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 1-3 of 3 results.

A237616 a(n) = n*(n + 1)*(5*n - 4)/2.

Original entry on oeis.org

0, 1, 18, 66, 160, 315, 546, 868, 1296, 1845, 2530, 3366, 4368, 5551, 6930, 8520, 10336, 12393, 14706, 17290, 20160, 23331, 26818, 30636, 34800, 39325, 44226, 49518, 55216, 61335, 67890, 74896, 82368, 90321, 98770, 107730, 117216, 127243, 137826, 148980, 160720
Offset: 0

Views

Author

Bruno Berselli, Feb 10 2014

Keywords

Comments

Also 17-gonal (or heptadecagonal) pyramidal numbers.
This sequence is related to A226489 by 2*a(n) = n*A226489(n) - Sum_{i=0..n-1} A226489(i).

Examples

			After 0, the sequence is provided by the row sums of the triangle:
   1;
   2,  16;
   3,  32,  31;
   4,  48,  62,  46;
   5,  64,  93,  92,  61;
   6,  80, 124, 138, 122,  76;
   7,  96, 155, 184, 183, 152,  91;
   8, 112, 186, 230, 244, 228, 182, 106;
   9, 128, 217, 276, 305, 304, 273, 212, 121;
  10, 144, 248, 322, 366, 380, 364, 318, 242, 136; etc.,
where (r = row index, c = column index):
T(r,r) = T(c,c) = 15*r-14 and T(r,c) = T(r-1,c)+T(r,r) = (r-c+1)*T(r,r), with r>=c>0.
		

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93 (fifteenth row of the table).

Crossrefs

Cf. sequences with formula n*(n+1)*(k*n-k+3)/6: A000217 (k=0), A000292 (k=1), A000330 (k=2), A002411 (k=3), A002412 (k=4), A002413 (k=5), A002414 (k=6), A007584 (k=7), A007585 (k=8), A007586 (k=9), A007587 (k=10), A050441 (k=11), A172073 (k=12), A177890 (k=13), A172076 (k=14), this sequence (k=15), A172078(k=16), A237617 (k=17), A172082 (k=18), A237618 (k=19), A172117(k=20), A256718 (k=21), A256716 (k=22), A256645 (k=23), A256646(k=24), A256647 (k=25), A256648 (k=26), A256649 (k=27), A256650(k=28).

Programs

  • GAP
    List([0..40], n-> n*(n+1)*(5*n-4)/2); # G. C. Greubel, Aug 30 2019
  • Magma
    [n*(n+1)*(5*n-4)/2: n in [0..40]];
    
  • Magma
    I:=[0,1,18,66]; [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..50]]; // Vincenzo Librandi, Feb 12 2014
    
  • Maple
    seq(n*(n+1)*(5*n-4)/2, n=0..40); # G. C. Greubel, Aug 30 2019
  • Mathematica
    Table[n(n+1)(5n-4)/2, {n, 0, 40}]
    CoefficientList[Series[x (1+14x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
    LinearRecurrence[{4,-6,4,-1},{0,1,18,66},50] (* Harvey P. Dale, Jan 11 2015 *)
  • PARI
    a(n)=n*(n+1)*(5*n-4)/2 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Sage
    [n*(n+1)*(5*n-4)/2 for n in (0..40)] # G. C. Greubel, Aug 30 2019
    

Formula

G.f.: x*(1 + 14*x)/(1 - x)^4.
For n>0, a(n) = Sum_{i=0..n-1} (n-i)*(15*i+1). More generally, the sequence with the closed form n*(n+1)*(k*n-k+3)/6 is also given by Sum_{i=0..n-1} (n-i)*(k*i+1) for n>0.
a(n) = A104728(A001844(n-1)) for n>0.
Sum_{n>=1} 1/a(n) = (2*sqrt(5*(5 + 2*sqrt(5)))*Pi + 10*sqrt(5)*arccoth(sqrt(5)) + 25*log(5) - 16)/72 = 1.086617842136293176... . - Vaclav Kotesovec, Dec 07 2016
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n >= 4. - Wesley Ivan Hurt, Dec 18 2020
E.g.f.: exp(x)*x*(2 + 16*x + 5*x^2)/2. - Elmo R. Oliveira, Aug 04 2025

A180223 a(n) = (11*n^2 - 7*n)/2.

Original entry on oeis.org

0, 2, 15, 39, 74, 120, 177, 245, 324, 414, 515, 627, 750, 884, 1029, 1185, 1352, 1530, 1719, 1919, 2130, 2352, 2585, 2829, 3084, 3350, 3627, 3915, 4214, 4524, 4845, 5177, 5520, 5874, 6239, 6615, 7002, 7400, 7809, 8229, 8660
Offset: 0

Views

Author

Graziano Aglietti (mg5055(AT)mclink.it), Aug 16 2010

Keywords

Comments

This sequence is related to A050441 by n*a(n) - Sum_{i=0..n-1} a(i) = 2*A050441(n). - Bruno Berselli, Aug 19 2010
Sum of n-th heptagonal number (A000566) and n-th octagonal number (A000567). - Bruno Berselli, Jun 11 2013
Create a triangle with T(r,1) = r^2 and T(r,c) = r^2 + r*c + c^2. The difference of the sum of the terms in row n and those in row n-1 is a(n). - J. M. Bergot, Jun 17 2013

Crossrefs

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=11). - Bruno Berselli, Jun 10 2013

Programs

Formula

G.f.: x*(2+9*x)/(1-x)^3. - Bruno Berselli, Aug 19 2010 - corrected in Apr 18 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with n>2. - Bruno Berselli, Aug 19 2010
a(n) = n + A226492(n). - Bruno Berselli, Jun 11 2013
E.g.f.: x*(4 + 11*x)*exp(x)/2. - G. C. Greubel, Aug 24 2015

A055268 a(n) = (11*n + 4)*C(n+3, 3)/4.

Original entry on oeis.org

1, 15, 65, 185, 420, 826, 1470, 2430, 3795, 5665, 8151, 11375, 15470, 20580, 26860, 34476, 43605, 54435, 67165, 82005, 99176, 118910, 141450, 167050, 195975, 228501, 264915, 305515, 350610, 400520, 455576, 516120, 582505, 655095, 734265
Offset: 0

Views

Author

Barry E. Williams, May 10 2000

Keywords

Comments

a(n) is the number of compositions of n when there are 9 types of each natural number. - Milan Janjic, Aug 13 2010
Convolution of A000027 with A051865 (excluding 0). - Bruno Berselli, Dec 07 2012

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Crossrefs

Partial sums of A050441.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.

Programs

  • GAP
    List([0..30], n-> (11*n+4)*Binomial(n+3,3)/4 ); # G. C. Greubel, Jan 17 2020
  • Magma
    /* A000027 convolved with A051865 (excluding 0): */ A051865:=func; [&+[(n-i+1)*A051865(i): i in [1..n]]: n in [1..35]]; // Bruno Berselli, Dec 07 2012
    
  • Maple
    seq( (11*n+4)*binomial(n+3,3)/4, n=0..30); # G. C. Greubel, Jan 17 2020
  • Mathematica
    Table[11*Binomial[n+4,4] -10*Binomial[n+3,3], {n,0,30}] (* G. C. Greubel, Jan 17 2020 *)
  • PARI
    a(n) = (11*n+4)*binomial(n+3, 3)/4; \\ Michel Marcus, Sep 07 2017
    
  • Python
    A055268_list, m = [], [11, 1, 1, 1, 1]
    for _ in range(10**2):
        A055268_list.append(m[-1])
        for i in range(4):
            m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016
    
  • Sage
    [(11*n+4)*binomial(n+3,3)/4 for n in (0..30)] # G. C. Greubel, Jan 17 2020
    

Formula

G.f.: (1 + 10*x)/(1-x)^5. - R. J. Mathar, Oct 26 2011
From G. C. Greubel, Jan 17 2020:(Start)
a(n) = 11*binomial(n+4,4) - 10*binomial(n+3,3).
E.g.f.: (24 + 336*x + 432*x^2 + 136*x^3 + 11*x^4)*exp(x)/24. (End)
Showing 1-3 of 3 results.