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.

A134153 a(n) = 15*n^2 + 9*n + 1.

Original entry on oeis.org

1, 25, 79, 163, 277, 421, 595, 799, 1033, 1297, 1591, 1915, 2269, 2653, 3067, 3511, 3985, 4489, 5023, 5587, 6181, 6805, 7459, 8143, 8857, 9601, 10375, 11179, 12013, 12877, 13771, 14695, 15649, 16633, 17647, 18691, 19765, 20869, 22003, 23167
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A119617 is union of A134153 and A134154. A000538(n) is divisible by A000330(n) if and only n is congruent to {1, 3} mod 5 (see A047219). A134154(n) is case when n is congruent to 1 mod 5 see cases 2)

Crossrefs

Programs

  • Mathematica
    Table[1 + 9 n + 15 n^2, {n, 0, 50}]
    Table[Sum[k^4, {k, 1, 5m + 1}]/Sum[k^2, {k, 1, 5m + 1}], {m, 0, 10}]
  • PARI
    a(n)=15*n^2+9*n+1 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 15*n^2 + 9*n + 1.
a(n) = (3*(5*n + 1)^2 + 3*(5*n + 1) - 1)/5.
a(n) = (Sum_{k=1..5*n+1} k^4) / (Sum_{k=1..5*n+1} k^2).
G.f.: -(1+22*x+7*x^2)/(-1+x)^3. - R. J. Mathar, Nov 14 2007

Extensions

Offset corrected and some punctuation added by R. J. Mathar, Jul 09 2009

A134154 a(n) = 15*n^2 - 9*n + 1.

Original entry on oeis.org

1, 7, 43, 109, 205, 331, 487, 673, 889, 1135, 1411, 1717, 2053, 2419, 2815, 3241, 3697, 4183, 4699, 5245, 5821, 6427, 7063, 7729, 8425, 9151, 9907, 10693, 11509, 12355, 13231, 14137, 15073, 16039, 17035, 18061, 19117, 20203, 21319, 22465, 23641
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A119617 is union of A134153 and A134154 A000538(n) is divisible by A000330(n) if and only n is congruent to {1, 3} mod 5 (see A047219) A134154(n) is case when n is congruent to 3 mod 5 see cases 2)

Crossrefs

Programs

  • Mathematica
    Table[1 - 9 n + 15 n^2, {n, 0, 50}]
    Table[Sum[k^4, {k, 1, 5m + 3}]/Sum[k^2, {k, 1, 5m + 3}], {m, 0, 30}]
  • PARI
    a(n)=15*n^2-9*n+1 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 15*n^2 - 9*n + 1.
a(n+1) = (3*(5*n + 3)^2 + 3*(5*n + 3) - 1)/5.
a(n+1) = (Sum_{k=1..5*n+3} k^4) / (Sum_{k=1..5*n+3} k^2).
G.f.: -(1+4*x+25*x^2)/(-1+x)^3. - R. J. Mathar, Nov 14 2007

A134158 a(n) = 1 + 27n + 252n^2 + 882n^3 + 1029n^4.

Original entry on oeis.org

1, 2191, 24583, 109513, 324013, 759811, 1533331, 2785693, 4682713, 7414903, 11197471, 16270321, 22898053, 31369963, 42000043, 55126981, 71114161, 90349663, 113246263, 140241433, 171797341, 208400851, 250563523, 298821613, 353736073, 415892551, 485901391
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000540(n) is divisible by A000330(n) if and only if n is congruent to {1,2,4,5} mod 7 (see A047380).
This sequence is the case when n is congruent to 1 mod 7.
A134159 is the case when n is congruent to 2 mod 7.
A134160 is the case when n is congruent to 4 mod 7.
A134161 is the case when n is congruent to 5 mod 7.
A133180 is the union of this sequence, A134159, A134160, and A134161.

Crossrefs

Programs

  • Mathematica
    Table[(3(7n + 1)^4 + 6(7n + 1)^3 - 3 (7n + 1) + 1)/7, {n, 0, 100}] (* or *) Table[Sum[k^6, {k, 1, 7n + 1}]/Sum[k^2, {k, 1, 7n + 1}], {n, 0, 100}] (* Artur Jasinski *)
  • PARI
    Vec((1 + 2186*x + 13638*x^2 + 8498*x^3 + 373*x^4) / (1 - x)^5 + O(x^30)) \\ Colin Barker, Aug 12 2017

Formula

a(n) = (3(7n + 1)^4 + 6(7n + 1)^3 - 3 (7n + 1) + 1)/7.
a(n) = (Sum_{k=1..7n+1} k^6) / (Sum_{k=1..7n+1} k^2).
G.f.: -(1 + 2186*x + 13638*x^2 + 8498*x^3 + 373*x^4)/(-1+x)^5. - R. J. Mathar, Nov 14 2007
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4. - Colin Barker, Aug 12 2017

A134159 a(n) = 13 + 165*n + 756*n^2 + 1470*n^3 + 1029*n^4.

Original entry on oeis.org

13, 3433, 31591, 130351, 370273, 846613, 1679323, 3013051, 5017141, 7885633, 11837263, 17115463, 23988361, 32748781, 43714243, 57226963, 73653853, 93386521, 116841271, 144459103, 176705713, 214071493, 257071531, 306245611
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000540(n) is divisible by A000330(n) if and only if n is congruent to {1,2,4,5} mod 7 (see A047380). A134158 is the case when n is congruent to 1 mod 7. A134159 is the case when n is congruent to 2 mod 7. A134160 is the case when n is congruent to 4 mod 7. A134161 is the case when n is congruent to 5 mod 7. A133180 is the union of A134158 and A134159 and A134160 and A134161.

Crossrefs

Programs

  • Mathematica
    Table[(3(7n + 2)^4 + 6(7n + 2)^3 - 3 (7n + 2) + 1)/7, {n, 0, 100}]
    Table[Sum[k^6, {k, 1, 7n + 2}]/Sum[k^2, {k, 1, 7n + 2}], {n, 0, 100}] (* Artur Jasinski *)

Formula

a(n) = (3*(7*n + 2)^4 + 6*(7*n + 2)^3 - 3*(7*n + 2) + 1)/7.
a(n) = (Sum_{k=1..7n+2} k^6) / (Sum_{k=1..7n+2} k^2).
G.f.: -(13+3368*x+14556*x^2+6596*x^3+163*x^4)/(-1+x)^5. - R. J. Mathar, Nov 14 2007

A134160 a(n) = 163 + 1053*n + 2520*n^2 + 2646*n^3 + 1029*n^4.

Original entry on oeis.org

163, 7411, 49981, 180793, 477463, 1042303, 2002321, 3509221, 5739403, 8893963, 13198693, 18904081, 26285311, 35642263, 47299513, 61606333, 78936691, 99689251, 124287373, 153179113, 186837223, 225759151, 270467041, 321507733
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000540(n) is divisible by A000330(n) if and only n is congruent to {1,2,4,5} mod 7 (see A047380) A134158 is case when n is congruent to 1 mod 7 A134159 is case when n is congruent to 2 mod 7 A134160 is case when n is congruent to 4 mod 7 A134161 is case when n is congruent to 5 mod 7 A133180 is union of A134158 and A134159 and A134160 and A134161

Crossrefs

Programs

  • Mathematica
    Table[(3(7n + 4)^4 + 6(7n + 4)^3 - 3 (7n + 4) + 1)/7, {n, 0, 100}] (*Artur Jasinski*)
    Table[Sum[k^6, {k, 1, 7n + 4}]/Sum[k^2, {k, 1, 7n + 4}], {n, 0, 100}] (*Artur Jasinski*)
    LinearRecurrence[{5,-10,10,-5,1},{163,7411,49981,180793,477463},30] (* Harvey P. Dale, Jul 20 2024 *)
  • PARI
    a(n)=163+1053*n+2520*n^2+2646*n^3+1029*n^4 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = (3*(7*n + 4)^4 + 6*(7*n + 4)^3 - 3*(7*n + 4) + 1)/7.
a(n) = sum(k=1..7*n+4, k^6) / sum(k=1..7*n+4, k^2).
G.f.: (163+6596*x+14556*x^2+3368*x^3+13*x^4)/(1-x)^5. - Colin Barker, May 25 2012

A133180 a(n) = (Sum_{k=1..A047380(n)} k^6) / (Sum_{k=1..A047380(n)} k^2).

Original entry on oeis.org

1, 13, 163, 373, 2191, 3433, 7411, 10363, 24583, 31591, 49981, 61723, 109513, 130351, 180793, 210901, 324013, 370273, 477463, 539041, 759811, 846613, 1042303, 1151983, 1533331, 1679323, 2002321, 2180263, 2785693, 3013051, 3509221
Offset: 1

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000540(n) is divisible by A000330(n) if and only if n is congruent to {1,2,4,5} mod 7 (see A047380).
This sequence is the union of A134158 and A134159 and A134160 and A134161.

Crossrefs

Programs

  • Mathematica
    a = {}; Do[j = Sum[k^6, {k, 1, n}]/Sum[k^2, {k, 1, n}]; If[IntegerQ[j], AppendTo[a, j]], {n, 1, 100}] ; a (*Artur Jasinski*)
    Select[Table[Sum[k^6,{k,n}]/Sum[k^2,{k,n}],{n,100}],IntegerQ] (* Harvey P. Dale, Nov 26 2019 *)

Formula

a(n) = A000540(A047380(n)) / A000330(A047380(n)). - Jason Yuen, Sep 23 2024

Extensions

Offset corrected by Jason Yuen, Sep 23 2024

A134161 a(n) = 373 + 1947*n + 3780*n^2 + 3234*n^3 + 1029*n^4.

Original entry on oeis.org

373, 10363, 61723, 210901, 539041, 1151983, 2180263, 3779113, 6128461, 9432931, 13921843, 19849213, 27493753, 37158871, 49172671, 63887953, 81682213, 102957643, 128141131, 157684261, 192063313, 231779263, 277357783, 329349241
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000540(n) is divisible by A000330(n) if and only n is congruent to {1,2,4,5} mod 7 (see A047380) A134158 is case when n is congruent to 1 mod 7 A134159 is case when n is congruent to 2 mod 7 A134160 is case when n is congruent to 4 mod 7 A134161 is case when n is congruent to 5 mod 7 A133180 is union of A134158 and A134159 and A134160 and A134161.

Crossrefs

Programs

  • Mathematica
    Table[(3(7n + 5)^4 + 6(7n + 5)^3 - 3 (7n + 5) + 1)/7, {n, 0, 100}]
    Table[Sum[k^6, {k, 1, 7n + 5}]/Sum[k^2, {k, 1, 7n + 5}], {n, 0, 100}]
    LinearRecurrence[{5,-10,10,-5,1},{373,10363,61723,210901,539041},100] (* Harvey P. Dale, Nov 25 2012 *)
  • PARI
    a(n)=373+1947*n+3780*n^2+3234*n^3+1029*n^4 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = (3*(7*n + 5)^4 + 6*(7*n + 5)^3 - 3*(7*n + 5) + 1)/7.
a(n) = (Sum_{k=1..7*n+5} k^6) / (Sum_{k=1..7*n+5} k^2).
G.f.: -(373+8498*x+13638*x^2+2186*x^3+x^4)/(-1+x)^5. - R. J. Mathar, Nov 14 2007
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) with a(0)=373, a(1)=10363, a(2)=61723, a(3)=210901, and a(4)=539041. - Harvey P. Dale, Nov 25 2012

A134155 a(n) = 1 + 21 n + 168 n^2 + 588 n^3 + 1029 n^4.

Original entry on oeis.org

1, 1807, 21883, 100801, 303829, 720931, 1466767, 2680693, 4526761, 7193719, 10895011, 15868777, 22377853, 30709771, 41176759, 54115741, 69888337, 88880863, 111504331, 138194449, 169411621, 205640947, 247392223, 295199941
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

All terms == 1 (mod 21). - Robert Israel, Aug 11 2017

Crossrefs

Programs

  • Maple
    seq( 1 + 21*n + 168*n^2 + 588*n^3 + 1029*n^4,n=0..30); # Robert Israel, Aug 11 2017
  • Mathematica
    Table[1 + 21 n + 168 n^2 + 588 n^3 + 1029 n^4,{n,0,50}]
    LinearRecurrence[{5,-10,10,-5,1},{1,1807,21883,100801,303829},30] (* Harvey P. Dale, Aug 29 2021 *)
  • PARI
    a(n)=1+21*n+168*n^2+588*n^3+1029*n^4 \\ Charles R Greathouse IV, Oct 21 2022

Formula

a(n) = (3*(7*n + 1)^4 + 6*(7*n + 1)^2 - 3*(7*n + 1) + 1)/7.
G.f.: -(1+1802*x+12858*x^2+9446*x^3+589*x^4)/(-1+x)^5. - R. J. Mathar, Nov 14 2007

A134163 1 + 12*n + 81*n^3 + n*(105*n + 81*n^3)/2.

Original entry on oeis.org

1, 187, 1531, 5977, 16441, 36811, 71947, 127681, 210817, 329131, 491371, 707257, 987481, 1343707, 1788571, 2335681, 2999617, 3795931, 4741147, 5852761, 7149241, 8650027, 10375531, 12347137, 14587201, 17119051, 19966987, 23156281, 26713177
Offset: 0

Views

Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000541(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777).

Crossrefs

Programs

  • Magma
    [1 + 12*n + 81*n^3 + n*(105*n+ 81*n^3)/2: n in [0..30]]; // Vincenzo Librandi, May 09 2011
  • Maple
    A134163:=n->1 + 12*n + 81*n^3 + n*(105*n + 81*n^3)/2: seq(A134163(n), n=0..30); # Wesley Ivan Hurt, Oct 23 2014
  • Mathematica
    Table[(3(3n + 1)^4 + 6(3n + 1)^3 - (3n + 1)^2 - 4 (3n + 1) + 2)/6, {n, 0, 100}] (* or *) Table[Sum[k^7, {k, 1, 3n + 1}]/Sum[k^3, {k, 1, 3n + 1}], {n, 0, 100}]

Formula

a(n) = (3(3n + 1)^4 + 6(3n + 1)^3 - (3n + 1)^2 - 4 (3n + 1) + 2)/6.
a(n) = ( sum_{k=1..3n+1} k^7 ) / ( sum_{k=1..3n+1} k^3 ).
G.f.: (1+182*x+606*x^2+182*x^3+x^4)/(1-x)^5. - R. J. Mathar, Nov 14 2007
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Wesley Ivan Hurt, Oct 23 2014

A192328 Numbers of the form 20*k+7 which are three times a square.

Original entry on oeis.org

27, 147, 507, 867, 1587, 2187, 3267, 4107, 5547, 6627, 8427, 9747, 11907, 13467, 15987, 17787, 20667, 22707, 25947, 28227, 31827, 34347, 38307, 41067, 45387, 48387, 53067, 56307, 61347, 64827, 70227, 73947, 79707, 83667, 89787, 93987
Offset: 1

Views

Author

Bruno Berselli, Jun 28 2011

Keywords

Comments

Text of the theorem in the paper mentioned in References: The necessary and sufficient condition so that a number of the form 20*k+7 is three times a square is that k is of the form 3*h*(5*h+3)+1 or 3*h*(5*h+7)+7.
A119617 gives the values of k.
A080512*120 gives the first differences.

References

  • "Supplemento al Periodico di Matematica", Raffaello Giusti Editore (Livorno) - Mar 1901 - p. 75 (Problem 286 and its generalization, G. Cardoso-Laynes).

Crossrefs

Programs

  • Magma
    [m: m in [7..10^5 by 20] | IsSquare(m/3)];
    
  • Maple
    select(t -> issqr(t/3), [seq(20*i+7,i=1..10000,3)]); # Robert Israel, Apr 28 2023
  • Mathematica
    Select[20 Range[5000] + 7, IntegerQ[Sqrt[#/3]] &] (* or *) LinearRecurrence[{1,2,-2,-1,1}, {27,147,507,867,1587}, 40] (* Harvey P. Dale, Jul 06 2011 *)
    CoefficientList[Series[3 (9 + 40 x + 102 x^2 + 40 x^3 + 9 x^4) / ((1 + x)^2 (1 - x)^3), {x, 0, 35}], x] (* Vincenzo Librandi, Aug 19 2013 *)
  • PARI
    for(k=0, 5*10^3, m=20*k+7; if(issquare(m/3), print1(m",")));
    
  • PARI
    a(n)=my(m=n--\4); 1200*m^2+[360*m+27, 840*m+147, 1560*m+507, 2040*m+867][n%4+1] \\ Charles R Greathouse IV, Jun 29 2011

Formula

G.f.: 3*x*(9 + 40*x + 102*x^2 + 40*x^3 + 9*x^4)/((1 + x)^2*(1 - x)^3).
a(n) = 3*((10*(n-1) + (-1)^(n-1) + 5)/2)^2.
a(n) = a(-n-1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
Sum_{i=1..n} a(i) = n*(50*(n-1)*(n+1) + 15*(-1)^(n-1) + 39)/2.
a(n) = 3*A020742(A047218(n))^2.

Extensions

Offset corrected by Mohammed Yaseen, Apr 27 2023
Showing 1-10 of 10 results.