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.

A154560 a(n) = (n+3)^2*n/2 + 1.

Original entry on oeis.org

1, 9, 26, 55, 99, 161, 244, 351, 485, 649, 846, 1079, 1351, 1665, 2024, 2431, 2889, 3401, 3970, 4599, 5291, 6049, 6876, 7775, 8749, 9801, 10934, 12151, 13455, 14849, 16336, 17919, 19601, 21385, 23274, 25271, 27379, 29601, 31940, 34399, 36981
Offset: 0

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Author

Klaus Brockhaus, Jan 12 2009

Keywords

Comments

8*a(n) is the y value of a solution (x, y) to the Diophantine equation 2*x^3+12*x^2 = y^2. The corresponding x value is A152811(n+1).

Examples

			a(5) = (5+3)^2*5/2+1 = 64*5/2+1 = 161.

Links

Crossrefs

Cf. A058794 (row 3 of A007754), A117560 (n*(n^2-1)/2-1), A144129 (4*n^3-3*n), A141530, A152811 (2*(n^2+2*n-2)).

Programs

  • Magma
    [(n+3)^2*n/2 + 1: n in [0..50]]; // Vincenzo Librandi, Sep 06 2011
  • PARI
    {for(n=0,40,print1((n+3)^2*n/2+1,","))}

Formula

G.f.: (1+5*x-4*x^2+x^3)/(1-x)^4.
a(n) = A058794(n)/2.
a(n) = A117560(n+2) - n - 1.
a(2*n) = A144129(n+1).
a(2*n-1) = A141530(n+1). a(n) = -a(-n-4). - Bruno Berselli, Sep 05 2011
a(n) = ((n+2-i)^3+(n+2+i)^3)/4, where i is the imaginary unit. - Nicolas Bělohoubek, Jul 03 2025

A117561 a(n) = floor(n*(n^3-n-3)/(2*(n-1))).

Original entry on oeis.org

3, 15, 38, 73, 124, 194, 286, 403, 548, 724, 934, 1181, 1468, 1798, 2174, 2599, 3076, 3608, 4198, 4849, 5564, 6346, 7198, 8123, 9124, 10204, 11366, 12613, 13948, 15374, 16894, 18511, 20228, 22048, 23974, 26009, 28156, 30418, 32798, 35299, 37924
Offset: 2

Views

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), Mar 29 2006

Keywords

Comments

a[n-1] is one approximation for the upper bound of the "antimagic constant" of an antimagic square of order n. The antimagic constant here is defined as the least integer in the set of consecutive integers to which the rows, columns and diagonals of the square sum. By analogy with the magic constant. This approximation follows from the observation that Sum[m + k, {k, 0, 2*n + 1}] <= (2*Sum[k, {k, 1, n^2}]) + (2*m) + (2*m + 1) where m is the antimagic constant for an antimagic square of order n. Stricter bounds seem likely to exist. See A117560 for the lower bounds. Note there exist no antimagic squares of order two or three, but the values are indexed here for completeness.

Examples

			a[3] = 38 because the antimagic constant of an antimagic square of order 4 cannot exceed 38 (see comments)

Links

Crossrefs

Cf. A117560, A050257, A049475.

Programs

  • Mathematica
    Table[Floor[n(n^3-n-3)/(2*(n-1))], {n, 2, 50}]

Formula

a(n) = floor(n*(n^3-n-3)/(2*(n-1))).
G.f.: x^2*(3+3*x-4*x^2-x^3+3*x^4-x^5)/(1-x)^4. - Colin Barker, Mar 29 2012

A337130 a(n) is the sum of all products of pairs of numbers joined by the diagonals of an n-gon when its vertices are numbered from 1 to n in order.

Original entry on oeis.org

0, 0, 0, 11, 40, 99, 203, 370, 621, 980, 1474, 2133, 2990, 4081, 5445, 7124, 9163, 11610, 14516, 17935, 21924, 26543, 31855, 37926, 44825, 52624, 61398, 71225, 82186, 94365, 107849, 122728, 139095, 157046, 176680, 198099, 221408, 246715, 274131, 303770
Offset: 1

Views

Author

Mohammed Yaseen, Aug 17 2020

Keywords

Comments

For n < 4, no n-gon has a diagonal and thus a(n)=0.

Examples

			The diagonals of 4-gon would be numbered (1,3) and (2,4). So a(4) = 1*3 + 2*4 = 11.
The diagonals of 5-gon would be numbered (1,3), (1,4), (2,4), (2,5) and (3,5). So a(5) = 1*3 + 1*4 + 2*4 + 2*5 + 3*5 = 40.

Links

Crossrefs

Partial sums of A117560. Cf. A000914 (products including sides), A007569, A007678.

Programs

  • PARI
    concat([0,0,0],Vec(x^4*(11 - 15*x + 9*x^2 - 2*x^3) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Aug 19 2020

Formula

a(n) = 3*binomial(n+1, 4) - n = (n-2)*(n-1)*n*(n+1)/8 - n for n>=3; a(1) = a(2) = 0.
a(n) = A000914(n-1) - A006527(n).
From Colin Barker, Aug 19 2020: (Start)
G.f.: x^4*(11 - 15*x + 9*x^2 - 2*x^3) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>7.
(End)
E.g.f.: x + x^2 + exp(x)*x*(-8 + 4*x^2 + x^3)/8. - Stefano Spezia, Aug 19 2020
Showing 1-3 of 3 results.

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