A000584 -id:A000584 - cp's OEIS frontend

A068236 First differences of (n+1)^5-n^5.

30, 180, 570, 1320, 2550, 4380, 6930, 10320, 14670, 20100, 26730, 34680, 44070, 55020, 67650, 82080, 98430, 116820, 137370, 160200, 185430, 213180, 243570, 276720, 312750, 351780, 393930, 439320, 488070, 540300, 596130, 655680, 719070, 786420, 857850, 933480

Offset: 0

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Mar 25 2002

For n>=0, a(n) is equal to the number of functions f:{1,2,3,4,5}->{1,2,...,n+2} such that Im(f) contains 2 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Feb 24 2007

Colin Barker, Table of n, a(n) for n = 0..1000
O. Bagdasar, On some functions involving the lcm and gcd of integer tuples, Scientific Publications of the State University of Novi Pazar, Appl. Maths. Inform. and Mech., Vol. 6, 2 (2014), 91--100.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A022521 ((n+1)^5-n^5), A000584 (5th powers), A005900 (octahedral numbers).

Table[20*n^3 + 10*n, {n, 1, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
Differences[#[[2]]-#[[1]]&/@Partition[Range[0,40]^5,2,1]] (* or *) LinearRecurrence[{4,-6,4,-1},{30,180,570,1320},40] (* Harvey P. Dale, Jun 05 2019 *)

Vec(30*(x+1)^2 / (x-1)^4 + O(x^100)) \\ Colin Barker, Dec 13 2014

a(n) = (n+2)^5-2*(n+1)^5+n^5.
a(n) = 30*A005900(n+1). - R. J. Mathar, Sep 02 2008
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Colin Barker, Dec 13 2014
G.f.: 30*(x+1)^2 / (x-1)^4. - Colin Barker, Dec 13 2014

A113850 Numbers whose prime factors are raised to the fifth power.

Original entry on oeis.org

32, 243, 3125, 7776, 16807, 100000, 161051, 371293, 537824, 759375, 1419857, 2476099, 4084101, 5153632, 6436343, 11881376, 20511149, 24300000, 28629151, 39135393, 45435424, 52521875, 69343957, 79235168, 90224199, 115856201

Offset: 1

Cino Hilliard, Jan 25 2006

			7776 = 32*243 = 2^5*3^5 so the prime factors, 2 and 3, are raised to the fifth power.

Proper subset of A000584.
Nonunit terms of A329332 column 5 in ascending order.
Cf. A005117, A157291.

Select[ Range@41^5, Union[Last /@ FactorInteger@# ] == {5} &] (* Robert G. Wilson v *)
Rest[Select[Range[100], SquareFreeQ]^5] (* Vaclav Kotesovec, May 22 2020 *)

allpwrfact(n,p) = \All prime factors are raised to the power p { local(x,j,ln,y,flag); for(x=4,n, y=Vec(factor(x)); ln = length(y[1]); flag=0; for(j=1,ln, if(y[2][j]==p,flag++); ); if(flag==ln,print1(x",")); ) }

from math import isqrt
from sympy import mobius
def A113850(n):
    def f(x): return int(n+x+1-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1)))
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m**5 # Chai Wah Wu, Sep 13 2024

Sum_{k>=1} 1/a(k) = zeta(5)/zeta(10) - 1 = A157291 - 1. - Amiram Eldar, May 22 2020

a(n) = A005117(n+1)^5. - Chai Wah Wu, Sep 13 2024

More terms from Robert G. Wilson v, Jan 26 2006

Offset corrected by Chai Wah Wu, Sep 13 2024

A300656 Triangle read by rows: T(n,k) = 30k^2(n-k)^2 + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 31, 1, 1, 121, 121, 1, 1, 271, 481, 271, 1, 1, 481, 1081, 1081, 481, 1, 1, 751, 1921, 2431, 1921, 751, 1, 1, 1081, 3001, 4321, 4321, 3001, 1081, 1, 1, 1471, 4321, 6751, 7681, 6751, 4321, 1471, 1, 1, 1921, 5881, 9721, 12001, 12001, 9721, 5881, 1921, 1

Offset: 0

Kolosov Petro, Mar 10 2018

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(2, n, k).
Fifth power can be expressed as row sum of triangle T(n, k).
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
--------------------------------------------------------------------------
k=    0     1     2      3      4      5      6      7     8     9    10
--------------------------------------------------------------------------
n=0:  1;
n=1:  1,    1;
n=2:  1,   31,    1;
n=3:  1,  121,  121,     1;
n=4:  1,  271,  481,   271,     1;
n=5:  1,  481, 1081,  1081,   481,     1;
n=6:  1,  751, 1921,  2431,  1921,   751,     1;
n=7:  1, 1081, 3001,  4321,  4321,  3001,  1081,     1;
n=8:  1, 1471, 4321,  6751,  7681,  6751,  4321,  1471,    1;
n=9:  1, 1921, 5881,  9721, 12001, 12001,  9721,  5881, 1921,    1;
n=10: 1, 2431, 7681, 13231, 17281, 18751, 17281, 13231, 7681, 2431,   1;

Kolosov Petro, Rows n = 0..2078 of triangle, flattened.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Various cases of L(m, n, k): A287326(m=1), This sequence (m=2), A300785(m=3). See comments for L(m, n, k).
Row sums give the nonzero terms of A002561.
Cf. A000584, A287326, A007318, A077028, A294317, A068236, A302971, A304042, A002561, A258807, A158558, A094053, A024003, A316349.

T:=Flat(List([0..9],n->List([0..n],k->30*k^2*(n-k)^2+1))); # Muniru A Asiru, Oct 24 2018

[[30*k^2*(n-k)^2+1: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Dec 14 2018

a:=(n,k)->30*k^2*(n-k)^2+1: seq(seq(a(n,k),k=0..n),n=0..9); # Muniru A Asiru, Oct 24 2018

T[n_, k_] := 30 k^2 (n - k)^2 + 1; Column[
Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Apr 12 2020 *)
f[n_]:=Table[SeriesCoefficient[(1 + 26 y + 336 y^2 + 326 y^3 + 31 y^4 + x^2 (1 + 116 y + 486 y^2 + 116 y^3 + y^4) + x (-2 - 82 y - 882 y^2 - 502 y^3 + 28 y^4))/((-1 + x)^3 (-1 + y)^5), {x, 0, i}, {y, 0, j}], {i, n, n}, {j, 0, n}]; Flatten[Array[f, 11, 0]] (* Stefano Spezia, Oct 30 2018 *)

t(n, k) = 30*k^2*(n-k)^2+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */ trianglerows(9)

[[30*k^2*(n-k)^2+1 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 14 2018

From Kolosov Petro, Apr 12 2020: (Start)
T(n, k) = 30 * k^2 * (n-k)^2 + 1.
T(n, k) = 30 * A094053(n,k)^2 + 1.
T(n, k) = A158558((n-k) * k).
T(n+2, k) = 3*T(n+1, k) - 3*T(n, k) + T(n-1, k), for n >= k.
Sum_{k=1..n} T(n, k) = A000584(n).
Sum_{k=0..n-1} T(n, k) = A000584(n).
Sum_{k=0..n} T(n, k) = A002561(n).
Sum_{k=1..n-1} T(n, k) = A258807(n).
Sum_{k=1..n-1} T(n, k) = -A024003(n), n > 1.
Sum_{k=1..r} T(n, k) = A316349(2,r,0)*n^0 - A316349(2,r,1)*n^1 + A316349(2,r,2)*n^2. (End)
G.f.: (1 + 26*y + 336*y^2 + 326*y^3 + 31*y^4 + x^2*(1 + 116*y + 486*y^2 + 116*y^3 + y^4) + x*(-2 - 82*y - 882*y^2 - 502*y^3 + 28*y^4))/((-1 + x)^3*(-1 + y)^5). - Stefano Spezia, Oct 30 2018

A300785 Triangle read by rows: T(n,k) = 140k^3(n-k)^3 - 14k(n-k) + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 127, 1, 1, 1093, 1093, 1, 1, 3739, 8905, 3739, 1, 1, 8905, 30157, 30157, 8905, 1, 1, 17431, 71569, 101935, 71569, 17431, 1, 1, 30157, 139861, 241753, 241753, 139861, 30157, 1, 1, 47923, 241753, 472291, 573217, 472291, 241753, 47923, 1, 1, 71569, 383965, 816229, 1119721, 1119721, 816229, 383965, 71569, 1

Offset: 0

Kolosov Petro, Mar 12 2018

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(3, n, k).
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
--------------------------------------------------------------------
k=   0      1       2       3       4       5       6      7     8
--------------------------------------------------------------------
n=0: 1;
n=1: 1,     1;
n=2: 1,   127,      1;
n=3: 1,  1093,   1093,      1;
n=4: 1,  3739,   8905,   3739,      1;
n=5: 1,  8905,  30157,  30157,   8905,      1;
n=6: 1, 17431,  71569, 101935,  71569,  17431,      1;
n=7: 1, 30157, 139861, 241753, 241753, 139861,  30157,     1;
n=8: 1, 47923, 241753, 472291, 573217, 472291, 241753, 47923,    1;

Muniru A Asiru, Rows n=0..100 of triangle, flattened.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Various cases of L(m, n, k): A287326 (m=1), A300656 (m=2), This sequence (m=3). See comments for L(m, n, k).
Row sums give A258806.
Cf. A000584, A287326, A007318, A077028, A294317, A068236, A300656, A302971, A304042, A001015, A094053, A258808, A024005, A316387.

T:=Flat(List([0..9], n->List([0..n], k->140*k^3*(n-k)^3 - 14*k*(n-k)+1))); # G. C. Greubel, Dec 14 2018

/* As triangle */ [[140*k^3*(n-k)^3-14*k*(n-k)+1: k in [0..n]]: n in [0..10]]; // Bruno Berselli, Mar 21 2018

T:=(n,k)->140*k^3*(n-k)^3-14*k*(n-k)+1: seq(seq(T(n,k),k=0..n),n=0..9); # Muniru A Asiru, Dec 14 2018

T[n_, k_] := 140*k^3*(n - k)^3 - 14*k*(n - k) + 1; Column[
Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* From Kolosov Petro, Apr 12 2020 *)

t(n, k) = 140*k^3*(n-k)^3-14*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */ trianglerows(9)

[[140*k^3*(n-k)^3 - 14*k*(n-k)+1 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 14 2018

From Kolosov Petro, Apr 12 2020: (Start)
T(n, k) = 140*k^3*(n-k)^3 - 14*k*(n-k) + 1.
T(n, k) = 140*A094053(n, k)^3 + 0*A094053(n, k)^2 - 14*A094053(n, k)^1 + 1.
T(n+3, k) = 4*T(n+2, k) - 6*T(n+1, k) + 4*T(n, k) - T(n-1, k), for n >= k.
Sum_{k=1..n} T(n, k) = A001015(n).
Sum_{k=0..n} T(n, k) = A258806(n).
Sum_{k=0..n-1} T(n, k) = A001015(n).
Sum_{k=1..n-1} T(n, k) = A258808(n).
Sum_{k=1..n-1} T(n, k) = -A024005(n).
Sum_{k=1..r} T(n, k) = -A316387(3, r, 0)*n^0 + A316387(3, r, 1)*n^1 - A316387(3, r, 2)*n^2 + A316387(3, r, 3)*n^3. (End)
G.f.: (1 + 127*x^6*y^3 - 3*x*(1 + y) + 585*x^5*y^2*(1 + y) + 129*x^4*y*(1 + 17*y + y^2) + 3*x^2*(1 + 45*y + y^2) - x^3*(1 - 579*y - 579*y^2 + y^3))/((1 - x)^4*(1 - x*y)^4). - Stefano Spezia, Sep 14 2024

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A284926 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^5.

Original entry on oeis.org

1, 31, 244, 991, 3126, 7564, 16808, 31711, 59293, 96906, 161052, 241804, 371294, 521048, 762744, 1014751, 1419858, 1838083, 2476100, 3097866, 4101152, 4992612, 6436344, 7737484, 9768751, 11510114, 14408200, 16656728, 20511150, 23645064, 28629152, 32472031, 39296688

Offset: 1

Seiichi Manyama, Apr 06 2017

Multiplicative because this sequence is the Dirichlet convolution of A000584 and A062157 which are both multiplicative. - Andrew Howroyd, Jul 20 2018

Seiichi Manyama, Table of n, a(n) for n = 1..10000
J. W. L. Glaisher, On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
Index entries for sequences mentioned by Glaisher.

Sum_{d|n} (-1)^(n/d+1)*d^k: A000593 (k=1), A078306 (k=2), A078307 (k=3), A284900 (k=4), this sequence (k=5), A284927 (k=6), A321552 (k=7), A321553 (k=8), A321554 (k=9), A321555 (k=10), A321556 (k=11), A321557 (k=12).

Cf. A000584, A013664, A062157.

Table[Sum[(-1)^(n/d + 1)*d^5, {d, Divisors[n]}], {n, 50}] (* Indranil Ghosh, Apr 06 2017 *)
f[p_, e_] := (p^(5*e + 5) - 1)/(p^5 - 1); f[2, e_] := (15*2^(5*e + 1) + 1)/31; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 11 2022 *)

a(n) = sumdiv(n, d, (-1)^(n/d + 1)*d^5); \\ Indranil Ghosh, Apr 06 2017

from sympy import divisors
print([sum((-1)**(n//d + 1)*d**5 for d in divisors(n)) for n in range(1, 51)]) # Indranil Ghosh, Apr 06 2017

G.f.: Sum_{k>=1} k^5*x^k/(1 + x^k). - Ilya Gutkovskiy, Apr 07 2017
From Amiram Eldar, Nov 11 2022: (Start)
Multiplicative with a(2^e) = (15*2^(5*e+1)+1)/31, and a(p^e) = (p^(5*e+5) - 1)/(p^5 - 1) if p > 2.
Sum_{k=1..n} a(k) ~ c * n^6, where c = 31*zeta(6)/192 = 0.164258... . (End)

Keyword:mult added by Andrew Howroyd, Jul 23 2018

A010803 15th powers: a(n) = n^15.

Original entry on oeis.org

0, 1, 32768, 14348907, 1073741824, 30517578125, 470184984576, 4747561509943, 35184372088832, 205891132094649, 1000000000000000, 4177248169415651, 15407021574586368, 51185893014090757, 155568095557812224

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (16, -120, 560, -1820, 4368, -8008, 11440, -12870, 11440, -8008, 4368, -1820, 560, -120, 16, -1).

Cf. A013673 (zeta(15)).

Cf. A000290 (squares), A000578 (cubes), A000583 (4th powers), A000584 (5th powers), A001015 (7th powers), A008455 (11th powers).

[n^15: n in [0..15]]; // Vincenzo Librandi, Jun 19 2011

Table[n^15,{n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2010 *)

for(n=0,15,print1(n^15,", ")) \\ Derek Orr, Feb 27 2017

A010803(n)=n^15 \\ M. F. Hasler, Jul 03 2025

A010803 = lambda n: n**15 # M. F. Hasler, Jul 03 2025

Totally multiplicative with a(p) = p^15 for prime p. Multiplicative with a(p^e) = p^(15e). - Jaroslav Krizek, Nov 01 2009
From Ilya Gutkovskiy, Feb 27 2017: (Start)
Dirichlet g.f.: zeta(s-15).
Sum_{n>=1} 1/a(n) = zeta(15) = A013673. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 16383*zeta(15)/16384. - Amiram Eldar, Oct 08 2020

A016793 a(n) = (3*n + 2)^5.

Original entry on oeis.org

32, 3125, 32768, 161051, 537824, 1419857, 3200000, 6436343, 11881376, 20511149, 33554432, 52521875, 79235168, 115856201, 164916224, 229345007, 312500000, 418195493, 550731776, 714924299, 916132832, 1160290625, 1453933568, 1804229351, 2219006624, 2706784157, 3276800000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A016789, A016790, A016791, A016792.

Subsequence of A000584.

[(3*n+2)^5 : n in [0..30]]; // Vincenzo Librandi, Sep 29 2011

Table[(3n+2)^5,{n,0,100}] (* Mohammad K. Azarian, Jun 15 2016 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{32,3125,32768,161051,537824,1419857},30] (* Harvey P. Dale, May 10 2024 *)

From Ilya Gutkovskiy, Jun 16 2016: (Start)
G.f.: (32 + 2933*x + 14498*x^2 + 10678*x^3 + 1018*x^4 + x^5)/(1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). (End)
From Amiram Eldar, Mar 31 2022: (Start)
a(n) = A016789(n)^5.
Sum_{n>=0} 1/a(n) = 121*zeta(5)/243 - 2*Pi^5/(729*sqrt(3)). (End)

A016949 a(n) = (6*n + 3)^5.

Original entry on oeis.org

243, 59049, 759375, 4084101, 14348907, 39135393, 90224199, 184528125, 345025251, 601692057, 992436543, 1564031349, 2373046875, 3486784401, 4984209207, 6956883693, 9509900499, 12762815625, 16850581551, 21924480357, 28153056843, 35723051649, 44840334375, 55730836701

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A016757, A016945, A016946, A016947, A016948.

Subsequence of A000584.

[(6*n+3)^5: n in [0..50]]; // Vincenzo Librandi, May 05 2011

a[n_] := (6*n + 3)^5; Array[a, 50, 0] (* Amiram Eldar, Mar 30 2022 *)

From Amiram Eldar, Mar 30 2022: (Start)
a(n) = A016945(n)^5.
a(n) = 3^5*A016757(n).
Sum_{n>=0} 1/a(n) = 31*zeta(5)/7776.
Sum_{n>=0} (-1)^n/a(n) = 5*Pi^5/373248. (End)

A016961 a(n) = (6*n + 4)^5.

Original entry on oeis.org

1024, 100000, 1048576, 5153632, 17210368, 45435424, 102400000, 205962976, 380204032, 656356768, 1073741824, 1680700000, 2535525376, 3707398432, 5277319168, 7339040224, 10000000000, 13382255776, 17623416832, 22877577568, 29316250624, 37129300000, 46525874176

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A016793, A016957, A016958, A016959, A016960.

Subsequence of A000584.

[(6*n+4)^5: n in [0..30]]; // Vincenzo Librandi, May 06 2011

a[n_] := (6*n + 4)^5; Array[a, 20, 0] (* Amiram Eldar, Mar 31 2022 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1024,100000,1048576,5153632,17210368,45435424},30] (* Harvey P. Dale, Jan 01 2025 *)

From Amiram Eldar, Mar 31 2022: (Start)
a(n) = A016957(n)^5.
a(n) = 32*A016793(n).
Sum_{n>=0} 1/a(n) = 121*zeta(5)/7776 - Pi^5/(11664*sqrt(3)). (End)

cp's OEIS Frontend

A068236 First differences of (n+1)^5-n^5.

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A113850 Numbers whose prime factors are raised to the fifth power.

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A300656 Triangle read by rows: T(n,k) = 30*k^2*(n-k)^2 + 1; n >= 0, 0 <= k <= n.

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A300785 Triangle read by rows: T(n,k) = 140*k^3*(n-k)^3 - 14*k*(n-k) + 1; n >= 0, 0 <= k <= n.

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A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

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A284926 a(n) = Sum_{d|n} (-1)^(n/d+1)*d^5.

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A300656 Triangle read by rows: T(n,k) = 30k^2(n-k)^2 + 1; n >= 0, 0 <= k <= n.

A300785 Triangle read by rows: T(n,k) = 140k^3(n-k)^3 - 14k(n-k) + 1; n >= 0, 0 <= k <= n.