A047866 -id:A047866 - cp's OEIS frontend

A047837 Honaker's triangle problem: form a triangle with base of length n, all entries different, all row sums equal; a(n) gives minimal row sum.

1, 3, 8, 15, 27, 43, 65, 94, 130, 175, 229, 294, 369, 456, 557, 671, 800, 944, 1105, 1283, 1479, 1695, 1930, 2187, 2465, 2765, 3090, 3439, 3813, 4213, 4641, 5096, 5580, 6095, 6639, 7216, 7825, 8466, 9143, 9855

Offset: 1

Jud McCranie

Suggested by G. L. Honaker, Jr.

Agrees with A047873 at least for n < 365, conjectured to always agree.

			a(1)..a(4), 1 // 3; 1 2 // 8; 2 6; 1 3 4 // 15; 7 8; 4 5 6; 1 2 3 9.
a(6) = 43, 21 22; 8 16 19; 5 9 12 17; 3 4 7 14 15; 1 2 6 10 11 13.
a(7) = 65, 32 33; 20 21 24; 14 15 17 19; 9 10 11 12 23; 5 6 7 13 16 18; 1 2 3 4 8 22 25.

Pickover, C. A., The Zen of Magic Squares, Circles and Stars: An Exhibition Of Surprising Structures Across Dimensions, Princeton University Press, 2002 (pp. 289-292).

Cf. A047866.

Appears to obey a 16-term linear recurrence. - Ralf Stephan, May 06 2004

Empirical g.f.: -x*(x^15 - 3*x^14 + 3*x^13 - 5*x^12 + 5*x^11 - 9*x^10 + 7*x^9 - 10*x^8 + 7*x^7 - 9*x^6 + 5*x^5 - 6*x^4 + 2*x^3 - 3*x^2 - 1) / ((x-1)^4*(x^2-x+1)*(x^2+1)*(x^2+x+1)^2*(x^4-x^2+1)). - Colin Barker, Jan 18 2013

A109900 The (n,r)-th term of the following triangle is T(n)-T(r) for r = 0 to n. The n-th row contains n+1 terms. T(n) = the n-th triangular number = n(n+1)/2. Sequence contains the sum of terms at a 45-degree angle.

Original entry on oeis.org

0, 1, 3, 8, 15, 27, 42, 64, 90, 125, 165, 216, 273, 343, 420, 512, 612, 729, 855, 1000, 1155, 1331, 1518, 1728, 1950, 2197, 2457, 2744, 3045, 3375, 3720, 4096, 4488, 4913, 5355, 5832, 6327, 6859, 7410, 8000, 8610, 9261, 9933, 10648, 11385, 12167, 12972

Offset: 0

Amarnath Murthy, Jul 13 2005

Initial terms match those of A047866 with a difference of +1 or -1 in some cases. A047866: 0, 1, 3, 8, 15, 27, 42, 63, 90, 124, 165, 215, ...

			The (n,r)-th term of the following triangle is T(n)-T(r) for r = 0 to n. The n-th row contains n+1 terms.
   0
   1  0
   3  2  0
   6  5  3  0
  10  9  7  4  0
  15 14 12  9  5  0
  21 20 18 15 11  6  0
  28 27 ...
  36 ...
Sequence contains the sum of terms at a 45-degree angle.
a(5) = 15 + 9 + 3 = 27.

Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A000217, A001318, A034828, A047866.

A109900 := proc(n) if n mod 2 = 1 then ( (n+1)/2)^3 ; else (n+1)*(n/2+1)*(n/2)/2 ; fi ; end: seq(A109900(n),n=0..80) ; # R. J. Mathar, Feb 11 2008

LinearRecurrence[{2, 1, -4, 1, 2, -1}, {0, 1, 3, 8, 15, 27}, 50] (* Amiram Eldar, Sep 17 2022 *)

a(2n+1) = (n+1)^3; a(2n) = (2n+1)*T(n) = (2n+1)*(n+1)*n/2, where T=A000217. - R. J. Mathar, Feb 11 2008
a(n) = A034828(n+1). - R. J. Mathar, Aug 18 2008
G.f.: x*(1+x+x^2)/(1-2*x-x^2+4*x^3-x^4-2*x^5+x^6). - Colin Barker, Jan 04 2012
a(n) = (2*n^3+6*n^2+5*n+1-(n+1)*(-1)^n)/16. - Luce ETIENNE, May 12 2015
a(n) = Sum_{k=0..n} A001318(k). - Jacob Szlachetka, Dec 20 2021
Sum_{n>=1} 1/a(n) = 6 - 8*log(2) + zeta(3). - Amiram Eldar, Sep 17 2022

Corrected and extended by R. J. Mathar, Feb 11 2008

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

Bruno Berselli, Jun 27 2018

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).

			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.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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).

GAP
```
List([0..40], n -> n*(2*n+1)*(4*n+1));
```

[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]];
```

seq(n*(2*n+1)*(4*n+1),n=0..40); # Muniru A Asiru, Jun 27 2018

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)]
```

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

A047873 a(n) = max_{r=1..n-1} ceiling(t(t(n)-t(r-1))/(n-r+1)), where t() = triangular numbers A000217.

Original entry on oeis.org

Offset: 1

Mike Keith

Another lower bound for Honaker triangle problem (A047837); conjectured to be exact value.

Cf. A047837, A047866.

cp's OEIS Frontend

A047837 Honaker's triangle problem: form a triangle with base of length n, all entries different, all row sums equal; a(n) gives minimal row sum.

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Formula

A109900 The (n,r)-th term of the following triangle is T(n)-T(r) for r = 0 to n. The n-th row contains n+1 terms. T(n) = the n-th triangular number = n(n+1)/2. Sequence contains the sum of terms at a 45-degree angle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Julia

Magma

Maple

Mathematica

Maxima

PARI

Python

Sage

Formula

A047873 a(n) = max_{r=1..n-1} ceiling(t(t(n)-t(r-1))/(n-r+1)), where t() = triangular numbers A000217.

Views

Author

Keywords

Comments

Crossrefs

Formula

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