A162624 -id:A162624 - cp's OEIS frontend

A000584 Fifth powers: a(n) = n^5.

0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049, 100000, 161051, 248832, 371293, 537824, 759375, 1048576, 1419857, 1889568, 2476099, 3200000, 4084101, 5153632, 6436343, 7962624, 9765625, 11881376, 14348907, 17210368, 20511149

Offset: 0

N. J. A. Sloane

Totally multiplicative sequence with a(p) = p^5 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A059338. The inverse binomial transform yields the (finite) 0, 1, 30, 150, 240, 120, the 5th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Equals sum of odd numbers from n^2*(n-1)+1 (A100104) to n^2*(n+1)-1 (A003777). - Bruno Berselli, Mar 14 2014
a(n) mod 10 = n mod 10. - Reinhard Zumkeller, May 10 2014
Numbers of the form a(n) + a(n+1) + ... + a(n+k) are nonprime for all n, k>=0; this can be proved by the method indicated in the comment in A256581. - Vladimir Shevelev and Peter J. C. Moses, Apr 04 2015

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Franklin T. Adams-Watters, Table of n, a(n) for n = 0..500
Brady Haran and Simon Pampena, Fifth Root Trick, Numberphile video (2014)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Partial sums give A000539.
Cf. A000012, A001477, A000290, A000578, A000583, A062392, A022521 (first differences), A008292, A173018, A123125.
Cf. A013663, A162624, A267316.

a000584 = (^ 5)  -- Reinhard Zumkeller, Nov 11 2012

Range[0,50]^5 (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)

makelist(n^5, n, 0, 30); /* Martin Ettl, Nov 12 2012 */

a(n)=n^5 \\ Charles R Greathouse IV, Jul 28 2015

[n**5 for n in range(30)]  # Zerinvary Lajos, Jun 03 2009

G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (x-1)^6. [Simon Plouffe in his 1992 dissertation]
Multiplicative with a(p^e) = p^(5e). - David W. Wilson, Aug 01 2001
E.g.f.: exp(x)*(x+15*x^2+25*x^3+10*x^4+x^5). - Geoffrey Critzer, Jun 12 2013
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 120. - Ant King, Sep 23 2013
a(n) = n + Sum_{j=0..n-1}{k=1..4}binomial(5,k)*j^(5-k). - Patrick J. McNab, Mar 28 2016
From Kolosov Petro, Oct 22 2018: (Start)
a(n) = Sum_{k=1..n} A300656(n,k).
a(n) = Sum_{k=0..n-1} A300656(n,k). (End)
a(n) = Sum_{k=1..5} Eulerian(5, k)*binomial(n+5-k, 5), with Eulerian(5, k) = A008292(5, k), the numbers 1, 26, 66, 26, 1, for n >= 0. Worpitzki's identity for powers of 5. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(5) (A013663).
Sum_{n>=1} (-1)^(n+1)/a(n) = 15*zeta(5)/16 (A267316). (End)

More terms from Henry Bottomley, Jun 21 2001

A162614 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^3 - 1.

Original entry on oeis.org

0, 1, 1, 2, 9, 16, 3, 29, 55, 81, 4, 67, 130, 193, 256, 5, 129, 253, 377, 501, 625, 6, 221, 436, 651, 866, 1081, 1296, 7, 349, 691, 1033, 1375, 1717, 2059, 2401, 8, 519, 1030, 1541, 2052, 2563, 3074, 3585, 4096, 9, 737, 1465, 2193, 2921, 3649, 4377, 5105, 5833

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the fourth power of n, A000583(n).

See also the triangles of A162615 and A162616.

			Triangle begins:
  0;
  1,   1;
  2,   9,  16;
  3,  29,  55,  81;
  4,  67, 130, 193, 256;
  5, 129, 253, 377, 501,  625;
  6, 221, 436, 651, 866, 1081, 1296;
  ...

Cf. A000583, A068601, A159797, A162609, A162610, A162611, A162612, A162613, A162615, A162616, A162622, A162623, A162624.

def A162614(n,k):
    return n+k*(n**3-1)
print([A162614(n,k) for n in range(20) for k in range(n+1)])
# R. J. Mathar, Oct 20 2009

Sum_{k=0..n} T(n,k) = n*(n^2-n+1)*(n+1)^2/2 (row sums). - R. J. Mathar, Jul 20 2009

T(n,k) = n + k*(n^3-1). - R. J. Mathar, Oct 20 2009

More terms from R. J. Mathar, Oct 20 2009

A162622 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^4 - 1.

Original entry on oeis.org

0, 1, 1, 2, 17, 32, 3, 83, 163, 243, 4, 259, 514, 769, 1024, 5, 629, 1253, 1877, 2501, 3125, 6, 1301, 2596, 3891, 5186, 6481, 7776, 7, 2407, 4807, 7207, 9607, 12007, 14407, 16807, 8, 4103, 8198, 12293, 16388, 20483, 24578, 28673, 32768, 9, 6569, 13129

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the 5th power of n, A000584(n).

See also the triangles of A162623 and A162624.

			Triangle begins:
  0;
  1,    1;
  2,   17,    32;
  3,   83,   163,   243;
  4,  259,   514,   769,  1024;
  5,  629,  1253,  1877,  2501,  3125;
  6, 1301,  2596,  3891,  5186,  6481,  7776;
  7, 2407,  4807,  7207,  9607, 12007, 14407, 16807;
  8, 4103,  8198, 12293, 16388, 20483, 24578, 28673, 32768;
  9, 6569, 13129, 19689, 26249, 32809, 39369, 45929, 52489, 59049; etc.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A000583, A000584, A123865, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162615, A162616, A162623, A162624.

/* Triangle: */ [[n+k*(n^4-1): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Dec 14 2012

A162622 := proc(n,k) n+k*(n^4-1) ; end proc: seq(seq( A162622(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Feb 11 2010

Flatten[Table[NestList[#+n^4-1&,n,n],{n,0,9}]] (* Harvey P. Dale, Jun 23 2013 *)

Sum_{k=0..n} T(n,k) = n*(n+1)*(1+n^4)/2 (row sums). [R. J. Mathar, Jul 20 2009]

7th and later rows from R. J. Mathar, Feb 11 2010

A177342 a(n) = (4n^3-3n^2+5*n-3)/3.

Original entry on oeis.org

1, 9, 31, 75, 149, 261, 419, 631, 905, 1249, 1671, 2179, 2781, 3485, 4299, 5231, 6289, 7481, 8815, 10299, 11941, 13749, 15731, 17895, 20249, 22801, 25559, 28531, 31725, 35149, 38811, 42719, 46881, 51305, 55999, 60971, 66229, 71781, 77635

Offset: 1

Bruno Berselli, May 06 2010 - Nov 27 2010

This sequence is related to the fourth powers (A000583) by n^4 = n*a(n) - Sum_{i=1..n-1} a(i) - (n-1), with n>1.

Also, n*a(n) - Sum_{i=1..n-1} a(i) provides the first column of A162624 and the second column of A162622 (or A162623). - Bruno Berselli, revised Dec 14 2012

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

First differences: 2*A084849.
Partial sums: A178073.
Cf. A174723, A162622-A162624.

[(4*n^3-3*n^2+5*n-3)/3: n in [1..39]]; // Bruno Berselli, Aug 24 2011

I:=[1,9,31,75]; [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..50]]; // Vincenzo Librandi, Aug 19 2013

CoefficientList[Series[(1 + 5 x + x^2 + x^3) / (1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
Table[(4 n^3 - 3 n^2 + 5 n - 3)/3, {n, 1, 40}] (* Bruno Berselli, Feb 17 2015 *)
LinearRecurrence[{4,-6,4,-1},{1,9,31,75},40] (* Harvey P. Dale, Jul 31 2021 *)

a(n)=(4*n^3-3*n^2+5*n-3)/3 \\ Charles R Greathouse IV, Jun 23 2011

G.f.: x*(1 + 5*x + x^2 + x^3)/(1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) - a(-n) = 2*A004006(2n).
a(n) + a(-n) = -A002522(n).
a(n) = 1 + (n-1)*(4*n^2+n+6)/3 = 2*A174723(n)-1.

Formulae added and revised by Bruno Berselli, Feb 17 2015

A162623 Triangle read by rows in which row n lists n terms, starting with n, such that the difference between successive terms is equal to n^4 - 1 = A123865(n).

Original entry on oeis.org

1, 2, 17, 3, 83, 163, 4, 259, 514, 769, 5, 629, 1253, 1877, 2501, 6, 1301, 2596, 3891, 5186, 6481, 7, 2407, 4807, 7207, 9607, 12007, 14407, 8, 4103, 8198, 12293, 16388, 20483, 24578, 28673, 9, 6569, 13129, 19689, 26249, 32809, 39369, 45929, 52489, 10

Offset: 1

Omar E. Pol, Jul 12 2009

See also the triangles of A162622 and A162624.

			Triangle begins:
  1;
  2,   17;
  3,   83,  163;
  4,  259,  514,  769;
  5,  629, 1253, 1877, 2501;
  6, 1301, 2596, 3891, 5186, 6481;

Cf. A000583, A000584, A123865, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162615, A162616, A162622, A162624.

A162623 := proc(n,k) n+k*(n^4-1) ; end: seq(seq(A162623(n,k),k=0..n-1),n=1..15) ; # R. J. Mathar, Sep 27 2009

dst[n_]:=Module[{c=n^4-1},Range[n,n*c,c]]; Flatten[Join[{1},Table[dst[n],{n,2,10}]]] (* Harvey P. Dale, Jul 29 2014 *)

Row sums: n*(n^5 - n^4 + n + 1)/2. - R. J. Mathar, Jul 20 2009

More terms from R. J. Mathar, Sep 27 2009

cp's OEIS Frontend

A000584 Fifth powers: a(n) = n^5.

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A162614 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^3 - 1.

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A162622 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^4 - 1.

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A177342 a(n) = (4*n^3-3*n^2+5*n-3)/3.

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A162623 Triangle read by rows in which row n lists n terms, starting with n, such that the difference between successive terms is equal to n^4 - 1 = A123865(n).

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Maple

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A177342 a(n) = (4n^3-3n^2+5*n-3)/3.