A016743 -id:A016743 - cp's OEIS frontend

A254371 Sum of cubes of the first n even numbers (A016743).

0, 8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800, 352800, 426888, 512072, 609408, 720000, 845000, 985608, 1143072, 1318688, 1513800, 1729800, 1968128, 2230272, 2517768, 2832200, 3175200

Offset: 0

Luciano Ancora, Mar 16 2015

Property: for n >= 2, each (a(n), a(n)+1, a(n)+2) is a triple of consecutive terms that are the sum of two nonzero squares; precisely: a(n) = (n*(n + 1))^2 + (n*(n + 1))^2, a(n)+1 = (n^2+2n)^2 + (n^2-1)^2 and a(n)+2 = (n^2+n+1)^2 + (n^2+n-1)^2 (see Diophante link). - Bernard Schott, Oct 05 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Luciano Ancora, The Square Pyramidal Number and other figurate numbers, ch. 3.
Diophante, Une miniature avec trois entiers consécutifs (in French).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000537 (sum of first n cubes); A002593 (sum of first n odd cubes).
Cf. A060300 (2*a(n)).
First bisection of A105636; second bisection of A212892.
Cf. A000217, A000415, A002378, A035287, A163102.

List([0..35],n->2*(n*(n+1))^2); # Muniru A Asiru, Oct 24 2018

[2*n^2*(n+1)^2: n in [0..40]]; // Bruno Berselli, Mar 23 2015

A254371:=n->2*n^2*(n + 1)^2: seq(A254371(n), n=0..50); # Wesley Ivan Hurt, Apr 28 2017

Table[2 n^2 (n+1)^2, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 72, 288, 800}, 40]
Accumulate[Range[0,80,2]^3] (* Harvey P. Dale, Jun 26 2017 *)

a(n)=sum(i=0, n, 8*i^3); \\ Michael B. Porter, Mar 16 2015

G.f.: 8*x*(1 + 4*x + x^2)/(1 - x)^5.
a(n) = 2*n^2*(n + 1)^2.
a(n) = 2*A035287(n+1) = 2*A002378(n)^2 = 8*A000217(n)^2. - Bruce J. Nicholson, Apr 23 2017
a(n) = 8*A000537(n). - Michel Marcus, Apr 23 2017
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/6 - 3/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/2 - 2*log(2). (End)
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: 2*x*(2 + x)*(2 + 6*x + x^2)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 4*A163102(n) = A060300(n)/2. (End)

A016755 Odd cubes: a(n) = (2*n + 1)^3.

Original entry on oeis.org

1, 27, 125, 343, 729, 1331, 2197, 3375, 4913, 6859, 9261, 12167, 15625, 19683, 24389, 29791, 35937, 42875, 50653, 59319, 68921, 79507, 91125, 103823, 117649, 132651, 148877, 166375, 185193, 205379, 226981, 250047, 274625, 300763, 328509, 357911, 389017, 421875

Offset: 0

N. J. A. Sloane

Partial sums of A010014. - Jani Melik, May 20 2013

Terms end in the repeating sequence 1, 7, 5, 3, 9, ... - Melvin Peralta, Jul 08 2015

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Marc Chamberland and Armin Straub, On gamma quotients and infinite products, Advances in Applied Mathematics, Vol. 51, No. 5 (2013), pp. 546-562.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A005408, A010014, A016743, A153071, A154537.

Cf. A000330, A005917, A069074.

[(2*n+1)^3: n in [0..50]]; // Vincenzo Librandi, Sep 05 2011

Range[1,101,2]^3 (* Harvey P. Dale, Nov 18 2013 *)

a(n)=(2*n+1)^3 \\ Charles R Greathouse IV, Jan 02 2012

def a(n): return (2*n+1)**3
print([a(n) for n in range(38)]) # Michael S. Branicky, Jan 27 2021

Sum_{n >= 0} 1/a(n) = 7 * zeta(3) / 8.
G.f.: (1+23*x+23*x^2+x^3)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 02 2012
a(n) = A000578(A005408(n)). - Michel Marcus, Jul 09 2015
E.g.f.: exp(x)*(1 + 26*x + 36*x^2 + 8*x^3). See A154537, row n=3. - Wolfdieter Lang, Mar 12 2017
From Bruce J. Nicholson, Dec 08 2019: (Start)
a(n) = 24 * A000330(n)   + A005408(n).
a(n) = 2  * A005917(n+1) - A005408(n). (End)
Sum_{n>=0} (-1)^n/a(n) = Pi^3/32 (A153071). - Amiram Eldar, Oct 10 2020
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/12)*(1 + sqrt(2)*cosh(sqrt(3)*Pi/4)) (Chamberland and Straub, 2013). - Amiram Eldar, Jan 26 2024

A244725 a(n) = 5*n^3.

Original entry on oeis.org

0, 5, 40, 135, 320, 625, 1080, 1715, 2560, 3645, 5000, 6655, 8640, 10985, 13720, 16875, 20480, 24565, 29160, 34295, 40000, 46305, 53240, 60835, 69120, 78125, 87880, 98415, 109760, 121945, 135000, 148955, 163840, 179685, 196520, 214375, 233280, 253265

Offset: 0

Vincenzo Librandi, Jul 05 2014

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. similar sequences of the type k*n^3: A000578 (k=1), A033431 (k=2), A117642 (k=3), A033430 (k=4), this sequence (k=5), A244726 (k=6), A244727 (k=7), A016743 (k=8), A244728 (k=9), A244729 (k=10), A016767 (k=27), A016803 (k=64), A016851 (k=125), A016911 (k=216), A016983 (k=343), A017067 (k=512), A017163 (k=729), A017271 (k=1000), A017391 (k=1331), A017523 (k=1728).

Magma
```
[5*n^3: n in [0..40]];
```

I:=[0,5,40,135]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]];

Table[5 n^3, {n, 0, 40}] (* or *) CoefficientList[Series[5 x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 40}], x]

a(n)=5*n^3 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: 5*x*(1 + 4*x + x^2)/(1 - x)^4.

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.

A164897 a(n) = 4n(n+1) + 3.

Original entry on oeis.org

3, 11, 27, 51, 83, 123, 171, 227, 291, 363, 443, 531, 627, 731, 843, 963, 1091, 1227, 1371, 1523, 1683, 1851, 2027, 2211, 2403, 2603, 2811, 3027, 3251, 3483, 3723, 3971, 4227, 4491, 4763, 5043, 5331, 5627, 5931, 6243, 6563, 6891, 7227, 7571, 7923, 8283, 8651, 9027, 9411

Offset: 0

Paul Curtz, Aug 30 2009

One-fourth the sum of the three terms produced by the division of complex numbers (2*n-3+(2*n-1)*i)/(2*n+1+(2*n+3)*i). For (b+c*i)/(d+e*i) the three terms in parentheses are ((b*d+c*e)+(c*d-b*e)*i)/(d^2+e^2). By substituting b=2*n-3, c=2*n-1, d=2*n+1, and e=2*n+3 one gets a(n). - J. M. Bergot, Sep 10 2015

The continued fraction expansion of sqrt(a(n)) is [2n+1; {2n+1, 4n+2}]. - Magus K. Chu, Sep 08 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..900
R. M. Green and Tianyuan Xu, 2-roots for simply laced Weyl groups, arXiv:2204.09765 [math.RT], 2022.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000124, A008590, A010731, A016743, A069894.

Odd-indexed terms of A059100.

[4*n*(n+1)+3: n in [0..50]]; // Vincenzo Librandi, Apr 24 2011

A164897:=n->4*n*(n+1)+3: seq(A164897(n), n=0..100); # Wesley Ivan Hurt, Sep 10 2015

Table[4 n (n + 1) + 3, {n, 0, 50}] (* Harvey P. Dale, Jan 23 2011 *)

a(n)=4*n*(n+1)+3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A000124(2*n) + A000124(2*n+1) = A069894(n)+1.
a(n+1) - a(n) = 8n+8 = A008590(n+1) (first differences).
a(n+1) - 2*a(n) + a(n-1) = 8 = A010731(n) (second differences).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n>2.
G.f.: (3+2*x+3*x^2) / (1-x)^3.
Sum_{k=n+1..2*n+1} a(k) - Sum_{k=0..n} a(k) = (2*n+2)^3. - Bruno Berselli, Jan 24 2011
E.g.f.: (4x^2 + 8x + 1)*exp(x). - G. C. Greubel, Sep 22 2015
a(n)^2 = A222465(n)*A222465(n+1) - 12. - Ezhilarasu Velayutham, Mar 18 2020
Sum_{n>=0} 1/a(n) = tanh(Pi/sqrt(2))*Pi/(4*sqrt(2)). - Amiram Eldar, Aug 21 2022
a(n) = A059100(2*n+1). - Dimitri Papadopoulos, Nov 21 2023

Definition simplified by R. J. Mathar, Sep 16 2009

A016827 a(n) = (4n+2)^3.

Original entry on oeis.org

8, 216, 1000, 2744, 5832, 10648, 17576, 27000, 39304, 54872, 74088, 97336, 125000, 157464, 195112, 238328, 287496, 343000, 405224, 474552, 551368, 636056, 729000, 830584, 941192, 1061208, 1191016

Offset: 0

N. J. A. Sloane

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Odd bisection of A016743.

Subsequence of A000578.

Table[(4n+2)^3,{n,0,100}] (* Mohammad K. Azarian, Jun 20 2016 *)

G.f.: 8*(1 + 23*x + 23*x^2 + x^3)/(1 - x)^4. - Ilya Gutkovskiy, Jun 21 2016
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=0} 1/a(n) = 7*zeta(3)/64.
Sum_{n>=0} (-1)^n/a(n) = Pi^3/256. (End)
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Wesley Ivan Hurt, Mar 08 2022

A331988 Table T(n,k) read by antidiagonals. T(n,k) is the maximum value of Product_{i=1..n} Sum_{j=1..k} r_j[i] where each r_j is a permutation of {1..n}.

Original entry on oeis.org

1, 2, 2, 6, 9, 3, 24, 64, 20, 4, 120, 625, 216, 36, 5, 720, 7776, 3136, 512, 56, 6, 5040, 117649, 59049, 10000, 1000, 81, 7, 40320, 2097152, 1331000, 248832, 24336, 1728, 110, 8, 362880, 43046721, 35831808, 7529536, 759375, 50625, 2744, 144, 9, 3628800, 1000000000, 1097199376, 268435456, 28652616, 1889568, 93636, 4096, 182, 10

Offset: 1

Chai Wah Wu, Feb 23 2020

A dual sequence to A260355. See arXiv link for sets of permutations that achieve the value of T(n,k). The minimum value of Product_{i=1..n} Sum_{j=1..k} r_j[i] is equal to n!*k^n.

			T(n,k)
   k    1    2     3      4      5      6      7      8      9     10     11     12
  ---------------------------------------------------------------------------------
n  1|   1    2     3      4      5      6      7      8      9     10     11     12
   2|   2    9    20     36     56     81    110    144    182    225    272    324
   3|   6   64   216    512   1000   1728   2744   4096   5832   8000  10648  13824
   4|  24  625  3136  10000  24336  50625  93636 160000 256036 390625 571536 810000

Chai Wah Wu, Table of n, a(n) for n = 1..70
Chai Wah Wu, Permutations r_j such that ∑i∏j r_j(i) is maximized or minimized, arXiv:1508.02934 [math.CO], 2015-2020.
Chai Wah Wu, On rearrangement inequalities for multiple sequences, arXiv:2002.10514 [math.CO], 2020.

Cf. A000027, A000142, A000169, A001504, A016743, A016766, A061711, A061718, A260355.

from itertools import permutations, combinations_with_replacement
def A331988(n,k): # compute T(n,k)
    if k == 1:
        count = 1
        for i in range(1,n):
            count *= i+1
        return count
    ntuple, count = tuple(range(1,n+1)), 0
    for s in combinations_with_replacement(permutations(ntuple,n),k-2):
        t = list(ntuple)
        for d in s:
            for i in range(n):
                t[i] += d[i]
        t.sort()
        w = 1
        for i in range(n):
            w *= (n-i)+t[i]
        if w > count:
            count = w
    return count

T(n,n) = (n*(n+1)/2)^n = A061718(n).
T(n,k) <= (k(n+1)/2)^n.
T(1,k) = k = A000027(k).
T(n,1) = n! = A000142(n).
T(2,2m) = 9m^2 = A016766(m).
T(2,2m+1) = (3m+1)*(3m+2) = A001504(m).
T(n,2) = (n+1)^n = A000169(n+1).
T(3,k) = 8k^3 = A016743(k) for k > 1.
If n divides k then T(n,k) = (k*(n+1)/2)^n.
If k is even then T(n,k) = (k*(n+1)/2)^n.
If n is odd and k >= n-1 then T(n,k) = (k*(n+1)/2)^n.
If n is even and k is odd such that k >= n-1, then T(n,k) = ((k^2*(n+1)^2-1)/4)^(n/2).

A353550 Primes having cube prime gaps to both neighbor primes.

Original entry on oeis.org

89689, 107441, 367957, 368021, 725209, 803749, 832583, 919511, 1070753, 1315151, 1333027, 1353487, 1414913, 1843357, 2001911, 2038039, 2201273, 2207783, 2269537, 2356699, 2356763, 2670817, 2696843, 2715071, 2717929, 2731493, 2906887, 2971841, 3032467, 3184177, 3252217

Offset: 1

Karl-Heinz Hofmann, Apr 25 2022

nonn

Up to prime 669763117 all gaps are 8 and 64 or 64 and 8. Prime 669763117 is the first one with gaps 8 and 216. Possible gaps must be in A016743.

			a(2) = 107441; previous prime is 107377 and the gap is 64 (a cube); next prime is 107449 and the gap is 8 (a cube too).

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Cf. A000040, A000578, A016743, A353088 (square gaps), A163112 (gaps > 20).
Cf. A353137 (gaps are a power of 2), A353135 (Fibonacci gaps).
Cf. A353136 (triangular numbers gaps).

iscube:= proc(n) option remember; is(n=iroot(n, 3)^3) end:
q:= n-> isprime(n) and andmap(iscube, [n-prevprime(n), nextprime(n)-n]):
select(q, [$3..3500000])[];  # Alois P. Heinz, Apr 25 2022

p = Prime[Range[3*10^5]]; pos = Position[Differences[p], ?(IntegerQ@Surd[#, 3] &)] // Flatten; p[[pos[[Position[Differences[pos], 1] // Flatten]] + 1]] (* _Amiram Eldar, Apr 26 2022 *)

from sympy import sieve as p
def A016743(totest): return (totest % 2 == 0 and round(totest**(1/3))**3 == totest)
print([p[n] for n in range(2,235000) if A016743(p[n]-p[n-1]) and A016743(p[n+1]-p[n])])

A047363 Numbers that are congruent to {0, 2, 3, 4, 5} mod 7.

Original entry on oeis.org

0, 2, 3, 4, 5, 7, 9, 10, 11, 12, 14, 16, 17, 18, 19, 21, 23, 24, 25, 26, 28, 30, 31, 32, 33, 35, 37, 38, 39, 40, 42, 44, 45, 46, 47, 49, 51, 52, 53, 54, 56, 58, 59, 60, 61, 63, 65, 66, 67, 68, 70, 72, 73, 74, 75, 77, 79

Offset: 1

N. J. A. Sloane

nonn

Conjecture: Apart from 0, and the further exclusions noted below, the sequence gives the values of c/6 such that an infinite number of primes, p, result in both p^3+c and p^3-c being positive primes. Taking the complement we say: the excluded c/6 values are {1,6} mod 7. See A005097 for a conjecture on the modulo patterns of excluded c/6 values for the general case of p^q + c and p^q - c both prime, for any q > 0, and see A047222 for q=2. Note that polynomial factorization also excludes a few c/6 values. This occurs here when c is an even cube (A016743), which requires a further exclusion of certain c/6 values in this sequence when (6c)^3/6 == 0 (mod 7), or c/6 = {0, 12348, 98784, ...}. - Richard R. Forberg, Jun 28 2016

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A047222, A005097.

Table[7 n + {0, 2, 3, 4, 5}, {n, 0, 12}] // Flatten (* or *)
Select[Range[0, 79], ! MemberQ[{1, 6}, Mod[#, 7]] &] (* or *)
Rest@ CoefficientList[Series[x^2 (2 x^2 + 3 x + 2) (x^2 - x + 1)/((x^4 + x^3 + x^2 + x + 1) (x - 1)^2), {x, 0, 57}], x] (* Michael De Vlieger, Jul 25 2016 *)

G.f.: x^2*(2*x^2 + 3*x + 2)*(x^2 - x + 1) / ( (x^4 + x^3 + x^2 + x + 1)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011

a(n) = a(n-1) + a(n-5) - a(n-6). - Wesley Ivan Hurt, Sep 03 2022

A201262 Primes of the form n^3 + 9.

Original entry on oeis.org

17, 73, 521, 1009, 2753, 8009, 10657, 21961, 39313, 54881, 85193, 140617, 195121, 262153, 314441, 512009, 681481, 778697, 941201, 1404937, 3241801, 3511817, 4410953, 4913009, 6028577, 6229513, 6644681, 6859009, 8000009, 8998921

Offset: 1

Vincenzo Librandi, Nov 29 2011

Old name was "Primes of the form 8n^3 + 9".

			Since 17 is prime and equal to 2^3 + 9, it is in the sequence.
Since 73 is prime and equal to 4^3 + 9, it is in the sequence.
225 is not in the sequence, because, although it is 6^3 + 9, it is divisible by 5.

Vincenzo Librandi, Table of n, a(n) for n = 1..6200

Cf. A016743.

[a: n in [0..300] | IsPrime(a) where a is 8*n^3+9]

Select[Table[n^3 + 9, {n, 0, 248, 2}], PrimeQ]

Name simplified by Alex Ratushnyak, Apr 06 2013

A379852 a(n) = floor(8*n^3/27).

Original entry on oeis.org

0, 0, 2, 8, 18, 37, 64, 101, 151, 216, 296, 394, 512, 650, 813, 1000, 1213, 1455, 1728, 2032, 2370, 2744, 3154, 3605, 4096, 4629, 5207, 5832, 6504, 7226, 8000, 8826, 9709, 10648, 11645, 12703, 13824, 15008, 16258, 17576, 18962, 20421, 21952, 23557, 25239

Offset: 0

Gonzalo Martínez and Javier Astudillo, Jan 04 2025

a(n) is the integer part of the area of the largest triangle that can be inscribed in the region bounded by the parabola y = x^2, the x-axis, and the line x = n.

To estimate the integral int_{x = 0..n} x^2 dx by means of a triangle, we find that the triangle with the largest area that can be inscribed in the region bounded by the parabola y = x^2, the x-axis and the line x = n is the right triangle with vertices (n/3, 0), (n, 0) and (n, (8/9)*n^2), whose area is (2n/3)^3 and a(n) has been defined as floor((2n/3)^3).

			If n = 2, the largest triangle that can be inscribed in the region bounded by the parabola y = x^2, the x-axis, and the line x = n is the right triangle with vertices (2/3,0),(2,0) and (2,32/9), whose area is 64/27. Since floor(64/27) = 2, it follows that a(2) = 2.

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,0,0,0,1,-3,3,-1).

Cf. A016743, A375473.

Floor[8/27*Range[0, 50]^3] (* Paolo Xausa, Jan 30 2025 *)

a(n) = floor(A016743(n)/27).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-9) - 3*a(n-10) + 3*a(n-11) - a(n-12) for n >= 12. - Pontus von Brömssen, Jan 14 2025

cp's OEIS Frontend

A254371 Sum of cubes of the first n even numbers (A016743).

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A016755 Odd cubes: a(n) = (2*n + 1)^3.

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A244725 a(n) = 5*n^3.

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A164897 a(n) = 4*n*(n+1) + 3.

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A016827 a(n) = (4n+2)^3.

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A331988 Table T(n,k) read by antidiagonals. T(n,k) is the maximum value of Product_{i=1..n} Sum_{j=1..k} r_j[i] where each r_j is a permutation of {1..n}.

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A353550 Primes having cube prime gaps to both neighbor primes.

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A164897 a(n) = 4n(n+1) + 3.