^13 +16*x^12 +10*x^11 +47*x^10 -22*x^9 +61*x^8 +10*x^7 +88*x^6 +8*x^5 +43*x^4 -14*x^3 +28*x^2 +7*x +4) / ((x -1)^4*(x +1)^3*(x^2 -x +1)^3*(x^2 +x +1)^2). - Colin Barker, Oct 27 2014

Extensions

Edited by R. J. Mathar, Aug 20 2009
Showing 1-6 of 6 results.

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