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 1-5 of 5 results.

A063986 Numbers k that divide Sum_{j=1..k} A051953(j) where A051953(j) = j - Phi(j). Arithmetic mean of first k cototient values is an integer.

Original entry on oeis.org

1, 4, 5, 24, 25, 249, 600, 617, 12272, 13763, 21332, 25228, 783665, 15748051, 41846733, 195853251, 2488541984, 14399065016, 21119309213, 22430204140, 43787603128, 157825075944, 206651865067, 271605149320, 374049315076, 650288309748
Offset: 1

Views

Author

Labos Elemer, Sep 06 2001

Keywords

Comments

The odd terms of A048290 and A063986 are the same. - Jud McCranie, Jun 26 2005
a(27) > 10^12. - Donovan Johnson, Dec 09 2011

Examples

			k=5: (1 + 1 + 2 + 2 + 4)/5 = 2.
		

Crossrefs

Programs

  • Mathematica
    s = 0; Do[s = s + n - EulerPhi[n]; If[ IntegerQ[s/n], Print[n]], {n, 1, 10^7} ]

Extensions

More terms from Dean Hickerson, Sep 07 2001
One more term from Robert G. Wilson v, Sep 07 2001
a(16) and a(17) from Jud McCranie, Jun 22 2005
a(18)-a(21) from Donovan Johnson, May 11 2010
a(22)-a(26) from Donovan Johnson, Dec 09 2011

A048919 Indices of 9-gonal numbers which are also heptagonal.

Original entry on oeis.org

1, 88, 12445, 1767052, 250908889, 35627295136, 5058825000373, 718317522757780, 101996029406604337, 14482717858215058024, 2056443939837131635021, 292000556739014477114908
Offset: 1

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Author

Keywords

Comments

As n increases, this sequence is approximately geometric with common ratio r = lim_{n->oo} a(n)/a(n-1) = (6 + sqrt(35))^2 = 71 + 12*sqrt(35). - Ant King, Jan 01 2012

Crossrefs

Programs

  • Magma
    I:=[1, 88, 12445]; [n le 3 select I[n] else 143*Self(n-1)-143*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 21 2011
    
  • Mathematica
    LinearRecurrence[{143,-143,1},{1,88,12445},30] (* Vincenzo Librandi, Dec 21 2011 *)
  • Maxima
    makelist(expand((50+(25-3*sqrt(35))*(6+sqrt(35))^(2*n-1)+(25+3*sqrt(35))*(6-sqrt(35))^(2*n-1))/140), n, 1, 12); /* Bruno Berselli, Dec 21 2011 */

Formula

G.f.: -x*(1 - 55*x + 4*x^2) / ( (x-1)*(x^2 - 142*x + 1) ). - R. J. Mathar, Dec 21 2011
a(n) = (50 + (25-3r)*(6+r)^(2n-1) + (25+3r)*(6-r)^(2n-1))/140, where r=sqrt(35). - Bruno Berselli, Dec 21 2011
From Ant King, Jan 01 2012: (Start)
a(n) = 142*a(n-1) - a(n-2) - 50.
a(n) = ceiling(1/140*(45 + 7*sqrt(35))*(6 + sqrt(35))^(2*n - 2)). (End)

A048921 9-gonal heptagonal numbers (A000566).

Original entry on oeis.org

1, 26884, 542041975, 10928650279834, 220343446399977901, 4442564555387704166896, 89570986345383445012986019, 1805930222253056462964119954950, 36411165051495138060899141518722649, 734121907962314751330792028336366100956
Offset: 1

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Author

Keywords

Comments

As n increases, this sequence is approximately geometric with common ratio r = lim(n->Infinity,a(n)/a(n-1)) = (6+sqrt(35))^4 = 10081+1704*sqrt(35). - Ant King, Dec 31 2011

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{20163, -20163, 1}, {1, 26884, 542041975}, 9]; (* Ant King, Dec 31 2011 *)
  • PARI
    Vec(x*(1+6721*x+46*x^2)/((1-x)*(1-20162*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 22 2015

Formula

From Ant King, Dec 31 2011: (Start)
a(n) = 20163*a(n-1)-20163*a(n-2)+a(n-3).
a(n) = 20162*a(n-1)-a(n-2)+6768.
a(n) = 1/560*((39+4*sqrt(35))*(6+sqrt(35))^(4*n-3)+(39-4*sqrt(35))*(6-sqrt(35))^(4*n-3)-188).
a(n) = floor(1/560*(39+4*sqrt(35))*(6+sqrt(35))^(4*n-3)).
G.f.: x(1+6721*x+46*x^2) / ((1-x)(1-20162*x+x^2)).
(End)

A073128 Integer quotients arising in A063986.

Original entry on oeis.org

0, 1, 1, 5, 5, 49, 118, 121, 2406, 2698, 4182, 4946, 153627, 3087192, 8203485, 38394376, 487844934, 2822741576, 4140154385, 4397137572, 8583966231
Offset: 1

Views

Author

Labos Elemer, Jul 16 2002

Keywords

Examples

			n=15, A063986(15)=41846733, sum of first 41846733 cototients is s=343289046464505, a(15)=s/41846733. Only in knowledge of either the quotients or of partial sums of cototients is it possible to continue A063986 without recomputing previous terms!
		

Crossrefs

Formula

a(n)=Sum[cototient[j], j=1..A063986(n)]/A063986(n)

Extensions

Changed A063896 to A063986 and a(16)-a(21) from Donovan Johnson, May 11 2010

A073129 Partial sums of cototients arising in A063986.

Original entry on oeis.org

0, 4, 5, 120, 125, 12201, 70800, 74657, 29526432, 37132574, 89210424, 124777688, 120392102955, 48617257062792, 343289046464505, 7519663359716376, 1214022599940709056, 40644839476190305216, 87437200646372849005, 98628693371623948080, 375871306587181970568
Offset: 1

Views

Author

Labos Elemer, Jul 16 2002

Keywords

Examples

			n=15, a(15)=343289046464505, sum of first 41846733 cototients A063986(15)*A073128(15)=a(15) To continue A063986, A073128 or A073129 without recomputing previous terms, corresponding entries from 2 of above sequences is required.
		

Crossrefs

Formula

a(n)=Sum[cototient[j], j=1..A063986(n)]

Extensions

Changed A063896 to A063986 and a(16)-a(21) from Donovan Johnson, May 11 2010
Showing 1-5 of 5 results.