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 1-4 of 4 results.

A130423 Main diagonal of array A[k,n] = n-th sum of 3 consecutive k-gonal numbers, k>2.

Original entry on oeis.org

4, 14, 39, 88, 170, 294, 469, 704, 1008, 1390, 1859, 2424, 3094, 3878, 4785, 5824, 7004, 8334, 9823, 11480, 13314, 15334, 17549, 19968, 22600, 25454, 28539, 31864, 35438, 39270, 43369, 47744, 52404, 57358, 62615, 68184, 74074, 80294, 86853
Offset: 1

Views

Author

Jonathan Vos Post, May 26 2007

Keywords

Comments

The first row of the array is the sum of 3 consecutive triangular numbers = A000217(n) + A000217(n+1) + A000217(n+2) = Centered triangular numbers: 3*n*(n-1)/2 + 1, for n>1. The second row of the array is the sum of 3 consecutive squares = Number of points on surface of square pyramid: 3*n^2 + 2 (n>1). The first column of the array is k+1 = 4, 5, 6, 7, 8, 9, ... The second column of the array is A016825 = 4*n + 2 (for n>2). The third column of the array is A017377 = 10*n + 9 (for n>0).

Examples

			The array begins:
k / A[k,n]
3.|.4.10.19.31..46..64..85.109.136.166....=A005448(n+1).
4.|.5.14.29..50..77.110.149.194.245.302...=A005918(n).
5.|.6.18.39..69.108.156.213.279.354.438...=A129863(n).
6.|.7.22.49..88.139.202.277.364.463.574...
7.|.8.26.59.107.170.248.341.449.572.710...
8.|.9.30.69.126.201.294.405.534.681.846...
		

Crossrefs

Programs

  • Magma
    I:=[4, 14, 39, 88]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012
  • Maple
    P := proc(k,n) n*((k-2)*n-k+4)/2 ; end: A := proc(k,n) add( P(k,i),i=n..n+2) ; end: A130423 := proc(n) A(n+3,n) ; end: seq(A130423(n),n=0..40) ; # R. J. Mathar, Jun 14 2007
  • Mathematica
    CoefficientList[Series[(4-2*x+7*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)
    Table[n (3n^2-3n+8)/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{4,14,39,88},40] (* Harvey P. Dale, Aug 15 2012 *)

Formula

a(n) = A[n+2,n] = P(k+2,n) + P(k+2,n+1) + P(k+2,n+2) where P(k,n) = k*((n-2)*k - (n-4))/2.
a(n) = n*(3*n^2-3*n+8)/2. G.f.: x*(4-2*x+7*x^2)/(1-x)^4. [Colin Barker, Apr 30 2012]
a(1)=4, a(2)=14, a(3)=39, a(4)=88, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Aug 15 2012

Extensions

More terms from R. J. Mathar, Jun 14 2007

A129109 Sums of three consecutive hexagonal numbers.

Original entry on oeis.org

7, 22, 49, 88, 139, 202, 277, 364, 463, 574, 697, 832, 979, 1138, 1309, 1492, 1687, 1894, 2113, 2344, 2587, 2842, 3109, 3388, 3679, 3982, 4297, 4624, 4963, 5314, 5677, 6052, 6439, 6838, 7249, 7672, 8107, 8554, 9013, 9484, 9967, 10462, 10969, 11488, 12019, 12562
Offset: 0

Views

Author

Jonathan Vos Post, May 24 2007

Keywords

Comments

Arises in hexagonal number analog to A129803 Triangular numbers which are the sum of three consecutive triangular numbers. What are the hexagonal numbers which are the sum of three consecutive hexagonal numbers? Prime for a(0) = 7, a(4) = 139, a(6) = 277, a(8) = 463, a(18) = 2113, a(22) = 3109, a(26) = 4297, a(38) = 9013, a(40) = 9967.

Examples

			a(0) = H(0) + H(1) + H(2) = 0 + 1 + 6 = 7 = 6*0^2 + 9*0 + 7.
a(1) = H(1) + H(2) + H(3) = 1 + 6 + 15 = 22 = 6*1^2 + 9*1 + 7.
a(2) = H(2) + H(3) + H(4) = 6 + 15 + 28 = 49 = 6*2^2 + 9*2 + 7.
		

Crossrefs

Programs

  • Magma
    I:=[7,22,49]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012
    
  • Mathematica
    LinearRecurrence[{3,-3,1},{7,22,49},50] (* Vincenzo Librandi, Feb 20 2012 *)
    Total/@Partition[PolygonalNumber[6,Range[0,50]],3,1] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 14 2020 *)
  • PARI
    a(n)=6*n^2+9*n+7 \\ Charles R Greathouse IV, Feb 20 2012

Formula

a(n) = H(n) + H(n+1) + H(n+2) where H(n) = A000384(n) = n*(2*n-1).
a(n) = 6*n^2 + 9*n + 7.
From Colin Barker, Feb 20 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (7 + x + 4*x^2)/(1-x)^3. (End)
E.g.f.: (7 + 15*x + 6*x^2)*exp(x). - Elmo R. Oliveira, Nov 16 2024

A129111 Sums of three consecutive heptagonal numbers.

Original entry on oeis.org

8, 26, 59, 107, 170, 248, 341, 449, 572, 710, 863, 1031, 1214, 1412, 1625, 1853, 2096, 2354, 2627, 2915, 3218, 3536, 3869, 4217, 4580, 4958, 5351, 5759, 6182, 6620, 7073, 7541, 8024, 8522, 9035, 9563, 10106, 10664, 11237, 11825, 12428, 13046, 13679, 14327, 14990
Offset: 0

Views

Author

Jonathan Vos Post, May 24 2007

Keywords

Comments

Arises in heptagonal number analog to A129803 (Triangular numbers which are the sum of three consecutive triangular numbers).
What are the heptagonal numbers which are the sum of three consecutive heptagonal numbers?
Prime for a(2) = 59, a(3) = 107, a(7) = 449, a(10) = 863, a(11) = 1031, a(23) = 4217, a(26) = 5351, a(31) = 7541, a(42) = 13679, a(43) = 14327, a(46) = 16361, a(51) = 20051.

Examples

			a(0) = Hep(0) + Hep(1) + Hep(2) = 0 + 1 + 7 = 8 = (15/2)*0^2 + (21/2)*0 + 8.
a(1) = Hep(1) + Hep(2) + Hep(3) = 1 + 7 + 18 = 26 = (15/2)*1^2 + (21/2)*1 + 8.
a(2) = Hep(2) + Hep(3) + Hep(4) = 7 + 18 + 34 = 59 = (15/2)*2^2 + (21/2)*2 + 8.
		

Crossrefs

Programs

  • Magma
    I:=[8,26,59]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012
    
  • Mathematica
    LinearRecurrence[{3,-3,1},{8,26,59},50] (* Vincenzo Librandi, Feb 12 2012 *)
  • PARI
    a(n)=3*n*(5*n+7)/2+8 \\ Charles R Greathouse IV, Jun 17 2017
    
  • Python
    def a(n): return 3*n*(5*n+7)//2 + 8
    print([a(n) for n in range(44)]) # Michael S. Branicky, Aug 26 2021

Formula

a(n) = Hep(n) + Hep(n+1) + Hep(n+2) where Hep(n) = A000566(n) = n*(5*n-3)/2.
a(n) = (15/2)*n^2 + (21/2)*n + 8.
From Colin Barker, Feb 20 2012: (Start)
G.f.: (8 + 2*x + 5*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
E.g.f.: exp(x)*(16 + 36*x + 15*x^2)/2. - Elmo R. Oliveira, Nov 16 2024

A130424 Main diagonal of array A[k,n] = n-th sum of k consecutive k-gonal numbers, k>2.

Original entry on oeis.org

4, 30, 125, 365, 854, 1724, 3135, 5275, 8360, 12634, 18369, 25865, 35450, 47480, 62339, 80439, 102220, 128150, 158725, 194469, 235934, 283700, 338375, 400595, 471024, 550354, 639305, 738625, 849090, 971504, 1106699, 1255535, 1418900
Offset: 1

Views

Author

Jonathan Vos Post, May 26 2007

Keywords

Comments

The first row of the array is the sum of 3 consecutive triangular numbers = A000217(n) + A000217(n+1) + A000217(n+2) = Centered triangular numbers: 3*n*(n-1)/2 + 1, for n>1. The second row of the array is the sum of 4 consecutive squares = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = A027575(n). The third row of the array is the sum of 5 consecutive pentagonal numbers.

Examples

			The array begins:
k / A[k,n]
3.|...4..10..19...31...46...64...85..109.136.166...=A005448(n+1).
4.|..14..30..54...86..126..174..230..294.366.446...=A027575(n).
5.|..40..75.125..190..270..365..475..600.740...
6.|..95.161.251..365..503..665..851.1061.1295...
7.|.196.308.455..637..854.1106.1393.1715.2072...
8.|.364.540.764.1036.1356.1724.2140.2604.3116...
		

Crossrefs

Programs

  • Maple
    P := proc(k,n) n*((k-2)*n-k+4)/2 ; end: A := proc(k,n) add( P(k,i),i=n..n+k-1) ; end: A130424 := proc(n) A(n+3,n) ; end: seq(A130424(n),n=0..40) ; # R. J. Mathar, Oct 28 2007
  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{4,30,125,365,854},50] (* Harvey P. Dale, Jun 23 2020 *)

Formula

a(n) = A[n+2,n] = P(k+2,n) + P(k+2,n+1) + P(k+2,n+2) + ... P(k+2,n+k-1) where P(k,n) = k*((n-2)*k - (n-4))/2.
a(n) = (n+2)*(7*n^3-8*n^2+12*n-3)/6. [R. J. Mathar, Oct 30 2008]
G.f.: x*(4+10*x+15*x^2-x^4)/(1-x)^5. [Colin Barker, Sep 08 2012]

Extensions

More terms from R. J. Mathar, Oct 28 2007
Showing 1-4 of 4 results.