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 1-10 of 10 results.

A170816 a(n)=A140475(n)-A141468(n).

Original entry on oeis.org

1, 4, 3, 5, 5, 8, 9, 11, 15, 16, 21, 23, 23, 26, 31, 35, 36, 41, 44, 45, 49, 51, 56, 63, 66, 67, 69, 70, 73, 85, 87, 92, 93, 101, 102, 107, 112, 115, 119, 124, 125, 134, 135, 137, 137, 148, 159, 162, 163, 165, 170, 171, 179, 183, 188, 193, 194, 199, 201, 202, 211, 223
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Dec 07 2009

Keywords

Examples

			a(1)=1-0=1, a(2)=5-1=4, a(3)=7-4=3, a(4)=11-6=5, a(5)=13-8=5, a(6)=17-9=8.

Crossrefs

Cf. A140475, A141468.

A171384 a(n) = A140475(n) - A167707(n).

Original entry on oeis.org

1, 4, 4, 6, 6, 9, 10, 13, 18, 18, 23, 26, 27, 30, 34, 39, 40, 45, 47, 48, 53, 56, 61, 68, 70, 71, 74, 75, 78, 91, 93, 98, 99, 108, 108, 113, 118, 121, 125, 130, 131, 140, 141, 143, 144, 155, 166, 169, 170, 172, 177, 178, 187, 192, 197, 201, 202, 207, 210, 210, 219
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Dec 07 2009

Keywords

Examples

			a(1)=1-0=1, a(2)=5-1=4, a(3)=7-3=4, a(4)=11-5=6, a(5)=13-7=6, a(6)=17-8=9.

Crossrefs

Cf. A140475, A167707.

Extensions

Edited by Omar E. Pol, Apr 14 2010

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)

A244329 a(n) = floor(antisigma(n) / sigma(n)) = floor(A024816(n) / A000203(n)).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 2, 1, 2, 2, 4, 1, 5, 3, 4, 3, 7, 3, 8, 4, 6, 6, 10, 4, 9, 7, 8, 6, 13, 5, 14, 7, 10, 10, 12, 6, 17, 11, 12, 8, 19, 8, 20, 10, 12, 14, 22, 8, 20, 12, 17, 13, 25, 11, 20, 12, 19, 18, 28, 9, 29, 19, 18, 15, 24, 14, 32, 17, 24, 16, 34, 12, 35, 23
Offset: 1

Views

Author

Jaroslav Krizek, Jul 08 2014

Keywords

Comments

RECORD transform of a(n) is A140475 (union of number 1 and primes >= 5).
Sequence of numbers n such that a(n) = floor(antisigma(n) / n) = A046022.
Sequence of numbers n such that a(n) = a(n+1) = A244666.

Examples

			For n = 10; a(10) = floor(A024816(10) / A000203(10)) = floor(37 / 18) = 2.

Links

Crossrefs

Cf. A000203, A000217, A024816, A046022, A140475, A244327, A244328, A244666.

Programs

  • Magma
    [Floor(((n*(n+1)div 2)-SumOfDivisors(n)) div (SumOfDivisors(n))) : n in [1..1000]];
  • Mathematica
    A244329[n_] := Floor[(n*(n + 1)/2 - #)/#] & [DivisorSigma[1, n]];
Array[A244329, 100] (* Paolo Xausa, Sep 01 2024 *)

Formula

a(n) = A244327(n) - A244328(n) for n >= 7.

A244327 a(n) = floor((n*(n+1)/2) / sigma(n)) = floor(A000217(n) / A000203(n)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 2, 3, 3, 5, 2, 6, 4, 5, 4, 8, 4, 9, 5, 7, 7, 11, 5, 10, 8, 9, 7, 14, 6, 15, 8, 11, 11, 13, 7, 18, 12, 13, 9, 20, 9, 21, 11, 13, 15, 23, 9, 21, 13, 18, 14, 26, 12, 21, 13, 20, 19, 29, 10, 30, 20, 19, 16, 25, 15, 33, 18, 25, 17, 35, 13, 36
Offset: 1

Views

Author

Jaroslav Krizek, Jul 08 2014

Keywords

Comments

RECORD transform of a(n) is A140475 (union of number 1 and primes >= 5).
Sequence of numbers n such that a(n) = a(n+1) = A244666.

Examples

			For n = 10; a(10) = floor(A000217(10) / A000203(10)) = floor(55 / 18) = 3.

Links

Crossrefs

Cf. A000203, A000217, A024816, A244328, A244329, A244666.

Programs

  • Magma
    [Floor((n*(n+1)div 2) div (SumOfDivisors(n))): n in [1..1000]];
  • Maple
    a:= n-> floor(n*(n+1)/(2*numtheory[sigma](n))):
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 28 2018
  • Mathematica
    A244327[n_] := Floor[n*(n + 1)/(2*DivisorSigma[1, n])];
Array[A244327, 100] (* Paolo Xausa, Sep 01 2024 *)

Formula

a(n) = A244328(n) + A244329(n) for n >= 7.

A279507 a(n) = floor(phi(n)/tau(n)).

Original entry on oeis.org

1, 0, 1, 0, 2, 0, 3, 1, 2, 1, 5, 0, 6, 1, 2, 1, 8, 1, 9, 1, 3, 2, 11, 1, 6, 3, 4, 2, 14, 1, 15, 2, 5, 4, 6, 1, 18, 4, 6, 2, 20, 1, 21, 3, 4, 5, 23, 1, 14, 3, 8, 4, 26, 2, 10, 3, 9, 7, 29, 1, 30, 7, 6, 4, 12, 2, 33, 5, 11, 3, 35, 2, 36, 9, 6, 6, 15, 3, 39, 3
Offset: 1

Views

Author

Jaroslav Krizek, Dec 13 2016

Keywords

Comments

a(n) = floor(A000010(n)/A000005(n)).
There are 11 numbers n such that phi(n) <= tau(n) and 7 numbers n such that phi(n) = tau(n); see A020490 and A020488.
Sequences b(k) of numbers n such that a(n) = k are finite for all k >=0; see A279508 (the smallest numbers n such that a(n) = k for k>=0) and A279509 (the largest numbers n such that a(n) = k for k>=0).
See A140475 (numbers n such that floor(phi(n)/tau(n)) > floor(phi(m)/tau(m)) for all m < n).

Examples

			For n=5; a(5) = floor(phi(5)/tau(5)) = floor(4/2) = 2.

Links

Crossrefs

Cf. A000005, A000010, A020488, A020490, A140475, A279289, A279508, A279509.

Programs

  • Magma
    [Floor(EulerPhi(n)/NumberOfDivisors(n)): n in[1..100]]
  • Mathematica
    Table[Floor[EulerPhi[n]/DivisorSigma[0, n]], {n,1,25}] (* G. C. Greubel, Dec 13 2016 *)
  • PARI
    for(n=1, 25, print1(floor(eulerphi(n)/numdiv(n)), ", ")) \\ G. C. Greubel, Dec 13 2016

Formula

a(n) > 1 for numbers in A279289.

A174027 Alternating triangle (version 2) read by rows: composites k such that k=6*m-+1=r*j r>=j and n>=q>0 where r=6*n-1 or r=6*n+1 and j=6*q-1 or j=6*q+1.

Original entry on oeis.org

35, 25, 49, 65, 77, 143, 55, 91, 121, 169, 95, 119, 209, 221, 323, 65, 133, 187, 247, 289, 361, 125, 161, 275, 299, 425, 437, 575, 115, 175, 253, 325, 391, 475, 529, 625, 155, 203, 341, 377, 527, 589, 713, 725, 899, 145, 217, 319, 403, 493, 569, 667, 775, 841
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Mar 06 2010, Mar 17 2010

Keywords

Comments

Composites of form 6*m-1 are in even (0,2,4,..) rows of alternating triangle only. Composites of form 6*m+1 are in odd (1,3,5,..) rows of alternating triangle only. 1 UNION nontrivial primes UNION A174027(without repetition) = A140475 U A174027(without repetition) = A007310 = numbers of form 6*n+-1, where alternating triangle (version 1) is A173865.

Examples

			Triangle begins:
35(=7*5) in even 0th row;
25(=5*5) and 49(=7*7) in odd 1st row;
65(=13*5), 77(=11*7) and 143(=13*11) in even 2nd row.

Crossrefs

Cf. A007310, A140475, A173825, A173865.

A175212 Numbers n such that A000975(n-1)/n is an integer. Also numbers n such that arithmetic mean of the first n Jacobsthal numbers is an integer.

Original entry on oeis.org

1, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293
Offset: 1

Views

Author

Ctibor O. Zizka, Mar 06 2010

Keywords

Comments

Contains the subsequence A066488 and differs therefore from A140475. [From R. J. Mathar, Aug 09 2010]

Examples

			Arithmetic mean of partial sums of Jacobsthal numbers is an integer : 0/1=0 n=1; (0+1+1+3+5)/5=2 n=5; (0+1+1+3+5+11+21)/7=6 n=7; (0+1+1+3+5+11+21+43+85+171+341)/11=62 n=11 etc.

Links

Crossrefs

Cf. A000975, A001045

Extensions

Term 34 removed, sequence extended and index in definition corrected by R. J. Mathar, Mar 29 2010

A171577 0, 1 and primes > 3.

Original entry on oeis.org

0, 1, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Dec 12 2009, Dec 16 2009

Keywords

Comments

0 together with A140475. [Arkadiusz Wesolowski, Aug 26 2011]

Programs

A174052 The sum of the numbers in n-th row of alternant triangle A173865.

Original entry on oeis.org

1, 40, 81, 296, 449, 984, 1301, 2320, 2913, 4558, 5441
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Mar 06 2010

Keywords

Crossrefs

Cf. A007310, A140475, A173825, A173865.
Showing 1-10 of 10 results.

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