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.

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A264850 a(n) = n*(n + 1)*(n + 2)*(7*n - 5)/12.

Original entry on oeis.org

0, 1, 18, 80, 230, 525, 1036, 1848, 3060, 4785, 7150, 10296, 14378, 19565, 26040, 34000, 43656, 55233, 68970, 85120, 103950, 125741, 150788, 179400, 211900, 248625, 289926, 336168, 387730, 445005, 508400, 578336, 655248, 739585, 831810, 932400, 1041846
Offset: 0

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Author

Ilya Gutkovskiy, Nov 26 2015

Keywords

Comments

Partial sums of 16-gonal (or hexadecagonal) pyramidal numbers. Therefore, this is the case k=7 of the general formula n*(n + 1)*(n + 2)*(k*n - k + 2)/12, which is related to 2*(k+1)-gonal pyramidal numbers.

Links

Crossrefs

Cf. A172076.
Cf. similar sequences with formula n*(n+1)*(n+2)*(k*n-k+2)/12: A000292 (k=0), A002415 (which arises from k=1), A002417 (k=2), A002419 (k=3), A051797 (k=4), A051799 (k=5), A220212 (k=6), this sequence (k=7), A264851 (k=8), A264852 (k=9).

Programs

  • Magma
    [n*(n+1)*(n+2)*(7*n-5)/12: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015
  • Mathematica
    Table[n (n + 1) (n + 2) (7 n - 5)/12, {n, 0, 50}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,18,80,230},40] (* Harvey P. Dale, Sep 27 2018 *)
  • PARI
    a(n)=n*(n+1)*(n+2)*(7*n-5)/12 \\ Charles R Greathouse IV, Jul 26 2016

Formula

G.f.: x*(1 + 13*x)/(1 - x)^5.
a(n) = Sum_{k = 0..n} A172076(k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015

A346083 Triangle, read by rows, defined by recurrence: T(n,k) = T(n-1,k-1) + (-1)^k * (2 * k + 1) * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0 and T(i,j) = 0 if j < 0 or j > i.

Original entry on oeis.org

1, 1, 1, 1, -2, 1, 1, 7, 3, 1, 1, -20, 22, -4, 1, 1, 61, 90, 50, 5, 1, 1, -182, 511, -260, 95, -6, 1, 1, 547, 2373, 2331, 595, 161, 7, 1, 1, -1640, 12412, -13944, 7686, -1176, 252, -8, 1, 1, 4921, 60420, 110020, 55230, 20622, 2100, 372, 9, 1, 1, -14762, 307021, -709720, 607090, -171612, 47922, -3480, 525, -10, 1
Offset: 0

Views

Author

Werner Schulte, Jul 04 2021

Keywords

Examples

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :  0      1      2       3      4      5     6    7  8  9
=============================================================
  0 :  1
  1 :  1      1
  2 :  1     -2      1
  3 :  1      7      3       1
  4 :  1    -20     22      -4      1
  5 :  1     61     90      50      5      1
  6 :  1   -182    511    -260     95     -6     1
  7 :  1    547   2373    2331    595    161     7    1
  8 :  1  -1640  12412  -13944   7686  -1176   252   -8  1
  9 :  1   4921  60420  110020  55230  20622  2100  372  9  1
  etc.

Crossrefs

Cf. A000012 (column 0 and main diagonal), A014983 (column 1), A181983 (1st subdiagonal), A002412 (2nd subdiagonal), A264851 (3rd subdiagonal without signs).
Cf. A051159.

Programs

  • Python
    from functools import cache
@cache
def T(n, k):
    if k == 0 or k == n: return 1
    return T(n-1, k-1) + (-1)**k*(2*k + 1)*T(n-1, k)
for n in range(10):
    print([T(n, k) for k in range(n+1)]) # Peter Luschny, Jul 22 2021

Formula

G.f. of column k >= 0: col(t,k) = Sum_{n >= k} T(n,k) * t^n = t^k / (Product_{i=0..k} (1 - (-1)^i * (2 * i + 1) * t)), i.e., col(t,k) = col(t,k-1) * t / (1 - (-1)^k * (2 * k + 1) * t) for k > 0.
Matrix inverse M = T^(-1) has row polynomials p(n,x) = Sum_{k=0..n} M(n,k) * x^k = Product_{i=1..n} (x + (-1)^i * (2 * i - 1)) for n >= 0 and empty product 1, i.e., p(n,x) = p(n-1,x) * (x + (-1)^n * (2 * n - 1)) for n > 0 with initial value p(0,x) = 1.
Conjecture: E.g.f. of column k >= 0: Sum_{n >= k} T(n,k) * t^n / (n!) = (Sum_{i=0..k} (-1)^(i * (i + 1 ) / 2) * binomial(k,floor((k - i) / 2)) * exp((-1)^i * (2 * i + 1) * t)) * (-1)^(k * (k - 1) / 2) / (4^k * (k!)), i.e., T(n,k) = (Sum_{i=0..k} (-1)^(i * (i + 1) / 2) * binomial(k,floor((k - i) / 2)) * ((-1)^i * (2 * i + 1))^n) * (-1)^(k * (k - 1) / 2) / (4^k * (k!)) for 0 <= k <= n.
Conjecture: E.g.f. of column k >= 0: Sum_{n >= k} T(n,k) * t^n / (n!) = exp(t) * (exp(4*t) - 1)^k / (4^k * (k!) * exp(4*t*floor((k+1)/2))), i.e., T(n,k) = (Sum_{i=0..k} (-1)^i * binomial(k,i) *(1 + 4*i - 4*floor((k+1)/2))^n) * (-1)^k / (4^k * (k!)) for 0 <= k <= n. Proved by Burkhard Hackmann and Werner Schulte (distinction of two cases: odd k, even k). - Werner Schulte, Aug 03 2021
Showing 1-2 of 2 results.

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