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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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A316224 a(n) = n*(2*n + 1)*(4*n + 1).

Original entry on oeis.org

0, 15, 90, 273, 612, 1155, 1950, 3045, 4488, 6327, 8610, 11385, 14700, 18603, 23142, 28365, 34320, 41055, 48618, 57057, 66420, 76755, 88110, 100533, 114072, 128775, 144690, 161865, 180348, 200187, 221430, 244125, 268320, 294063, 321402, 350385, 381060, 413475, 447678, 483717
Offset: 0

Views

Author

Bruno Berselli, Jun 27 2018

Keywords

Comments

Sums of the consecutive integers from A000384(n) to A000384(n+1)-1. This is the case s=6 of the formula n*(n*(s-2) + 1)*(n*(s-2) + 2)/2 related to s-gonal numbers.
The inverse binomial transform is 0, 15, 60, 48, 0, ... (0 continued).

Examples

			Row sums of the triangle:
|  0 |  ................................................................. 0
|  1 |  2  3  4  5  .................................................... 15
|  6 |  7  8  9 10 11 12 13 14  ........................................ 90
| 15 | 16 17 18 19 20 21 22 23 24 25 26 27  ........................... 273
| 28 | 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44  ............... 612
| 45 | 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65  .. 1155
...
where:
. first column is A000384,
. second column is A130883 (without 1),
. third column is A033816,
. diagonal is A014106,
. 0, 2, 8, 18, 32, 50, ... are in A001105.

Links

Crossrefs

First bisection of A059270 and subsequence of A034828, A047866, A109900, A290168.
Sums of the consecutive integers from P(s,n) to P(s,n+1)-1, where P(s,k) is the k-th s-gonal number: A027480 (s=3), A055112 (s=4), A228888 (s=5).

Programs

  • GAP
    List([0..40], n -> n*(2*n+1)*(4*n+1));
  • Julia
    [n*(2*n+1)*(4*n+1) for n in 0:40] |> println
  • Magma
    [n*(2*n+1)*(4*n+1): n in [0..40]];
  • Maple
    seq(n*(2*n+1)*(4*n+1),n=0..40); # Muniru A Asiru, Jun 27 2018
  • Mathematica
    Table[n (2 n + 1) (4 n + 1), {n, 0, 40}]
  • Maxima
    makelist(n*(2*n+1)*(4*n+1), n, 0, 40);
  • PARI
    vector(40, n, n--; n*(2*n+1)*(4*n+1))
  • Python
    [n*(2*n+1)*(4*n+1) for n in range(40)]
  • Sage
    [n*(2*n+1)*(4*n+1) for n in (0..40)]

Formula

O.g.f.: 3*x*(5 + 10*x + x^2)/(1 - x)^4.
E.g.f.: x*(15 + 30*x + 8*x^2)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 3*A258582(n).
a(n) = -3*A100157(-n).
Sum_{n>0} 1/a(n) = 2*(3 - log(4)) - Pi.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) + 2*sqrt(2)*log(1+sqrt(2)) + (sqrt(2)-1/2)*Pi - 6. - Amiram Eldar, Sep 17 2022

A269845 Irregular triangle read by rows: T(n,k) = (k/2+1/2)^2 if odd-k otherwise T(n,k) = (n-k/2)^2 where n >= 1, k = 0..2*n-1.

Original entry on oeis.org

1, 1, 4, 1, 1, 4, 9, 1, 4, 4, 1, 9, 16, 1, 9, 4, 4, 9, 1, 16, 25, 1, 16, 4, 9, 9, 4, 16, 1, 25, 36, 1, 25, 4, 16, 9, 9, 16, 4, 25, 1, 36, 49, 1, 36, 4, 25, 9, 16, 16, 9, 25, 4, 36, 1, 49, 64, 1, 49, 4, 36, 9, 25, 16, 16, 25, 9, 36, 4, 49, 1, 64, 81, 1, 64, 4, 49, 9, 36, 16, 25, 25, 16, 36, 9, 49, 4, 64, 1, 81, 100, 1, 81, 4, 64, 9, 49, 16, 36, 25, 25, 36, 16, 49
Offset: 1

Views

Author

Kival Ngaokrajang, Mar 06 2016

Keywords

Comments

Inspired by A268317, but change to n+1 X n instead of Fib(n+1) X Fib(n).
There are triangles appearing along main diagonal. If the area of the smallest triangles are defined as 1, then the areas of all other triangles seem to be square numbers. Conjectures: (i) Odd terms of row sum/2 is A100157. (ii) Even terms of row sum/2 is A258582. See illustration in links.

Examples

			Irregular triangle begins:
n\k 0  1   2  3   4  5   6   7   8   9  10 11 12  13 14  15 ...
1   1, 1
2   4, 1,  1, 4
3   9, 1,  4, 4,  1, 9
4  16, 1,  9, 4,  4, 9,  1, 16
5  25, 1, 16, 4,  9, 9,  4, 16,  1, 25
6  36, 1, 25, 4, 16, 9,  9, 16,  4, 25, 1, 36
7  49, 1, 36, 4, 25, 9, 16, 16,  9, 25, 4, 36, 1, 49
8  64, 1, 49, 4, 36, 9, 25, 16, 16, 25, 9, 36, 4, 49, 1, 64
...

Links

Crossrefs

Cf. A100157, A258582, A268317.

Programs

  • Mathematica
    Table[If[OddQ@ k, (k/2 + 1/2)^2, (n - k/2)^2], {n, 8}, {k, 0, 2 n - 1}] // Flatten (* Michael De Vlieger, Apr 01 2016 *)
  • PARI
    for (n = 1, 20, for (k = 0, 2*n-1, if (Mod(k,2)==0, t = (n-k/2)^2, t = (k/2+1/2)^2); print1(t, ", ")))

Formula

T(n,k) = (k/2+1/2)^2 if odd-k, T(n,k) = (n-k/2)^2 if even-k; n >= 1, k = 0..2*n-1.
Showing 1-2 of 2 results.

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