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 24 results. Next

A001108 a(n)-th triangular number is a square: a(n+1) = 6*a(n) - a(n-1) + 2, with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521
Offset: 0

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Author

Keywords

Comments

b(0)=0, c(0)=1, b(i+1)=b(i)+c(i), c(i+1)=b(i+1)+b(i); then a(i) (the number in the sequence) is 2b(i)^2 if i is even, c(i)^2 if i is odd and b(n)=A000129(n) and c(n)=A001333(n). - Darin Stephenson (stephenson(AT)cs.hope.edu) and Alan Koch
For n > 1 gives solutions to A007913(2x) = A007913(x+1). - Benoit Cloitre, Apr 07 2002
If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X are in the sequence.
For n >= 2, a(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is A001109. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
This is the r=8 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Also, 1^3 + 2^3 + 3^3 + ... + a(n)^3 = k(n)^4 where k(n) is A001109. - Anton Vrba (antonvrba(AT)yahoo.com), Nov 18 2006
If T_x = y^2 is a triangular number which is also a square, the least number which is both triangular and square and greater than T_x is T_(3*x + 4*y + 1) = (2*x + 3*y + 1)^2 (W. Sierpiński 1961). - Richard Choulet, Apr 28 2009
If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or (a+1) is a perfect square. If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or a/8 is a perfect square. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c, then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution, too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c < e, then (8*d^2, d*(f-b)) is a solution, too. - Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with p < r, then r = 3p + 4q + 1 and s = 2p + 3q + 1. - Mohamed Bouhamida, Sep 02 2009
Also numbers k such that (ceiling(sqrt(k*(k+1)/2)))^2 - k*(k+1)/2 = 0. - Ctibor O. Zizka, Nov 10 2009
From Lekraj Beedassy, Mar 04 2011: (Start)
Let x=a(n) be the index of the associated triangular number T_x=1+2+3+...+x and y=A001109(n) be the base of the associated perfect square S_y=y^2. Now using the identity S_y = T_y + T_{y-1}, the defining T_x = S_y may be rewritten as T_y = T_x - T_{y-1}, or 1+2+3+...+y = y+(y+1)+...+x. This solves the Strand Magazine House Number problem mentioned in A001109 in references from Poo-Sung Park and John C. Butcher. In a variant of the problem, solving the equation 1+3+5+...+(2*x+1) = (2*x+1)+(2*x+3)+...+(2*y-1) implies S_(x+1) = S_y - S_x, i.e., with (x,x+1,y) forming a Pythagorean triple, the solutions are given by pairs of x=A001652(n), y=A001653(n). (End)
If P = 8*n +- 1 is a prime, then P divides a((P-1)/2); e.g., 7 divides a(3) and 41 divides a(20). Also, if P = 8*n +- 3 is prime, then 4*P divides (a((P-1)/2) + a((P+1)/2) + 3). - Kenneth J Ramsey, Mar 05 2012
Starting at a(2), a(n) gives all the dimensions of Euclidean k-space in which the ratio of outer to inner Soddy hyperspheres' radii for k+1 identical kissing hyperspheres is rational. The formula for this ratio is (1+3k+2*sqrt(2k*(k+1)))/(k-1) where k is the dimension. So for a(3) = 49, the ratio is 6 in the 49th dimension. See comment for A010502. - Frank M Jackson, Feb 09 2013
Conjecture: For n>1 a(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
For n=2*k, k>0, a(n) is divisible by 8 (deficient), so since all proper divisors of deficient numbers are deficient, then a(n) is deficient. For n=2*k+1, k>0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. sigma(a(5)) = 1723 < 3362 = 2*a(5). In either case, a(n) is deficient. - Muniru A Asiru, Apr 14 2016
The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016
For n > 0, numbers for which the number of odd divisors of both n and of n + 1 is odd. - Gionata Neri, Apr 30 2018
a(n) will be solutions to some (A000217(k) + A000217(k+1))/2. - Art Baker, Jul 16 2019
For n >= 2, a(n) is the base for which A058331(A001109(n)) is a length-3 repunit. Example: for n=2, A001109(2)=6 and A058331(6)=73 and 73 in base a(2)=8 is 111. See Grantham and Graves. - Michel Marcus, Sep 11 2020

Examples

			a(1) = ((3 + 2*sqrt(2)) + (3 - 2*sqrt(2)) - 2) / 4 = (3 + 3 - 2) / 4 = 4 / 4 = 1;
a(2) = ((3 + 2*sqrt(2))^2 + (3 - 2*sqrt(2))^2 - 2) / 4 = (9 + 4*sqrt(2) + 8 + 9 - 4*sqrt(2) + 8 - 2) / 4 = (18 + 16 - 2) / 4 = (34 - 2) / 4 = 32 / 4 = 8, etc.
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.
  • M. S. Klamkin, "International Mathematical Olympiads 1978-1985," (Supplementary problem N.T.6)
  • W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, pp. 21-22 MR2002669
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.

Crossrefs

Partial sums of A002315. A000129, A005319.
a(n) = A115598(n), n > 0. - Hermann Stamm-Wilbrandt, Jul 27 2014

Programs

  • Haskell
    a001108 n = a001108_list !! n
    a001108_list = 0 : 1 : map (+ 2)
       (zipWith (-) (map (* 6) (tail a001108_list)) a001108_list)
    -- Reinhard Zumkeller, Jan 10 2012
    
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018
  • Maple
    A001108:=-(1+z)/(z-1)/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation, without the leading 0
  • Mathematica
    Table[(1/2)(-1 + Sqrt[1 + Expand[8(((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]))^2]]), {n, 0, 100}] (* Artur Jasinski, Dec 10 2006 *)
    Transpose[NestList[{#[[2]],#[[3]],6#[[3]]-#[[2]]+2}&,{0,1,8},20]][[1]] (* Harvey P. Dale, Sep 04 2011 *)
    LinearRecurrence[{7, -7, 1}, {0, 1, 8}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
  • PARI
    a(n)=(real((3+quadgen(32))^n)-1)/2
    
  • PARI
    a(n)=(subst(poltchebi(abs(n)),x,3)-1)/2
    
  • PARI
    a(n)=if(n<0,a(-n),(polsym(1-6*x+x^2,n)[n+1]-2)/4)
    
  • PARI
    x='x+O('x^99); concat(0, Vec(x*(1+x)/((1-x)*(1-6*x+x^2)))) \\ Altug Alkan, May 01 2018
    

Formula

a(0) = 0, a(n+1) = 3*a(n) + 1 + 2*sqrt(2*a(n)*(a(n)+1)). - Jim Nastos, Jun 18 2002
a(n) = floor( (1/4) * (3+2*sqrt(2))^n ). - Benoit Cloitre, Sep 04 2002
a(n) = A001653(k)*A001653(k+n) - A001652(k)*A001652(k+n) - A046090(k)*A046090(k+n). - Charlie Marion, Jul 01 2003
a(n) = A001652(n-1) + A001653(n-1) = A001653(n) - A046090(n) = (A001541(n)-1)/2 = a(-n). - Michael Somos, Mar 03 2004
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Antonio Alberto Olivares, Oct 23 2003
a(n) = Sum_{r=1..n} 2^(r-1)*binomial(2n, 2r). - Lekraj Beedassy, Aug 21 2004
If n > 1, then both A000203(n) and A000203(n+1) are odd numbers: n is either a square or twice a square. - Labos Elemer, Aug 23 2004
a(n) = (T(n, 3)-1)/2 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3) = A001541(n). - Wolfdieter Lang, Oct 18 2004
G.f.: x*(1+x)/((1-x)*(1-6*x+x^2)). Binet form: a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n - 2)/4. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
a(n) = floor(sqrt(2*A001110(n))) = floor(A001109(n)*sqrt(2)) = 2*(A000129(n)^2) - (n mod 2) = A001333(n)^2 - 1 + (n mod 2). - Henry Bottomley, Apr 19 2000, corrected by Eric Rowland, Jun 23 2017
A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006
A028982(a(n)) + 1 = A028982(a(n) + 1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n+1)^2 + a(n)^2 + 1 = 6*a(n+1)*a(n) + 2*a(n+1) + 2*a(n). - Charlie Marion, Sep 28 2011
a(n) = 2*A001653(m)*A053141(n-m-1) + A002315(m)*A046090(n-m-1) + a(m) with m < n; otherwise, a(n) = 2*A001653(m)*A053141(m-n) - A002315(m)*A001652(m-n) + a(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = A048739(2n-2), n > 0. - Richard R. Forberg, Aug 31 2013
From Peter Bala, Jan 28 2014: (Start)
A divisibility sequence: that is, a(n) divides a(n*m) for all n and m. Case P1 = 8, P2 = 12, Q = 1 of the 3-parameter family of linear divisibility sequences found by Williams and Guy.
a(2*n+1) = A002315(n)^2 = Sum_{k = 0..4*n + 1} Pell(n), where Pell(n) = A000129(n).
a(2*n) = (1/2)*A005319(n)^2 = 8*A001109(n)^2.
(2,1) entry of the 2 X 2 matrix T(n,M), where M = [0, -3; 1, 4] and T(n,x) is the Chebyshev polynomial of the first kind. (End)
E.g.f.: exp(x)*(exp(2*x)*cosh(2*sqrt(2)*x) - 1)/2. - Stefano Spezia, Oct 25 2024

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000
More terms from Lekraj Beedassy, Aug 21 2004

A074384 Solutions to mod(sigma(x), 6) = 5.

Original entry on oeis.org

2401, 9604, 21609, 28561, 38416, 60025, 86436, 114244, 130321, 153664, 194481, 240100, 257049, 290521, 345744, 456976, 521284, 540225, 614656, 693889, 714025, 777924, 923521, 960400, 1028196, 1162084, 1172889, 1270129, 1382976, 1500625
Offset: 1

Views

Author

Labos Elemer, Aug 22 2002

Keywords

Examples

			4th powers of primes of the form 6k+1 are here because sigma[p^4]=p^4+p^3+p^2+p+1 congruent 1+1+1+1+1=5 mod 6. There are also other fourth powers, like 38416=(2*7)^4, 194481=(3*7)^4, 456976=(2*13)^4, and solutions which are not fourth powers like 9604=2^2*7^4 and 21609=3^2*7^4.
		

Crossrefs

Programs

  • Mathematica
    Do[s=Mod[DivisorSigma[1, n], 6]; If[s==5, Print[n]], {n, 1, 1000000}]
    Select[Range[1600000],Mod[DivisorSigma[1,#],6]==5&] (* Harvey P. Dale, Jul 06 2014 *)

Formula

{n: A084301(n) = 5}. - R. J. Mathar, May 19 2020

A105824 a(n) = sigma(n) mod 4.

Original entry on oeis.org

1, 3, 0, 3, 2, 0, 0, 3, 1, 2, 0, 0, 2, 0, 0, 3, 2, 3, 0, 2, 0, 0, 0, 0, 3, 2, 0, 0, 2, 0, 0, 3, 0, 2, 0, 3, 2, 0, 0, 2, 2, 0, 0, 0, 2, 0, 0, 0, 1, 1, 0, 2, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 3, 0, 0, 0, 2, 0, 0, 0, 3, 2, 2, 0, 0, 0, 0, 0, 2, 1, 2, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 3, 0, 1, 2, 0, 0, 2, 0
Offset: 1

Views

Author

Shyam Sunder Gupta, May 05 2005

Keywords

Crossrefs

Sequences sigma(n) mod k: A053866 (k=2), A074941 (k=3), A105824 (k=4), A105825 (k=5), A084301 (k=6), A105826 (k=7), A105827 (k=8).

Programs

  • Maple
    A105824:= n-> (numtheory[sigma](n) mod 4):
    seq (A105824(n), n=1..105); # Jani Melik, Jan 26 2011
  • Mathematica
    Table[Mod[DivisorSigma[1, n], 4], {n, 100}] (* Wesley Ivan Hurt, Nov 07 2017 *)
  • PARI
    a(n)=sigma(n)%4

Formula

a(n) = A010873(A000203(n)). - Antti Karttunen, Nov 07 2017

A067051 The smallest k>1 such that k divides sigma(k*n) is equal to 3.

Original entry on oeis.org

2, 8, 18, 32, 49, 50, 72, 98, 128, 162, 169, 196, 200, 242, 288, 338, 361, 392, 441, 450, 512, 578, 648, 676, 722, 784, 800, 882, 961, 968, 1058, 1152, 1225, 1250, 1352, 1369, 1444, 1458, 1521, 1568, 1682, 1764, 1800, 1849, 1922, 2048, 2178, 2312, 2450, 2592
Offset: 1

Views

Author

Benoit Cloitre, Jul 26 2002

Keywords

Comments

The smallest m>1 such that m divides sigma(m*n) is 2, 3 or 6.
Appears to be the same sequence as A074629. - Ralf Stephan, Aug 18 2004. [Proof: Mathar link]
Square terms are in A074216. Nonsquare terms appear to be A001105 except {0}. - Michel Marcus, Dec 26 2013

Crossrefs

Subsequence of A087943.

Programs

  • Magma
    [n: n in [1..3*10^3] | (SumOfDivisors(n) mod 6) eq 3]; // Vincenzo Librandi, Dec 11 2015
  • Maple
    select(t -> numtheory:-sigma(t) mod 6 = 3, [$1..10000]); # Robert Israel, Dec 11 2015
  • Mathematica
    Select[Range@ 2600, Mod[DivisorSigma[1, #], 6] == 3 &] (* Michael De Vlieger, Dec 10 2015 *)
  • PARI
    isok(n) = (sigma(2*n) % 2) && !(sigma(3*n) % 3); \\ Michel Marcus, Dec 26 2013
    

Formula

{n: A000203(n) mod 6 = 3.} (Old definition of A074629) - Labos Elemer, Aug 26 2002
In the prime factorization of n, no odd prime has odd exponent, and 2 has odd exponent or at least one prime == 1 (mod 6) has exponent == 2 (mod 6). - Robert Israel, Dec 11 2015
{n: A049605(n) = 3}. - R. J. Mathar, May 19 2020
{n: A084301(n) = 3 }. - R. J. Mathar, May 19 2020
A087943 INTERSECT A028982. - R. J. Mathar, May 30 2020

A074627 Numbers n such that sigma(n) is divisible by 6.

Original entry on oeis.org

5, 6, 10, 11, 14, 15, 17, 20, 22, 23, 24, 26, 29, 30, 33, 34, 35, 38, 40, 41, 42, 44, 45, 46, 47, 51, 53, 54, 55, 56, 58, 59, 60, 62, 65, 66, 68, 69, 70, 71, 74, 77, 78, 80, 82, 83, 85, 86, 87, 88, 89, 90, 92, 94, 95, 96, 99, 101, 102, 104, 105, 106, 107, 110, 113, 114, 115
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Comments

n=10: sigma(10) = 1+2+5+10 = 18 = 3*6.

Crossrefs

Programs

  • Mathematica
    Select[Range@ 120, Divisible[DivisorSigma[1, #], 6] &] (* Michael De Vlieger, Feb 25 2017 *)
  • PARI
    isok(n) = !(sigma(n) % 6); \\ Michel Marcus, Dec 17 2013

Formula

A000203(n) modulo 6 = 0.
{n: A084301(n) = 0 }. - R. J. Mathar, May 19 2020
A087943 INTERSECT A028983. - R. J. Mathar, May 19 2020

A074628 Numbers k such that sigma(k) == 2 mod 6.

Original entry on oeis.org

7, 13, 19, 21, 28, 31, 37, 39, 43, 52, 57, 61, 63, 67, 73, 76, 79, 84, 93, 97, 103, 109, 111, 112, 117, 124, 127, 129, 139, 148, 151, 156, 157, 163, 171, 172, 175, 181, 183, 189, 193, 199, 201, 208, 211, 219, 223, 228, 229, 237, 241, 244, 252
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Examples

			For k=39: sigma(39) = 1+3+13+39 = 56 = 6*9 + 2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[300],Mod[DivisorSigma[1,#],6]==2&] (* Harvey P. Dale, Nov 14 2014 *)
  • PARI
    isok(n) = ((sigma(n) % 6) == 2); \\ Michel Marcus, Dec 19 2013

Formula

A000203(n) mod 6 = 2.
{n: A084301(n) = 2}. - R. J. Mathar, May 19 2020

A074630 Numbers k such that sigma(k) == 4 mod 6.

Original entry on oeis.org

3, 12, 27, 48, 75, 91, 108, 133, 192, 217, 243, 247, 259, 273, 300, 301, 343, 363, 364, 399, 403, 427, 432, 469, 481, 511, 532, 553, 559, 589, 651, 675, 679, 703, 721, 741, 763, 768, 777, 793, 817, 819, 867, 868, 871, 889, 903, 949, 972, 973, 988, 1027, 1029
Offset: 1

Views

Author

Labos Elemer, Aug 26 2002

Keywords

Examples

			k=48 is a term because sigma(48) = 1+2+3+4+6+8+12+16+24+48 = 124 = 6*20 + 4. [corrected by _Harvey P. Dale_, Jan 17 2013]
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1100],Mod[DivisorSigma[1,#],6]==4&] (* Harvey P. Dale, Jan 17 2013 *)

Formula

Mod(A000203(n), 6) = 4.
{n: A084301(n) = 4}. - R. J. Mathar, May 19 2020

A105827 a(n) = sigma(n) (mod 8).

Original entry on oeis.org

1, 3, 4, 7, 6, 4, 0, 7, 5, 2, 4, 4, 6, 0, 0, 7, 2, 7, 4, 2, 0, 4, 0, 4, 7, 2, 0, 0, 6, 0, 0, 7, 0, 6, 0, 3, 6, 4, 0, 2, 2, 0, 4, 4, 6, 0, 0, 4, 1, 5, 0, 2, 6, 0, 0, 0, 0, 2, 4, 0, 6, 0, 0, 7, 4, 0, 4, 6, 0, 0, 0, 3, 2, 2, 4, 4, 0, 0, 0, 2, 1, 6, 4, 0, 4, 4, 0, 4, 2, 2, 0, 0, 0, 0, 0, 4, 2, 3, 4, 1, 6, 0, 0, 2, 0
Offset: 1

Views

Author

Shyam Sunder Gupta, May 05 2005

Keywords

Crossrefs

Cf. A000203.
Sequences sigma(n) mod k: A053866 (k=2), A074941 (k=3), A105824 (k=4), A105825 (k=5), A084301 (k=6), A105826 (k=7), A105827 (k=8), A105852 (k=9), A105853 (k=10).

Programs

  • Maple
    A105827:= (n-> numtheory[sigma](n) mod 8):
    seq (A105827(n), n=1..105); # Jani Melik, Jan 26 2011
  • PARI
    a(n)=sigma(n)%8

A084300 a(n) = phi(n) mod 6.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 0, 4, 0, 4, 4, 4, 0, 0, 2, 2, 4, 0, 0, 2, 0, 4, 4, 2, 2, 0, 0, 0, 4, 2, 0, 4, 2, 4, 0, 0, 0, 0, 0, 4, 4, 0, 0, 2, 0, 4, 4, 4, 0, 2, 2, 0, 4, 0, 4, 0, 0, 4, 4, 4, 0, 0, 0, 2, 0, 2, 0, 2, 2, 0, 4, 0, 0, 0, 4, 0, 0, 0, 0, 2, 0, 4, 4, 0, 4, 0, 2, 4, 4, 0, 0, 2, 0, 4, 0, 2, 0, 0, 0, 4, 4, 2, 0, 0, 0
Offset: 1

Views

Author

Labos Elemer, Jun 02 2003

Keywords

Crossrefs

Programs

Formula

a(n) = A000010(n) mod 6.
a(n) = A010875(A000010(n)). - Amiram Eldar, Aug 17 2024

A105825 a(n) = sigma(n) (mod 5).

Original entry on oeis.org

1, 3, 4, 2, 1, 2, 3, 0, 3, 3, 2, 3, 4, 4, 4, 1, 3, 4, 0, 2, 2, 1, 4, 0, 1, 2, 0, 1, 0, 2, 2, 3, 3, 4, 3, 1, 3, 0, 1, 0, 2, 1, 4, 4, 3, 2, 3, 4, 2, 3, 2, 3, 4, 0, 2, 0, 0, 0, 0, 3, 2, 1, 4, 2, 4, 4, 3, 1, 1, 4, 2, 0, 4, 4, 4, 0, 1, 3, 0, 1, 1, 1, 4, 4, 3, 2, 0, 0, 0, 4, 2, 3, 3, 4, 0, 2, 3, 1, 1, 2, 2, 1, 4, 0, 2
Offset: 1

Views

Author

Shyam Sunder Gupta, May 05 2005

Keywords

Crossrefs

Cf. A000203.
Sequences sigma(n) mod k: A053866 (k=2), A074941 (k=3), A105824 (k=4), A105825 (k=5), A084301 (k=6), A105826 (k=7), A105827 (k=8).

Programs

  • Maple
    A105825:= n-> (numtheory[sigma](n) mod 5):
    seq (A105825(n), n=1..105); # Jani Melik, Jan 26 2011
  • Mathematica
    Mod[DivisorSigma[1,Range[110]],5] (* Harvey P. Dale, Oct 03 2013 *)
  • PARI
    a(n)=sigma(n)%5
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