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 1-9 of 9 results.

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

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 (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).

Crossrefs

Partial sums give A000539.

Programs

Formula

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)

Extensions

More terms from Henry Bottomley, Jun 21 2001

A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1
Offset: 0

Views

Author

Kolosov Petro, Aug 31 2017

Keywords

Comments

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(1, n, k), i.e T(n, k) is partial case of L(m, n, k) for m = 1.
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

Examples

			Triangle begins:
  ----------------------------------------
  k=    0   1   2   3   4   5   6   7   8
  ----------------------------------------
  n=0:  1;
  n=1:  1,  1;
  n=2:  1,  7,  1;
  n=3:  1, 13, 13,  1;
  n=4:  1, 19, 25, 19,  1;
  n=5:  1, 25, 37, 37, 25,  1;
  n=6:  1, 31, 49, 55, 49, 31,  1;
  n=7:  1, 37, 61, 73, 73, 61, 37,  1;
  n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;
		

Crossrefs

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.
Column sums k=0..4 give A000027, A000567, A051866, A051872, A255185.
Row sums give A001093.
Various cases of L(m, n, k): This sequence (m=1), A300656(m=2), A300785(m=3). See comments for L(m, n, k).
Differences of cubes n^3 are T(A000124(n), 1).

Programs

  • GAP
    Flat(List([0..11],n->List([0..n],k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018
    
  • Magma
    /* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018
    
  • Maple
    T := (n, k) -> 6*k*(n-k) + 1:
    seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018
  • Mathematica
    T[n_, k_] := 6 k (n - k) + 1; Column[Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jun 02 2019 *)
  • PARI
    t(n, k) = 6*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) \\ Felix Fröhlich, Jan 09 2018
    
  • SageMath
    def A287326(n,k): return 6*k*(n-k) + 1
    flatten([[A287326(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 25 2024

Formula

T(n, k) = 6*k*(n-k) + 1.
G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.
G.f.: (1 - x - x*y + 7*x^2*y)/((1 - x)^2*(1 - x*y)^2). - Stefano Spezia, Oct 09 2018 [Adapted by Stefano Spezia, Sep 25 2024]
From Kolosov Petro, Jun 05 2019: (Start)
T(n, k) = 1/2 * T(A294317(n, k), k) + 1/2.
T(n+1, k) = 2*T(n, k) - T(n-1, k), for n >= k.
T(n, k) = 6*A077028(n, k) - 5.
T(2n, n) = A227776(n).
T(2n+1, n) = A003154(n+1).
T(2n+3, n) = A166873(n+1).
Sum_{k=0..n-1} T(n, k) = Sum_{k=1..n} T(n, k) = A000578(n).
Sum_{k=1..n-1} T(n, k) = A068601(n).
(n+1)^3 - n^3 = T(A000124(n), 1). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = (-1/2)*(1 + (-1)^n)*A016969(floor(n/2) - 1). - G. C. Greubel, Sep 25 2024

A300785 Triangle read by rows: T(n,k) = 140*k^3*(n-k)^3 - 14*k*(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

Views

Author

Kolosov Petro, Mar 12 2018

Keywords

Comments

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)

Examples

			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;
		

Crossrefs

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.

Programs

  • GAP
    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
  • Magma
    /* 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
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    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)
    
  • Sage
    [[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
    

Formula

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

A302971 Triangle read by rows: T(n,k) is the numerator of R(n,k) defined implicitly by the identity Sum_{i=0..l-1} Sum_{j=0..m} R(m,j)*(l-i)^j*i^j = l^(2*m+1) holding for all l,m >= 0.

Original entry on oeis.org

1, 1, 6, 1, 0, 30, 1, -14, 0, 140, 1, -120, 0, 0, 630, 1, -1386, 660, 0, 0, 2772, 1, -21840, 18018, 0, 0, 0, 12012, 1, -450054, 491400, -60060, 0, 0, 0, 51480, 1, -11880960, 15506040, -3712800, 0, 0, 0, 0, 218790, 1, -394788954, 581981400, -196409840, 8817900, 0, 0, 0, 0, 923780, 1, -16172552880, 26003271294, -10863652800, 1031151660, 0, 0, 0, 0, 0, 3879876
Offset: 0

Views

Author

Kolosov Petro, Apr 16 2018

Keywords

Examples

			Triangle begins:
------------------------------------------------------------------------
k=   0          1         2         3    4     5      6      7       8
------------------------------------------------------------------------
n=0: 1;
n=1: 1,         6;
n=2: 1,         0,       30;
n=3: 1,       -14,        0,      140;
n=4: 1,      -120,        0,        0, 630;
n=5: 1,     -1386,      660,        0,   0, 2772;
n=6: 1,    -21840,    18018,        0,   0,    0, 12012;
n=7: 1,   -450054,   491400,   -60060,   0,    0,     0, 51480;
n=8: 1, -11880960, 15506040, -3712800,   0,    0,     0,     0, 218790;
		

Crossrefs

Items of second row are the coefficients in the definition of A287326.
Items of third row are the coefficients in the definition of A300656.
Items of fourth row are the coefficients in the definition of A300785.
T(n,n) gives A002457(n).
Denominators of R(n,k) are shown in A304042.
Row sums return A000079(2n+1) - 1.

Programs

  • Maple
    R := proc(n, k) if k < 0 or k > n then return 0 fi; (2*k+1)*binomial(2*k, k);
    if n = k then % else -%*add((-1)^j*R(n, j)*binomial(j, 2*k+1)*
    bernoulli(2*j-2*k)/(j-k), j=2*k+1..n) fi end: T := (n, k) -> numer(R(n, k)):
    seq(print(seq(T(n, k), k=0..n)), n=0..12);
    # Numerical check that S(m, n) = n^(2*m+1):
    S := (m, n) -> add(add(R(m, j)*(n-k)^j*k^j, j=0..m), k=0..n-1):
    seq(seq(S(m, n) - n^(2*m+1), n=0..12), m=0..12); # Peter Luschny, Apr 30 2018
  • Mathematica
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    T[n_, k_] := Numerator[R[n, k]];
    (* Print Fifteen Initial rows of Triangle A302971 *)
    Column[ Table[ T[n, k], {n, 0, 15}, {k, 0, n}], Center]
  • PARI
    T(n, k) = if ((n>k) || (n<0), 0, if (k==n, (2*n+1)*binomial(2*n, n), if (2*n+1>k, 0, if (n==0, 1, (2*n+1)*binomial(2*n, n)*sum(j=2*n+1, k+1, T(j, k)*binomial(j, 2*n+1)*(-1)^(j-1)/(j-n)*bernfrac(2*j-2*n))))));
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(numerator(T(k,n)), ", ")); print); \\ Michel Marcus, Apr 27 2018

Formula

Recurrence given by Max Alekseyev (see the MathOverflow link):
R(n, k) = 0 if k < 0 or k > n.
R(n, k) = (2k+1)*binomial(2k, k) if k = n.
R(n, k) = (2k+1)*binomial(2k, k)*Sum_{j=2k+1..n} R(n, j)*binomial(j, 2k+1)*(-1)^(j-1)/(j-k)*Bernoulli(2j-2k), otherwise.
T(n, k) = numerator(R(n, k)).

A304042 Triangle read by rows: T(n,k) is the denominator of R(n,k) defined implicitly by the identity Sum_{i=0..l-1} Sum_{j=0..m} R(m,j)*(l-i)^j*i^j = l^(2*m+1) holding for all l,m >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Kolosov Petro, May 05 2018

Keywords

Examples

			Triangle begins:
-----------------------------------------------------
k=    0  1  2  3  4  5  6  7  8  9  10 11 12 13 14 15
-----------------------------------------------------
n=0:  1;
n=1:  1, 1;
n=2:  1, 1, 1;
n=3:  1, 1, 1, 1;
n=4:  1, 1, 1, 1, 1;
n=5:  1, 1, 1, 1, 1, 1;
n=6:  1, 1, 1, 1, 1, 1, 1;
n=7:  1, 1, 1, 1, 1, 1, 1, 1;
n=8:  1, 1, 1, 1, 1, 1, 1, 1, 1;
n=9:  1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=10: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=11: 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1;
n=12: 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=13: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=14: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
n=15: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
		

Crossrefs

Programs

  • Mathematica
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    T[n_, k_] := Denominator[R[n, k]];
    (* Print Fifteen Initial rows of Triangle A304042 *)
    Column[ Table[ T[n, k], {n, 0, 15}, {k, 0, n}], Center]
  • PARI
    up_to = 1274; \\ = binomial(50+1,2)-1
    A304042aux(n, k) = if((k<0)||(k>n),0,(k+k+1)*binomial(2*k, k)*if(k==n,1,sum(j=k+k+1,n, A304042aux(n, j)*binomial(j, k+k+1)*((-1)^(j-1))/(j-k)*bernfrac(2*(j-k)))));
    A304042tr(n, k) = denominator(A304042aux(n, k));
    A304042list(up_to) = { my(v = vector(up_to), i=0); for(n=0,oo, for(k=0,n, if(i++ > up_to, return(v)); v[i] = A304042tr(n,k))); (v); };
    v304042 = A304042list(1+up_to);
    A304042(n) = v304042[1+n]; \\ Antti Karttunen, Nov 07 2018

Formula

Recurrence given by Max Alekseyev (see the MathOverflow link):
R(n, k) = 0 if k < 0 or k > n.
R(n, k) = (2k+1)*binomial(2k, k) if k = n.
R(n, k) = (2k+1)*binomial(2k, k)*Sum_{j=2k+1..n} R(n, j)*binomial(j, 2k+1)*(-1)^(j-1)/(j-k)*Bernoulli(2j-2k), otherwise.
T(n, k) = denominator(R(n, k)).

A258807 a(n) = n^5 - 1.

Original entry on oeis.org

0, 31, 242, 1023, 3124, 7775, 16806, 32767, 59048, 99999, 161050, 248831, 371292, 537823, 759374, 1048575, 1419856, 1889567, 2476098, 3199999, 4084100, 5153631, 6436342, 7962623, 9765624, 11881375, 14348906, 17210367, 20511148, 24299999, 28629150, 33554431
Offset: 1

Views

Author

Vincenzo Librandi, Jun 11 2015

Keywords

Crossrefs

Subsequence of A181124.
Sequences of the type n^k-1: A132411 (k=2), A068601 (k=3), A123865 (k=4), this sequence (k=5), A123866 (k=6), A258808 (k=7), A258809 (k=8), A258810 (k=9), A123867 (k=10), A258812 (k=11), A123868 (k=12).

Programs

  • GAP
    List([1..35],n->n^5-1); # Muniru A Asiru, Oct 28 2018
    
  • Magma
    [n^5-1: n in [1..50]];
    
  • Magma
    I:=[0,31,242,1023, 3124,7775]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+ 6*Self(n-5)-Self(n-6): n in [1..50]];
    
  • Maple
    seq(n^5-1,n=1..35); # Muniru A Asiru, Oct 28 2018
  • Mathematica
    Table[n^5 - 1, {n, 1, 50}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 31, 242, 1023, 3124, 7775}, 50]
  • PARI
    a(n)=n^5-1 \\ Charles R Greathouse IV, Jun 11 2015
    
  • Python
    for n in range(1, 50): print(n**5 - 1, end=', ') # Stefano Spezia, Oct 28 2018
  • Sage
    [n^5-1 for n in (1..50)] # Bruno Berselli, Jun 11 2015
    

Formula

G.f.: x^2*(31 + 56*x + 36*x^2 - 4*x^3 + x^4)/(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).
a(n) = -A024003(n). - Bruno Berselli, Jun 11 2015
Sum_{n>=2} 1/a(n) = Sum_{n>=1} (zeta(5*n) - 1) = 0.0379539032... - Amiram Eldar, Nov 06 2020

A316349 Consider coefficients U(m,L,k) defined by the identity Sum_{k=1..L} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,L,k) * T^k that holds for all positive integers L,m,T. This sequence gives 3-column table read by rows, where the n-th row lists coefficients U(2,n,k) for k = 0, 1, 2; n >= 1.

Original entry on oeis.org

31, 60, 30, 512, 540, 150, 2943, 2160, 420, 10624, 6000, 900, 29375, 13500, 1650, 68256, 26460, 2730, 140287, 47040, 4200, 263168, 77760, 6120, 459999, 121500, 8550, 760000, 181500, 11550, 1199231, 261360, 15180, 1821312, 365040, 19500, 2678143, 496860, 24570, 3830624, 661500, 30450
Offset: 1

Views

Author

Kolosov Petro, Jun 29 2018

Keywords

Comments

For L=T, the identity takes form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

Examples

			       column   column  column
   L     k=0      k=1     k=2
  --  -------  -------  ------
   1       31       60      30
   2      512      540     150
   3     2943     2160     420
   4    10624     6000     900
   5    29375    13500    1650
   6    68256    26460    2730
   7   140287    47040    4200
   8   263168    77760    6120
   9   459999   121500    8550
  10   760000   181500   11550
  11  1199231   261360   15180
  12  1821312   365040   19500
  ...
		

Crossrefs

The case m=1 is A320047.
The case m=3 is A316387.
Column k=0 is A316457.
Column k=1 is A316458.
Column k=2 is A316459.

Programs

  • Mathematica
    (* Define the R[n,k] := A302971(m,j)/A304042(m,j) *)
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    (* Define the U(m,l,t) coefficients *)
    U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];
    (* Define the value of the variable 'm', should be m = 2 for A316349 *)
    m = 2;
    (* Print first 10 rows of U(m,l,t) coefficients over l: 1 <= l <= 10 *)
    Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

Formula

U(2,n,0) = 6*n^5 + 15*n^4 + 10*n^3; U(2,n,1) = 15*n^4 + 30*n^3 + 15*n^2; U(2,n,2) = 10*n^3 + 15*n^2 + 5*n. - Max Alekseyev, Sep 06 2018
From Colin Barker, Jul 06 2018: (Start)
G.f.: x*(31 + 60*x + 30*x^2 + 326*x^3 + 180*x^4 - 30*x^5 + 336*x^6 - 180*x^7 - 30*x^8 + 26*x^9 - 60*x^10 + 30*x^11 + x^12) / ((1 - x)^6*(1 + x + x^2)^6).
a(n) = 6*a(n-3) - 15*a(n-6) + 20*a(n-9) - 15*a(n-12) + 6*a(n-15) - a(n-18) for n>18. (End)
U(m,L,t) = (-1)^m * Sum_{k=1..L} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, L >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after Max Alekseyev, see links.

Extensions

Edited by Max Alekseyev, Sep 06 2018

A316387 Consider coefficients U(m,L,k) defined by the identity Sum_{k=1..L} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,L,k) * T^k that holds for all positive integers L,m,T. This sequence gives 4-column table read by rows, where the n-th row lists coefficients U(3,n,k) for k = 0, 1, 2, 3; n >= 1.

Original entry on oeis.org

125, 406, 420, 140, 9028, 13818, 7140, 1260, 110961, 115836, 41160, 5040, 684176, 545860, 148680, 14000, 2871325, 1858290, 411180, 31500, 9402660, 5124126, 955500, 61740, 25872833, 12182968, 1963920, 109760, 62572096, 25945416, 3684240, 181440, 136972701, 50745870, 6439860, 283500, 276971300, 92745730, 10639860, 423500
Offset: 1

Views

Author

Kolosov Petro, Jul 01 2018

Keywords

Comments

For T=L, the identity takes form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

Examples

			            column      column      column   column
   L          k=0         k=1         k=2      k=3
  --  ------------  ----------  ----------  -------
   1           125         406         420      140
   2          9028       13818        7140     1260
   3        110961      115836       41160     5040
   4        684176      545860      148680    14000
   5       2871325     1858290      411180    31500
   6       9402660     5124126      955500    61740
   7      25872833    12182968     1963920   109760
   8      62572096    25945416     3684240   181440
   9     136972701    50745870     6439860   283500
  10     276971300    92745730    10639860   423500
  11     524988145   160386996    16789080   609840
  12     943023888   264896268    25498200   851760
  13    1618774781   420839146    37493820  1159340
  14    2672907076   646725030    53628540  1543500
  15    4267591425   965662320    74891040  2016000
  16    6616398080  1406064016   102416160  2589440
  17    9995653693  2002403718   137494980  3277260
  18   14757360516  2796022026   181584900  4093740
  19   21343778801  3835983340   236319720  5054000
  20   30303773200  5179983060   303519720  6174000
  ...
		

Crossrefs

The case m=1 is A320047.
The case m=2 is A316349.
Column k=0 is A317981.
Column k=1 is A317982.
Column k=2 is A317983.
Column k=3 is A317984.

Programs

  • Mathematica
    (* Define the R[n,k] := A302971(m,j)/A304042(m,j) *)
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    (* Define the U(m,l,t) coefficients *)
    U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];
    (* Define the value of the variable 'm' *)
    m = 3;
    (* Print first 10 rows of U(m,l,t) coefficients for 'm' defined above *)
    Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

Formula

U(3,n,0) = 20*n^7 + 70*n^6 + 70*n^5 - 28*n^3 - 7*n^2; U(3,n,1) = 70*n^6 + 210*n^5 + 175*n^4 - 42*n^2 - 7*n; U(3,n,2) = 84*n^5 + 210*n^4 + 140*n^3 - 14*n; U(3,n,3) = 35*n^4 + 70*n^3 + 35*n^2. - Max Alekseyev, Sep 06 2018
From Colin Barker, Jul 09 2018; corrected by Max Alekseyev, Sep 06 2018: (Start)
G.f.: x*(125 + 406*x + 420*x^2 + 140*x^3 + 8028*x^4 + 10570*x^5 + 3780*x^6 + 140*x^7 + 42237*x^8 + 16660*x^9 - 4200*x^10 - 1120*x^11 + 42272*x^12 - 16660*x^13 - 4200*x^14 + 1120*x^15 + 8007*x^16 - 10570*x^17 + 3780*x^18 - 140*x^19 + 132*x^20 - 406*x^21 + 420*x^22 - 140*x^23 - x^24) / ((1 - x)^8*(1 + x)^8*(1 + x^2)^8).
a(n) = 8*a(n-4) - 28*a(n-8) + 56*a(n-12) - 70*a(n-16) + 56*a(n-20) - 28*a(n-24) + 8*a(n-28) - a(n-32) for n>32. (End)
U(m,L,t) = (-1)^m * Sum_{k=1..L} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, L >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after Max Alekseyev, see links. - Kolosov Petro, Oct 04 2018

Extensions

Edited by Max Alekseyev, Sep 06 2018

A320047 Consider coefficients U(m,l,k) defined by the identity Sum_{k=1..l} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,l,k) * T^k that holds for all positive integers l,m,T. This sequence gives 2-column table read by rows, where n-th row lists coefficients U(1,n,k) for k = 0, 1 and n >= 1.

Original entry on oeis.org

5, 6, 28, 18, 81, 36, 176, 60, 325, 90, 540, 126, 833, 168, 1216, 216, 1701, 270, 2300, 330, 3025, 396, 3888, 468, 4901, 546, 6076, 630, 7425, 720, 8960, 816, 10693, 918, 12636, 1026, 14801, 1140, 17200, 1260, 19845, 1386, 22748, 1518, 25921, 1656
Offset: 1

Views

Author

Kolosov Petro, Oct 04 2018

Keywords

Comments

For l=T, the identity takes the form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

Examples

			        column  column
   l      k=0     k=1
  ---   ------  ------
   1       5       6
   2      28      18
   3      81      36
   4     176      60
   5     325      90
   6     540     126
   7     833     168
   8    1216     216
   9    1701     270
  10    2300     330
  11    3025     396
  12    3888     468
  ...
		

Crossrefs

The case m=2 is A316349.
The case m=3 is A316387.
Column k=0 is A275709.
Column k=1 is A028896.

Programs

  • Mathematica
    (* Define the R[n,k] := A302971(n,k)/A304042(n,k) *)
    R[n_, k_] := 0
    R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
       Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
       BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
    R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
    (* Define the U(m,l,t) coefficients *)
    U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];
    (* Define the value of the variable 'm' to be m = 1 for A320047 *)
    m = 1;
    (* Print first 10 rows of U(m,l,t) coefficients for 'm' defined above *)
    Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

Formula

U(m,l,t) = (-1)^m * Sum_{k=1..l} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, l >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after Max Alekseyev, see links.
Conjectures from Colin Barker, Aug 03 2019: (Start)
G.f.: x*(5 + 6*x + 8*x^2 - 6*x^3 - x^4) / ((1 - x)^4*(1 + x)^4).
a(n) = (4 - 4*(-1)^n - 3*(-5+(-1)^n)*n - 3*(-3+(-1)^n)*n^2 + (1+(-1)^(1+n))*n^3) / 8.
a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n>8.
(End)
Showing 1-9 of 9 results.