cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 26 results. Next

A259109 2*A000540.

Original entry on oeis.org

0, 2, 130, 1588, 9780, 41030, 134342, 369640, 893928, 1956810, 3956810, 7499932, 13471900, 23125518, 38184590, 60965840, 94520272, 142795410, 210819858, 304911620, 432911620, 604443862, 831203670, 1127275448, 1509481400, 1997762650, 2615594202, 3390435180
Offset: 0

Views

Author

N. J. A. Sloane, Jun 24 2015

Keywords

Crossrefs

Cf. A000540.

Programs

  • Mathematica
    LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{0,2,130,1588,9780,41030,134342,369640},30] (* Harvey P. Dale, Jul 18 2020 *)
  • PARI
    concat(0, Vec(2*x*(x+1)*(x^4+56*x^3+246*x^2+56*x+1)/(x-1)^8 + O(x^100))) \\ Colin Barker, Jun 28 2015

Formula

a(n) = n/21-n^3/3+n^5+n^6+(2*n^7)/7. - Colin Barker, Jun 28 2015
G.f.: 2*x*(x+1)*(x^4+56*x^3+246*x^2+56*x+1) / (x-1)^8. - Colin Barker, Jun 28 2015

A001014 Sixth powers: a(n) = n^6.

Original entry on oeis.org

0, 1, 64, 729, 4096, 15625, 46656, 117649, 262144, 531441, 1000000, 1771561, 2985984, 4826809, 7529536, 11390625, 16777216, 24137569, 34012224, 47045881, 64000000, 85766121, 113379904, 148035889, 191102976, 244140625, 308915776, 387420489, 481890304
Offset: 0

Views

Author

Keywords

Comments

Numbers both square and cubic. - Patrick De Geest
Totally multiplicative sequence with a(p) = p^6 for prime p. - Jaroslav Krizek, Nov 01 2009
Numbers n for which the order of the torsion subgroup of the elliptic curve y^2 = x^3 + n is t = 6, cf. Gebel link. - Artur Jasinski, Jun 30 2010
Note that Sum_{n>=1} 1/a(n) = Pi^6 / 945. - Mohammad K. Azarian, Nov 01 2011
The binomial transform yields A056468. The inverse binomial transform yields the (finite) 0, 1, 62, 540, ..., 720, the 6th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
For n > 0, a(n) is the largest number k such that k + n^3 divides k^2 + n^3. - Derek Orr, Oct 01 2014

Examples

			The 6th powers of the first few integers are: 0^6 = 0 = a(0), 1^6 = 1 = a(1), 2^6 = 64 = a(2), 3^6 = 9^3 = 729 = a(3), 4^6 = 2^12 = 4096 = a(4), 5^6 = 25^3 = 15625 = a(5), etc.
		

References

  • 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, eq. (6.37).
  • Granino A. Korn and Theresa M.Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
  • 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).

Crossrefs

Subsequence of A201217.
Cf. A000540 (partial sums), A022522 (first differences), A008292.
Intersection of A000290 (squares) and A000578 (cubes).
Cf. A002604 (n^6+1), A123866 (n^6-1), A013664 (zeta(6)), A275703 (eta(6)).
Cf. A003358 - A003368 (sums of 2, ..., 12 positive sixth powers).

Programs

Formula

a(n) = A123866(n) + 1 = A002604(n) - 1.
G.f.: -x*(1+x)*(x^4+56*x^3+246*x^2+56*x+1) / (x-1)^7. - Simon Plouffe in his 1992 dissertation
Multiplicative with a(p^e) = p^(6e). - David W. Wilson, Aug 01 2001
E.g.f.: (x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6)*exp(x). Generally, the e.g.f. for n^m is Sum_{k=1..m} A008277(m,k)*x^k*exp(x). - Geoffrey Critzer, Aug 25 2013
From Ant King, Sep 23 2013: (Start)
Signature {7, -21, 35, -35, 21, -7, 1}.
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) + 720. (End)
a(n) == 1 (mod 7) if gcd(n, 7) = 1, otherwise a(n) == 0 (mod 7). See A109720. - Jake Lawrence, May 28 2016
From Ilya Gutkovskiy, Jul 06 2016: (Start)
Dirichlet g.f.: zeta(s-6).
Sum_{n>=1} 1/a(n) = Pi^6/945 = A013664. (End)
a(n) = Sum_{k=1..6} Eulerian(6, k)*binomial(n+6-k, 6), with Eulerian(6, k) = A008292(6, k) (the numbers are 1, 57, 302, 302, 57, 1) for n >= 0. Worpitzki's identity for powers of 6. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = 31*zeta(6)/32 = 31*Pi^6/30240 (A275703). - Amiram Eldar, Oct 08 2020
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = (cosh(Pi)-cos(sqrt(3)*Pi))*sinh(Pi)/(2*Pi^3).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)^2/(6*Pi^2). (End)

Extensions

Comments from 2010 - 2011 edited by M. F. Hasler, Jul 05 2024

A000538 Sum of fourth powers: 0^4 + 1^4 + ... + n^4.

Original entry on oeis.org

0, 1, 17, 98, 354, 979, 2275, 4676, 8772, 15333, 25333, 39974, 60710, 89271, 127687, 178312, 243848, 327369, 432345, 562666, 722666, 917147, 1151403, 1431244, 1763020, 2153645, 2610621, 3142062, 3756718, 4463999, 5273999, 6197520, 7246096, 8432017, 9768353
Offset: 0

Views

Author

Keywords

Comments

This sequence is related to A000537 by the transform a(n) = n*A000537(n) - Sum_{i=0..n-1} A000537(i). - Bruno Berselli, Apr 26 2010
A formula for the r-th successive summation of k^4, for k = 1 to n, is ((12*n^2+(12*n-5)*r+r^2)*(2*n+r)*(n+r)!)/((r+4)!*(n-1)!), (H. W. Gould). - Gary Detlefs, Jan 02 2014
The number of four dimensional hypercubes in a 4D grid with side lengths n. This applies in general to k dimensions. That is, the number of k-dimensional hypercubes in a k-dimensional grid with side lengths n is equal to the sum of 1^k + 2^k + ... + n^k. - Alejandro Rodriguez, Oct 20 2020

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 222.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
  • 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).

Crossrefs

Programs

  • Haskell
    a000538 n = (3 * n * (n + 1) - 1) * (2 * n + 1) * (n + 1) * n `div` 30
    -- Reinhard Zumkeller, Nov 11 2012
    
  • Magma
    [n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30: n in [0..35]]; // Vincenzo Librandi, Apr 04 2015
  • Maple
    A000538 := n-> n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30;
  • Mathematica
    Accumulate[Range[0,40]^4] (* Harvey P. Dale, Jan 13 2011 *)
    CoefficientList[Series[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 07 2015 *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 17, 98, 354, 979}, 35] (* Jean-François Alcover, Feb 09 2016 *)
    Table[x^5/5+x^4/2+x^3/3-x/30,{x,40}] (* Harvey P. Dale, Jun 06 2021 *)
  • Maxima
    A000538(n):=n*(n+1)*(2*n+1)*(3*n^2+3*n-1)/30$
    makelist(A000538(n),n,0,30); /* Martin Ettl, Nov 12 2012 */
    
  • PARI
    a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30 \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    concat(0, Vec(x*(1+11*x+11*x^2+x^3)/(1-x)^6 + O(x^100))) \\ Altug Alkan, Dec 07 2015
    
  • Python
    A000538_list, m = [0], [24, -36, 14, -1, 0, 0]
    for _ in range(10**2):
        for i in range(5):
            m[i+1] += m[i]
        A000538_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Python
    def A000538(n): return n*(n**2*(n*(6*n+15)+10)-1)//30 # Chai Wah Wu, Oct 03 2024
    
  • Sage
    [bernoulli_polynomial(n,5)/5 for n in range(1, 35)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = n*(1+n)*(1+2*n)*(-1+3*n+3*n^2)/30.
The preceding formula is due to al-Kachi (1394-1437). - Juri-Stepan Gerasimov, Jul 12 2009
G.f.: x*(x+1)*(1+10*x+x^2)/(1-x)^6. Simon Plouffe in his 1992 dissertation. More generally, the o.g.f. for Sum_{k=0..n} k^m is x*E(m, x)/(1-x)^(m+2), where E(m, x) is the Eulerian polynomial of degree m (cf. A008292). The e.g.f. for these o.g.f.s is: x/(1-x)^2*(exp(y/(1-x))-exp(x*y/(1-x)))/(exp(x*y/(1-x))-x*exp(y/(1-x))). - Vladeta Jovovic, May 08 2002
a(n) = Sum_{i = 1..n} J_4(i)*floor(n/i), where J_4 is A059377. - Enrique Pérez Herrero, Feb 26 2012
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 24. - Ant King, Sep 23 2013
a(n) = -Sum_{j=1..4} j*Stirling1(n+1,n+1-j)*Stirling2(n+4-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = -30*(4 + 3/cos(sqrt(7/3)*Pi/2))*Pi/7. - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)*(n + 1/2)*n*(n + 1/2 + sqrt(7/12))*(n + 1/2 - sqrt(7/12))/5, see the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015

Extensions

The general V. Jovovic formula has been slightly changed after his approval by Wolfdieter Lang, Nov 03 2011

A103438 Square array T(m,n) read by antidiagonals: Sum_{k=1..n} k^m.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 5, 6, 4, 0, 1, 9, 14, 10, 5, 0, 1, 17, 36, 30, 15, 6, 0, 1, 33, 98, 100, 55, 21, 7, 0, 1, 65, 276, 354, 225, 91, 28, 8, 0, 1, 129, 794, 1300, 979, 441, 140, 36, 9, 0, 1, 257, 2316, 4890, 4425, 2275, 784, 204, 45, 10
Offset: 0

Views

Author

Ralf Stephan, Feb 11 2005

Keywords

Comments

For the o.g.f.s of the column sequences for this array, see A196837 and the link given there. - Wolfdieter Lang, Oct 15 2011
T(m,n)/n is the m-th moment of the discrete uniform distribution on {1,2,...,n}. - Geoffrey Critzer, Dec 31 2018
T(1,n) divides T(m,n) for odd m. - Franz Vrabec, Dec 23 2020

Examples

			Square array begins:
  0, 1,  2,   3,    4,     5,     6,      7,      8,      9, ... A001477;
  0, 1,  3,   6,   10,    15,    21,     28,     36,     45, ... A000217;
  0, 1,  5,  14,   30,    55,    91,    140,    204,    285, ... A000330;
  0, 1,  9,  36,  100,   225,   441,    784,   1296,   2025, ... A000537;
  0, 1, 17,  98,  354,   979,  2275,   4676,   8772,  15333, ... A000538;
  0, 1, 33, 276, 1300,  4425, 12201,  29008,  61776, 120825, ... A000539;
  0, 1, 65, 794, 4890, 20515, 67171, 184820, 446964, 978405, ... A000540;
Antidiagonal triangle begins as:
  0;
  0, 1;
  0, 1,  2;
  0, 1,  3,  3;
  0, 1,  5,  6,  4;
  0, 1,  9, 14, 10,  5;
  0, 1, 17, 36, 30, 15, 6;
		

References

  • J. Faulhaber, Academia Algebrae, Darinnen die miraculosische inventiones zu den höchsten Cossen weiters continuirt und profitirt werden, Augspurg, bey Johann Ulrich Schönigs, 1631.

Crossrefs

Diagonals include A076015 and A031971.
Antidiagonal sums are in A103439.
Antidiagonals are the rows of triangle A192001.

Programs

  • Magma
    T:= func< n,k | n eq 0 select k else (&+[j^n: j in [0..k]]) >;
    [T(n-k,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 22 2021
    
  • Maple
    seq(print(seq(Zeta(0,-k,1)-Zeta(0,-k,n+1),n=0..9)),k=0..6);
    # (Produces the square array from the example.) Peter Luschny, Nov 16 2008
    # alternative
    A103438 := proc(m,n)
        (bernoulli(m+1,n+1)-bernoulli(m+1))/(m+1) ;
        if m = 0 then
            %-1 ;
        else
            % ;
        end if;
    end proc: # R. J. Mathar, May 10 2013
    # simpler:
    A103438 := proc(m,n)
        (bernoulli(m+1,n+1)-bernoulli(m+1,1))/(m+1) ;
    end proc: # Peter Luschny, Mar 20 2024
  • Mathematica
    T[m_, n_]:= HarmonicNumber[m, -n]; Flatten[Table[T[m-n, n], {m, 0, 11}, {n, m, 0, -1}]] (* Jean-François Alcover, May 11 2012 *)
  • PARI
    T(m,n)=sum(k=0,n,k^m)
    
  • Python
    from itertools import count, islice
    from math import comb
    from fractions import Fraction
    from sympy import bernoulli
    def A103438_T(m,n): return sum(k**m for k in range(1,n+1)) if n<=m else int(sum(comb(m+1,i)*(bernoulli(i) if i!=1 else Fraction(1,2))*n**(m-i+1) for i in range(m+1))/(m+1))
    def A103438_gen(): # generator of terms
        for m in count(0):
            for n in range(m+1):
                yield A103438_T(m-n,n)
    A103438_list = list(islice(A103438_gen(),100)) # Chai Wah Wu, Oct 23 2024
  • SageMath
    def T(n,k): return (bernoulli_polynomial(k+1, n+1) - bernoulli_polynomial(1, n+1)) /(n+1)
    flatten([[T(n-k,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Dec 22 2021
    

Formula

E.g.f.: e^x*(e^(x*y)-1)/(e^x-1).
T(m, n) = Zeta(-n, 1) - Zeta(-n, m + 1), for m >= 0 and n >= 0, where Zeta(z, v) is the Hurwitz zeta function. - Peter Luschny, Nov 16 2008
T(m, n) = HarmonicNumber(m, -n). - Jean-François Alcover, May 11 2012
T(m, n) = (Bernoulli(m + 1, n + 1) - Bernoulli(m + 1, 1)) / (m + 1). - Peter Luschny, Mar 20 2024
T(m, n) = Sum_{k=0...m-n} B(k)*(-1)^k*binomial(m-n,k)*n^(m-n-k+1)/(m-n-k+1), where B(k) = Bernoulli number A027641(k) / A027642(k). - Robert B Fowler, Aug 20 2024
T(m, n) = Sum_{i=1..n} J_m(i)*floor(n/i), where J_m is the m-th Jordan totient function. - Ridouane Oudra, Jul 19 2025

A000541 Sum of 7th powers: 1^7 + 2^7 + ... + n^7.

Original entry on oeis.org

0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456, 8080425, 18080425, 37567596, 73399404, 136147921, 241561425, 412420800, 680856256, 1091194929, 1703414961, 2597286700, 3877286700, 5678375241, 8172733129, 11577558576, 16164030000, 22267545625
Offset: 0

Views

Author

Keywords

Comments

a(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777) - Artur Jasinski, Oct 10 2007
This sequence is related to A000540 by a(n) = n*A000540(n) - Sum_{i=0..n-1} A000540(i). - Bruno Berselli, Apr 26 2010

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 815.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • 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).

Crossrefs

Row 7 of array A103438.

Programs

  • Magma
    [n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24: n in [0..30]]; // Vincenzo Librandi, Feb 20 2016
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^7 od: seq(a[n], n=0..25); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    Table[Sum[k^7, {k, 1, n}], {n, 0, 100}] (* Artur Jasinski, Oct 10 2007 *)
    s = 0; lst = {s}; Do[s += n^7; AppendTo[lst, s], {n, 1, 30, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
    LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 3297456}, 35] (* Vincenzo Librandi, Feb 20 2016 *)
  • PARI
    a(n)=n^2*(n+1)^2*(3*n^4+6*n^3-n^2-4*n+2)/24 \\ Edward Jiang, Sep 10 2014
    
  • PARI
    a(n) = sum(i=1, n, i^7); \\ Michel Marcus, Sep 11 2014
    
  • Python
    A000541_list, m = [0], [5040, -15120, 16800, -8400, 1806, -126, 1, 0, 0]
    for _ in range(10**2):
        for i in range(8):
            m[i+1] += m[i]
        A000541_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    

Formula

a(n) = n^2*(n+1)^2*(3*n^4 + 6*n^3 - n^2 - 4*n + 2)/24.
a(n) = sqrt(Sum_{j=1..n} Sum_{i=1..n} (i*j)^7). - Alexander Adamchuk, Oct 26 2004
Jacobi formula: a(n) = 2(A000217(n))^4 - A000539(n). - Artur Jasinski, Oct 10 2007
G.f.: x*(1 + 120*x + 1191*x^2 + 2416*x^3 + 1191*x^4 + 120*x^5 + x^6)/(1-x)^9. - Colin Barker, May 25 2012
a(n) = 8*a(n-1) - 28* a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) + 5040. - Ant King, Sep 24 2013
a(n) = -Sum_{j=1..7} j*Stirling1(n+1,n+1-j)*Stirling2(n+7-j,n). - Mircea Merca, Jan 25 2014
a(n) = 2*A000217(n)^4 - (4/3)*A000217(n)^3 + (1/3)*A000217(n)^2. - Michael Raney, Feb 19 2016
a(n) = 72*A288876(n-2) + 48*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, Apr 27 2024
a(n) = Sum_{i=1..n} J_7(i)*floor(n/i), where J_7 is A069092. - Ridouane Oudra, Jul 17 2025

A323544 a(n) = Product_{k=0..n} (k^6 + (n-k)^6).

Original entry on oeis.org

0, 1, 8192, 2245338225, 1144394036019200, 2577023355527587890625, 13410804447068120796679372800, 172661401915668867785003701060950625, 4548909593429214367033270472265433088000000, 234845240509381890690238640158397433600579682850625
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 17 2019

Keywords

Crossrefs

Cf. 2*A000540 and A259109 (with sum instead of product).

Programs

  • Magma
    [(&*[(k^6 + (n-k)^6): k in [0..n]]): n in [0..10]]; // Vincenzo Librandi, Jan 18 2019
    
  • Mathematica
    Table[Product[k^6+(n-k)^6, {k, 0, n}], {n, 0, 10}]
  • PARI
    m=6; vector(10, n, n--; prod(k=0,n, k^m + (n-k)^m)) \\ G. C. Greubel, Jan 18 2019
    
  • Sage
    m=6; [product(k^m +(n-k)^m for k in (0..n)) for n in (0..10)] # G. C. Greubel, Jan 18 2019

Formula

a(n) ~ exp(((15 - 4*sqrt(3))*Pi/6 - 6)*n) * n^(6*n+6).

A101093 Second partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 66, 860, 5750, 26265, 93436, 278256, 725220, 1703625, 3682030, 7431996, 14167946, 25730705, 44823000, 75305920, 122566056, 193963761, 299373690, 451829500, 668285310, 970507241, 1386109076, 1949746800, 2704487500
Offset: 1

Views

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Keywords

Crossrefs

Cf. A000540.

Programs

  • GAP
    List([2..30], n-> n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168); # G. C. Greubel, Aug 28 2019
  • Magma
    [n*(1+n)^2*(2+n)*(-1+n*(2+n))*(-2+3*n*(2+n))/168: n in [1..40]]; // Vincenzo Librandi, Mar 24 2014
    
  • Maple
    f:=n->(3*n^8-14*n^6+21*n^4-10*n^2)/168;
    [seq(f(n),n=0..50)];  # N. J. A. Sloane, Mar 23 2014
  • Mathematica
    CoefficientList[Series[(x+1)(x^4+56x^3+246x^2+56x+1)/(1-x)^9, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
    Nest[Accumulate,Range[30]^6,2] (* or *) LinearRecurrence[{9,-36,84,-126, 126,-84,36,-9,1},{1,66,860,5750,26265,93436,278256,725220,1703625},30] (* Harvey P. Dale, Jun 05 2019 *)
  • PARI
    vector(30, n, m=n+1; m^2*(3*m^6 -14*m^4 +21*m^2 -10)/168) \\ G. C. Greubel, Aug 28 2019
    
  • Sage
    [n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168 for n in (2..30)] # G. C. Greubel, Aug 28 2019
    

Formula

a(n) = n*(1 + n)^2*(2 + n)*(-1 + n*(2 + n))*(-2 + 3*n*(2 + n))/168.
G.f.: x*(1+x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4)/(1-x)^9. - Colin Barker, Dec 18 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^6, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^6. - Luciano Ancora, Jan 08 2015

Extensions

Edited by Ralf Stephan, Dec 16 2004

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

Views

Author

Cecilia Rossiter, Dec 15 2004

Keywords

Comments

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

Crossrefs

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
Cf. A101095 for an expanded table and more about MagicNKZ.

Programs

  • Mathematica
    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 *)

Formula

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

Extensions

New name from Joerg Arndt, Nov 30 2014
Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A023002 Sum of 10th powers.

Original entry on oeis.org

0, 1, 1025, 60074, 1108650, 10874275, 71340451, 353815700, 1427557524, 4914341925, 14914341925, 40851766526, 102769130750, 240627622599, 529882277575, 1106532668200, 2206044295976, 4222038196425, 7792505423049, 13923571680850
Offset: 0

Views

Author

Keywords

Crossrefs

Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), this sequence (m=10), A123095 (m=11), A123094 (m=12), A181134 (m=13).
Row 10 of array A103438.

Programs

Formula

a(n) = n*(n+1)*(2*n+1)*(n^2+n-1)(3*n^6 +9*n^5 +2*n^4 -11*n^3 +3*n^2 +10*n -5)/66 (see MathWorld, Power Sum, formula 40). - Bruno Berselli, Apr 26 2010
a(n) = n*A007487(n) - Sum_{i=0..n-1} A007487(i). - Bruno Berselli, Apr 27 2010
From Bruno Berselli, Aug 23 2011: (Start)
a(n) = -a(-n-1).
G.f.: x*(1+x)*(1 +1012*x +46828*x^2 +408364*x^3 +901990*x^4 +408364*x^5 +46828*x^6 +1012*x^7 +x^8)/(1-x)^12. (End)
a(n) = (-1)*Sum_{j=1..10} j*Stirling1(n+1,n+1-j)*Stirling2(n+10-j,n). - Mircea Merca, Jan 25 2014
a(n) = Sum_{i=1..n} J_10(i)*floor(n/i), where J_10 is A069095. - Ridouane Oudra, Jul 17 2025

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280
Offset: 1

Views

Author

Cecilia Rossiter, Dec 15 2004

Keywords

Comments

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.
The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Crossrefs

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... | n = 1 | n = 2 | n = 3 | n = 4 | n = 5 | n = 6 | n = 7 | n = 8
--------------------------------------------------------------------------------------
z = 0 | A000007 | A019590 | ....... MagicNKZ(n,k,0) = T(n,k+1) from A008292 .......
z = 1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

Programs

  • Magma
    I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015
    
  • Mathematica
    MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
    CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
    Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)
  • PARI
    a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015
  • Sage
    [1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015
    

Formula

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

Extensions

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015
Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015
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