A185065 -id:A185065 - cp's OEIS frontend

A068601 a(n) = n^3 - 1.

0, 7, 26, 63, 124, 215, 342, 511, 728, 999, 1330, 1727, 2196, 2743, 3374, 4095, 4912, 5831, 6858, 7999, 9260, 10647, 12166, 13823, 15624, 17575, 19682, 21951, 24388, 26999, 29790, 32767, 35936, 39303, 42874, 46655, 50652, 54871, 59318, 63999, 68920

Offset: 1

Naohiro Nomoto, Mar 28 2002

a(n) is the least positive integer k such that k can only contain 'n-1' in exactly 2 different bases B, where 1 < B <= k.
Apart from the first term, the same as A135300. - R. J. Mathar, Apr 29 2008
A058895(n)^3 + a(n)^3 + A033562(n)^3 = A185065(n)^3. - Vincenzo Librandi, Mar 13 2012
Numbers k such that for every nonnegative integer m, k^(3*m+1) + k^(3*m) is a cube. - Arkadiusz Wesolowski, Aug 10 2013

			For n=6; 215 written in bases 6 and 42 is 555, 55 and (555, 55) are exactly 2 different bases.

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000217, A003215, A005448, A016921, A033562, A058895, A129294, A135300, A185065, A339604.

List([1..45],n->n^3-1); # Muniru A Asiru, Oct 23 2018

[n^3-1: n in [1..40]]; // Vincenzo Librandi, Mar 11 2012

A068601:=n->n^3-1: seq(A068601(n), n=1..50); # Wesley Ivan Hurt, Jul 23 2014

f[n_]:=n^3-1;f[Range[60]] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011*)
LinearRecurrence[{4,-6,4,-1},{0,7,26,63},50] (* Vincenzo Librandi, Mar 11 2012 *)
Range[50]^3 - 1 (* Wesley Ivan Hurt, Jul 23 2014 *)

PARI
```
a(n)=n^3-1
```

for n in range(1,50): print(n**3-1, end=', ') # Stefano Spezia, Nov 21 2018

Partial sums of A003215, hex (or centered hexagonal) numbers: 3*n(n+1)+1. - Jonathan Vos Post, Mar 16 2006
G.f.: x^2*(7-2*x+x^2)/(1-x)^4. - Colin Barker, Feb 12 2012
4*a(m^2-2*m+2) = (m^2-m+1)^3 + (m^2-m-1)^3 + (m^2-3*m+3)^3 + (m^2-3*m+1)^3. - Bruno Berselli, Jun 23 2014
a(n) = Sum_{i=1..n-1} (i+1)^3 - i^3. - Wesley Ivan Hurt, Jul 23 2014
Sum_{n>=2} 1/a(n) = Sum_{n>=1} (zeta(3*n) - 1) = A339604. - Amiram Eldar, Nov 06 2020
Product_{n>=2} (1 + 1/a(n)) = 3*Pi*sech(sqrt(3)*Pi/2). - Amiram Eldar, Jan 20 2021
E.g.f.: 1 + exp(x)*(x^3 + 3*x^2 + x - 1). - Stefano Spezia, Jul 06 2021

A058895 a(n) = n^4 - n.

Original entry on oeis.org

0, 0, 14, 78, 252, 620, 1290, 2394, 4088, 6552, 9990, 14630, 20724, 28548, 38402, 50610, 65520, 83504, 104958, 130302, 159980, 194460, 234234, 279818, 331752, 390600, 456950, 531414, 614628, 707252, 809970, 923490, 1048544, 1185888, 1336302, 1500590, 1679580

Offset: 0

Henry Bottomley, Jan 08 2001

a(n) is the number of ways to assign 4 different students to n different dorm rooms, each of which can hold at most 3 students. In other words, a(n) is the number of functions f:[4]->[n] with the size of the pre-image set of each element of the codomain at most 3. - Dennis P. Walsh, Mar 21 2013

a(n) are the values of m that yield integer solutions to this family of equations: x = sqrt(m + sqrt(x)), which may also be viewed as an infinitely recursive radical. The real solutions for x at each m = a(n) is n^2, except at n = 1 (m = 0) where x = 0 or 1 is a solution. - Richard R. Forberg, Oct 15 2014

Harry J. Smith, Table of n, a(n) for n = 0..2000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A002061, A002378, A024001, A027444, A027482, A068601, A033562, A185065, A339605.

[n^4-n: n in [0..40]]; // Vincenzo Librandi, Feb 23 2012

seq(n*(n^3-1),n=0..25) ; # R. J. Mathar, Dec 10 2015

Table[n^4 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = { n^4-n } \\ Harry J. Smith, Jun 23 2009

(2*x^2*(7+4*x+x^2)/(1-x)^5).series(x, 37).coefficients(x, sparse=False) # Stefano Spezia, Jul 09 2021

a(n) = n*(n-1)*(n^2+n+1) = A000583(n) - n = A002061(n+1) * A002378(n-1) = (n-1) * A027444(n) = -n * A024001(n).
a(n) = 2*A027482(n). - Zerinvary Lajos, Jan 28 2008
a(n) = floor(n^7/(n^3+1)). - Gary Detlefs, Feb 11 2010
a(n)^3 = (a(n)/n)^4 + (a(n)/n)^3. - Vincenzo Librandi, Feb 23 2012
a(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3, for n > 0. - Vincenzo Librandi, Mar 13 2012
G.f.: 2*x^2*(7 + 4*x + x^2)/(1 - x)^5. - Colin Barker, Apr 23 2012
a(n) = 14*C(n,2) + 36*C(n,3) + 24*C(n,4). - Dennis P. Walsh, Mar 21 2013
Sum_{n>=2} (-1)^n/a(n) = (Pi/3)*sech(Pi*sqrt(3)/2) + 4*log(2)/3 - 1 = 0.06147271494... . - Amiram Eldar, Jul 04 2020
Sum_{n>=2} 1/a(n) = A339605. - R. J. Mathar, Jan 08 2021
E.g.f.: exp(x)*x^2*(7 + 6*x + x^2). - Stefano Spezia, Jul 09 2021
a(n) = 12*A000332(n+2) + 2*A000537(n-1). - Yasser Arath Chavez Reyes, Apr 05 2024

A033562 a(n) = 2*n^3 + 1.

Original entry on oeis.org

1, 3, 17, 55, 129, 251, 433, 687, 1025, 1459, 2001, 2663, 3457, 4395, 5489, 6751, 8193, 9827, 11665, 13719, 16001, 18523, 21297, 24335, 27649, 31251, 35153, 39367, 43905, 48779, 54001, 59583, 65537, 71875, 78609, 85751, 93313, 101307, 109745, 118639, 128001

Offset: 0

N. J. A. Sloane

A058895(n)^3 + A068601(n)^3 + a(n)^3 = A185065(n)^3, for n>0. - Vincenzo Librandi, Mar 13 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A058895, A068601, A185065.

List([0..50], n-> 2*n^3+1); # G. C. Greubel, Oct 12 2019

[2*n^3+1: n in [0..50]]; // G. C. Greubel, Oct 12 2019

seq(2*n^3+1, n=0..50); # G. C. Greubel, Oct 12 2019

2*Range[0,50]^3+1 (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011*)
CoefficientList[Series[1+x*(3+5x+5x^2-x^3)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Mar 13 2012 *)
LinearRecurrence[{4,-6,4,-1},{1,3,17,55},50] (* Harvey P. Dale, Aug 14 2023 *)

a(n)=2*n^3+1 \\ Charles R Greathouse IV, Mar 11 2012

[2*n^3+1 for n in range(50)] # G. C. Greubel, Oct 12 2019

G.f.: 1 + x*(3 + 5*x + 5*x^2 - x^3)/(1-x)^4. - Vincenzo Librandi, Mar 13 2012

E.g.f.: (1 + 2*x + 6*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Oct 12 2019

Terms a(34) onward added by G. C. Greubel, Oct 12 2019

A117716 Triangle T(n,k) read by rows: the coefficient [x^n] of x^2/(1-(k+1)*x-x^3) in row n, columns 0 <= k <= n.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 3, 4, 1, 4, 9, 16, 25, 2, 9, 28, 65, 126, 217, 3, 20, 87, 264, 635, 1308, 2415, 4, 44, 270, 1072, 3200, 7884, 16954, 32960, 6, 97, 838, 4353, 16126, 47521, 119022, 264193, 534358, 9, 214, 2601, 17676, 81265, 286434, 835569, 2117656, 4815801, 10050030

Offset: 0

Roger L. Bagula, Apr 13 2006, corrected Apr 15 2006

			Triangle begins as:
  0;
  0,  0;
  1,  1,   1;
  1,  2,   3,    4;
  1,  4,   9,   16,   25;
  2,  9,  28,   65,  126,  217;
  3, 20,  87,  264,  635, 1308,  2415;
  4, 44, 270, 1072, 3200, 7884, 16954, 32960;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000930 (column 0), A008998 (column 1), A052541 (column 2), A052927 (column 3), A001093 (row 5), A185065 (row 6), A117715, A117724.

m:=12;
R:=PowerSeriesRing(Integers(), m+2);
A117716:= func< n,k | Coefficient(R!( x^2/(1-(k+1)*x-x^3) ), n) >;
[[A117716(n,k): k in [0..n]]: n in [0..m]]; // G. C. Greubel, Jul 23 2023

A117716 := proc(n,m)
        x^2/(1-(m+1)*x-x^3) ;
        if n < 0 then
                0;
        else
                coeftayl(%,x=0,n) ;
        end if;
end proc: # R. J. Mathar, May 14 2013

T[n_, k_]:= T[n, k]= Coefficient[Series[x^2/(1-(k+1)*x-x^3), {x,0,n+ 2}], x, n];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten

def A117716(n,k):
    P. = PowerSeriesRing(QQ)
    return P( x^2/(1-(k+1)*x-x^3) ).list()[n]
flatten([[A117716(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 23 2023

Edited by G. C. Greubel, Jul 23 2023

A204767 Quadruples (a,b,c,d) of the form ( n(n^3-1), n^3-1, 2n^3+1, n*(n^3+2) ).

Original entry on oeis.org

0, 0, 3, 3, 14, 7, 17, 20, 78, 26, 55, 87, 252, 63, 129, 264, 620, 124, 251, 635, 1290, 215, 433, 1308, 2394, 342, 687, 2415, 4088, 511, 1025, 4112, 6552, 728, 1459, 6579, 9990, 999, 2001, 10020, 14630, 1330, 2663, 14663, 20724, 1727, 3457, 20760, 28548, 2196, 4395, 28587

Offset: 1

Vincenzo Librandi, Mar 04 2012

Four consecutive (a,b,c,d) in the sequence are solutions to a^3+b^3+c^3 = d^3, that is a(4k+1)^3+a(4k+2)^3+a(4k+3)^3 = a(4k+4)^3.
Also, A058895(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3.
The sequence corresponds to the case m=1 in the identity (n*(n^3-m^3))^3+(m*(n^3-m^3))^3+(m*(2*n^3+m^3))^3 = (n*(n^3+2*m^3))^3.
G. H. Hardy and E. M. Wright gave this identity in their "An Introduction to the Theory of Numbers" together with (n*(n^3-2*m^3))^3+(m*(n^3+m^3))^3+(m*(2*n^3-m^3))^3 = (n*(n^3+m^3))^3 (see References). - Bruno Berselli, Mar 13 2012

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 2008 (Sixth edition), Par. 13.7.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A058895, A068601, A033562, A185065.

&cat[[n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2)]: n in [1..40]];

Flatten[Table[{n^4 - n, n^3 - 1, 2 n^3 + 1, n^4 + 2 n}, {n, 1, 40}]] (* Vincenzo Librandi, Jan 02 2014 *)

cp's OEIS Frontend

A068601 a(n) = n^3 - 1.

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A058895 a(n) = n^4 - n.

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A033562 a(n) = 2*n^3 + 1.

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A117716 Triangle T(n,k) read by rows: the coefficient [x^n] of x^2/(1-(k+1)*x-x^3) in row n, columns 0 <= k <= n.

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A204767 Quadruples (a,b,c,d) of the form ( n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2) ).

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A204767 Quadruples (a,b,c,d) of the form ( n(n^3-1), n^3-1, 2n^3+1, n*(n^3+2) ).