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.

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A213500 Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.

Original entry on oeis.org

1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90
Offset: 1

Views

Author

Clark Kimberling, Jun 14 2012

Keywords

Comments

Principal diagonal: A002412.
Antidiagonal sums: A002415.
Row 1: (1,2,3,...)**(1,2,3,...) = A000292.
Row 2: (1,2,3,...)**(2,3,4,...) = A005581.
Row 3: (1,2,3,...)**(3,4,5,...) = A006503.
Row 4: (1,2,3,...)**(4,5,6,...) = A060488.
Row 5: (1,2,3,...)**(5,6,7,...) = A096941.
Row 6: (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples: let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
h .......... h .......... A213500 . A002412 . A002415
h .......... h^2 ........ A212891 . A213436 . A024166
h^2 ........ h .......... A213503 . A117066 . A033455
h^2 ........ h^2 ........ A213505 . A213546 . A213547
h .......... h*(h+1)/2 .. A213548 . A213549 . A051836
h*(h+1)/2 .. h .......... A213550 . A002418 . A005585
h*(h+1)/2 .. h*(h+1)/2 .. A213551 . A213552 . A051923
h .......... h^3 ........ A213553 . A213554 . A101089
h^3 ........ h .......... A213555 . A213556 . A213547
h^3 ........ h^3 ........ A213558 . A213559 . A213560
h^2 ........ h*(h+1)/2 .. A213561 . A213562 . A213563
h*(h+1)/2 .. h^2 ........ A213564 . A213565 . A101094
2^(h-1) .... h .......... A213568 . A213569 . A047520
2^(h-1) .... h^2 ........ A213573 . A213574 . A213575
h .......... Fibo(h) .... A213576 . A213577 . A213578
Fibo(h) .... h .......... A213579 . A213580 . A053808
Fibo(h) .... Fibo(h) .... A067418 . A027991 . A067988
Fibo(h+1) .. h .......... A213584 . A213585 . A213586
Fibo(n+1) .. Fibo(h+1) .. A213587 . A213588 . A213589
h^2 ........ Fibo(h) .... A213590 . A213504 . A213557
Fibo(h) .... h^2 ........ A213566 . A213567 . A213570
h .......... -1+2^h ..... A213571 . A213572 . A213581
-1+2^h ..... h .......... A213582 . A213583 . A156928
-1+2^h ..... -1+2^h ..... A213747 . A213748 . A213749
h .......... 2*h-1 ...... A213750 . A007585 . A002417
2*h-1 ...... h .......... A213751 . A051662 . A006325
2*h-1 ...... 2*h-1 ...... A213752 . A100157 . A071238
2*h-1 ...... -1+2^h ..... A213753 . A213754 . A213755
-1+2^h ..... 2*h-1 ...... A213756 . A213757 . A213758
2^(n-1) .... 2*h-1 ...... A213762 . A213763 . A213764
2*h-1 ...... Fibo(h) .... A213765 . A213766 . A213767
Fibo(h) .... 2*h-1 ...... A213768 . A213769 . A213770
Fibo(h+1) .. 2*h-1 ...... A213774 . A213775 . A213776
Fibo(h) .... Fibo(h+1) .. A213777 . A001870 . A152881
h .......... 1+[h/2] .... A213778 . A213779 . A213780
1+[h/2] .... h .......... A213781 . A213782 . A005712
1+[h/2] .... [(h+1)/2] .. A213783 . A213759 . A213760
h .......... 3*h-2 ...... A213761 . A172073 . A002419
3*h-2 ...... h .......... A213771 . A213772 . A132117
3*h-2 ...... 3*h-2 ...... A213773 . A214092 . A213818
h .......... 3*h-1 ...... A213819 . A213820 . A153978
3*h-1 ...... h .......... A213821 . A033431 . A176060
3*h-1 ...... 3*h-1 ...... A213822 . A213823 . A213824
3*h-1 ...... 3*h-2 ...... A213825 . A213826 . A213827
3*h-2 ...... 3*h-1 ...... A213828 . A213829 . A213830
2*h-1 ...... 3*h-2 ...... A213831 . A213832 . A212560
3*h-2 ...... 2*h-1 ...... A213833 . A130748 . A213834
h .......... 4*h-3 ...... A213835 . A172078 . A051797
4*h-3 ...... h .......... A213836 . A213837 . A071238
4*h-3 ...... 2*h-1 ...... A213838 . A213839 . A213840
2*h-1 ...... 4*h-3 ...... A213841 . A213842 . A213843
2*h-1 ...... 4*h-1 ...... A213844 . A213845 . A213846
4*h-1 ...... 2*h-1 ...... A213847 . A213848 . A180324
[(h+1)/2] .. [(h+1)/2] .. A213849 . A049778 . A213850
h .......... C(2*h-2,h-1) A213853
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.

Examples

			Northwest corner (the array is read by southwest falling antidiagonals):
  1,  4, 10, 20,  35,  56,  84, ...
  2,  7, 16, 30,  50,  77, 112, ...
  3, 10, 22, 40,  65,  98, 140, ...
  4, 13, 28, 50,  80, 119, 168, ...
  5, 16, 34, 60,  95, 140, 196, ...
  6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
		

Crossrefs

Cf. A000027.

Programs

  • Mathematica
    b[n_] := n; c[n_] := n
    t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
    TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
    Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
    r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213500 *)
  • PARI
    t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
    tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
    tabl(12) \\ Indranil Ghosh, Mar 26 2017
    
  • Python
    def t(n, k): return sum((k - i) * (n + i) for i in range(k))
    for n in range(1, 13):
        print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

Formula

T(n,k) = 4*T(n,k-1) - 6*T(n,k-2) + 4*T(n,k-3) - T(n,k-4).
T(n,k) = 2*T(n-1,k) - T(n-2,k).
G.f. for row n: x*(n - (n - 1)*x)/(1 - x)^4.

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

A253636 Second partial sums of eighth powers (A001016).

Original entry on oeis.org

1, 258, 7076, 79430, 542409, 2685004, 10592400, 35277012, 103008345, 270739678, 652829892, 1464901802, 3092704433, 6196296120, 11862778432, 21824228040, 38761435089, 66718602714, 111659333380, 182200064046, 290563654073, 453803117636, 695353566480, 1046979329500
Offset: 1

Views

Author

Luciano Ancora, Jan 07 2015

Keywords

Comments

The general formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m)-F(m+1), where F(m) are the m-th Faulhaber’s formulas.

Crossrefs

Programs

  • GAP
    List([1..30], n-> n*(n+1)^2*(n+2)*(2*n^6 +12*n^5 +17*n^4 -12*n^3 -19*n^2 +18*n -3)/180); # G. C. Greubel, Aug 28 2019
  • Magma
    [n*(n+1)^2*(n+2)*(2*n^6+12*n^5+17*n^4-12*n^3-19*n^2+18*n-3)/180: n in [1..25]]; // Bruno Berselli, Jan 08 2015
    
  • Maple
    seq(n*(n+1)^2*(n+2)*(2*n^6 +12*n^5 +17*n^4 -12*n^3 -19*n^2 +18*n -3))/180, n=1..30); # G. C. Greubel, Aug 28 2019
  • Mathematica
    Table[n(n+1)^2(n+2)(2n^6 +12n^5 +17n^4 -12n^3 -19n^2 +18n -3)/180, {n,30}] (* Bruno Berselli, Jan 08 2015 *)
    Nest[Accumulate,Range[30]^8,2] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,258,7076,79430,542409,2685004,10592400, 35277012, 103008345,270739678,652829892},30] (* Harvey P. Dale, Jul 02 2017 *)
  • PARI
    a(n)=(2*n^10+20*n^9+75*n^8+120*n^7+42*n^6-84*n^5-50*n^4+40*n^3+21*n^2-6*n)/180 \\ Charles R Greathouse IV, Sep 08 2015
    
  • Sage
    [(2*n^10+20*n^9+75*n^8+120*n^7+42*n^6-84*n^5-50*n^4+40*n^3+21*n^2-6*n)/180 for n in [1..24]] # Tom Edgar, Jan 07 2015
    

Formula

a(n) = (2*n^10 + 20*n^9 + 75*n^8 + 120*n^7 + 42*n^6 - 84*n^5 - 50*n^4 + 40*n^3 + 21*n^2 - 6*n)/180.
a(n) = 2*a(n-1) - a(n-2) + n^8. - Robert Israel, Jan 07 2015
G.f.: x*(1 + x)*(1 + 246*x + 4047*x^2 + 11572*x^3 + 4047*x^4 + 246*x^5 + x^6) / (1 - x)^11. - Bruno Berselli, Jan 08 2015

A101090 Third partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 19, 135, 605, 2054, 5778, 14178, 31350, 63855, 121693, 219505, 378027, 625820, 1001300, 1555092, 2352732, 3477741, 5035095, 7155115, 9997801, 13757634, 18668870, 25011350, 33116850, 43375995, 56245761, 72257589, 92026135, 116258680, 145765224, 181469288
Offset: 1

Views

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004

Keywords

Comments

In general, the r-th successive summation of the fourth powers from 1 to n = (2*n+r)*(12*n^2+12*n*r+r^2-5*r)*(r+n)!/((r+4)!*(n-1)!). Here r = 3. - Gary Detlefs, Mar 01 2013

Crossrefs

Programs

  • Mathematica
    Nest[Accumulate, Range[50]^4, 3] (* Paolo Xausa, Jun 17 2024 *)

Formula

a(n) = (n*(1+n)*(2+n)*(3+n)*(3+2*n)*(-1+2*n*(3+n)))/840.
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^8. - Colin Barker, Apr 04 2012
a(n) = (2*n+3)*(12*n^2+36*n-6)*(n+3)!/(5040*(n-1)!). - Gary Detlefs, Mar 01 2013

Extensions

Edited by Ralf Stephan, Dec 16 2004

A254681 Fifth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 21, 176, 936, 3750, 12342, 35112, 89232, 207207, 446875, 906048, 1743248, 3206268, 5670588, 9690000, 16062144, 25912029, 40797009, 62837104, 94875000, 140670530, 205134930, 294610680, 417203280, 583171875, 805386231
Offset: 1

Views

Author

Luciano Ancora, Feb 12 2015

Keywords

Examples

			Fourth differences:  1, 12,  23,  24, (repeat 24)  ...   (A101104)
Third differences:   1, 13,  36,  60,   84,   108, ...   (A101103)
Second differences:  1, 14,  50, 110,  194,   302, ...   (A005914)
First differences:   1, 15,  65, 175,  369,   671, ...   (A005917)
-------------------------------------------------------------------------
The fourth powers:   1, 16,  81, 256,  625,  1296, ...   (A000583)
-------------------------------------------------------------------------
First partial sums:  1, 17,  98, 354,  979,  2275, ...   (A000538)
Second partial sums: 1, 18, 116, 470, 1449,  3724, ...   (A101089)
Third partial sums:  1, 19, 135, 605, 2054,  5778, ...   (A101090)
Fourth partial sums: 1, 20, 155, 760, 2814,  8592, ...   (A101091)
Fifth partial sums:  1, 21, 176, 936, 3750, 12342, ...   (this sequence)
		

Crossrefs

Programs

  • Magma
    [Binomial(n+5,6)*n*(n+5)*(2*n+5)/42: n in [1..30]]; // G. C. Greubel, Dec 01 2018
    
  • Maple
    seq(coeff(series((x+11*x^2+11*x^3+x^4)/(1-x)^10,x,n+1), x, n), n = 1 .. 30); # Muniru A Asiru, Dec 02 2018
  • Mathematica
    Table[n^2(1+n)(2+n)(3+n)(4+n)(5+n)^2(5+2n)/30240, {n,26}] (* or *)
    CoefficientList[Series[(1 + 11 x + 11 x^2 + x^3)/(1-x)^10, {x,0,25}], x]
    CoefficientList[Series[(1/30240)E^x (30240 + 604800 x + 2041200 x^2 + 2368800 x^3 + 1233540 x^4 + 326592 x^5 + 46410 x^6 + 3540 x^7 + 135 x^8 + 2 x^9), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* Stefano Spezia, Dec 02 2018 *)
    Nest[Accumulate[#]&,Range[30]^4,5] (* Harvey P. Dale, Jan 03 2022 *)
  • PARI
    my(x='x+O('x^30)); Vec((x+11*x^2+11*x^3+x^4)/(1-x)^10) \\ G. C. Greubel, Dec 01 2018
    
  • Sage
    [binomial(n+5,6)*n*(n+5)*(2*n+5)/42 for n in (1..30)] # G. C. Greubel, Dec 01 2018

Formula

G.f.:(x + 11*x^2 + 11*x^3 + x^4)/(1 - x)^10.
a(n) = n^2*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(5 + 2*n)/30240.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^4.
E.g.f.: (1/30240)*exp(x)*(30240 + 604800*x + 2041200*x^2 + 2368800*x^3 + 1233540*x^4 + 326592*x^5 + 46410*x^6 + 3540*x^7 + 135*x^8 + 2*x^9). - Stefano Spezia, Dec 02 2018
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 172032*log(2)/125 - 2382233/2500.
Sum_{n>=1} (-1)^(n+1)/a(n) = 42*Pi^2/25 - 43008*Pi/125 + 2663213/2500. (End)

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

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

A213553 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = (n-1+h)^3, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 10, 8, 46, 43, 27, 146, 142, 118, 64, 371, 366, 334, 253, 125, 812, 806, 766, 658, 466, 216, 1596, 1589, 1541, 1406, 1150, 775, 343, 2892, 2884, 2828, 2666, 2346, 1846, 1198, 512, 4917, 4908, 4844, 4655, 4271, 3646, 2782, 1753, 729, 7942
Offset: 1

Views

Author

Clark Kimberling, Jun 17 2012

Keywords

Comments

Principal diagonal: A213554
Antidiagonal sums: A101089
row 1, (1,2,3,...)**(1,8,27,...): A024166
row 2, (1,2,3,...)**(8,27,64,...): (3*k^5 + 30*k^4 + 115*k^3 + 210*k^2 + 122*k)/60
row 3, (1,2,3,...)**(27,64,125,...): (3*k^5 + 45*k^4 + 265*k^3 + 765*k^2 + 542*k)/120
For a guide to related arrays, see A213500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1.....10....46.....146....371
8.....43....142....366....806
27....118...334....766....1541
64....253...658....1406...2666
125...466...1150...2346...4271
		

Crossrefs

Cf. A213500.

Programs

  • GAP
    Flat(List([1..12], n-> List([1..n], k-> Binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30 ))); # G. C. Greubel, Jul 31 2019
  • Magma
    [Binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k + n*(3*n^2 +6*n +1))/30: k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 31 2019
    
  • Mathematica
    (* First program *)
    b[n_]:= n; c[n_]:= n^3;
    T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
    TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
    Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]]
    r[n_]:= Table[T[n, k], {k, 1, 60}]  (* A213553 *)
    d = Table[T[n, n], {n, 1, 40}] (* A213554 *)
    s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
    s1 = Table[s[n], {n, 1, 50}] (* A101089 *)
    (* Second program *)
    Table[Binomial[n-k+2, 2]*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 31 2019 *)
  • PARI
    t(n,k) = binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30;
    for(n=1,12, for(k=1,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Jul 31 2019
    
  • Sage
    [[binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k + n*(3*n^2 +6*n +1))/30 for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 31 2019
    

Formula

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) -T(n,k-6).
G.f. for row n: f(x)/g(x), where f(x) = n^3 + (-3*n^3 + 3*n^2 + 3*n + 1)*x + (3*n^3 - 6*n^2 + 4)*x^2 - ((n-1)^3)*x^3 and g(x) = (1 - x)^6.
T(n,k) = k*((3*k^4 - 5*k^2 + 2) + 15*k*(k^2 - 1)*n + 30*(k^2 - 1)*n^2 + 30*(k + 1)*n^3)/60. - G. C. Greubel, Jul 31 2019

A254470 Sixth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 22, 198, 1134, 4884, 17226, 52338, 141570, 348777, 795652, 1701700, 3444948, 6651216, 12321804, 22011804, 38073948, 63985977, 104782986, 167620090, 262495090, 403165620, 608300550, 902911230, 1320114510, 1903286385, 2708672616, 3808530792, 5294887048
Offset: 1

Views

Author

Luciano Ancora, Feb 15 2015

Keywords

Examples

			First differences:   1, 15,  65, 175,  369,   671, ... (A005917)
-------------------------------------------------------------------------
The fourth powers:   1, 16,  81, 256,  625,  1296, ... (A000583)
-------------------------------------------------------------------------
First partial sums:  1, 17,  98, 354,  979,  2275, ... (A000538)
Second partial sums: 1, 18, 116, 470, 1449,  3724, ... (A101089)
Third partial sums:  1, 19, 135, 605, 2054,  5778, ... (A101090)
Fourth partial sums: 1, 20, 155, 760, 2814,  8592, ... (A101091)
Fifth partial sums:  1, 21, 176, 936, 3750, 12342, ... (A254681)
Sixth partial sums:  1, 22, 198,1134, 4884, 17226, ... (this sequence)
		

Crossrefs

Programs

  • Magma
    [n*(1+n)*(2+n)*(3+n)^2*(4+n)*(5+n)*(6+n)*(1+12*n+ 2*n^2)/302400: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015
    
  • Mathematica
    Table[n (1 + n) (2 + n) (3 + n)^2 (4 + n) (5 + n) (6 + n) (1 + 12 n + 2 n^2)/302400,{n, 25}] (* or *) CoefficientList[Series[(- 1 - 11 x - 11 x^2 - x^3)/(- 1 + x)^11, {x, 0, 24}], x]
    Nest[Accumulate,Range[30]^4,6] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,22,198,1134,4884,17226,52338,141570,348777,795652,1701700},30] (* Harvey P. Dale, Apr 23 2016 *)
  • PARI
    vector(50,n,n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400) \\ Derek Orr, Feb 19 2015

Formula

G.f.: (-x - 11*x^2 - 11*x^3 - x^4)/(- 1 + x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400.
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) + n^4.
Sum_{n>=1} 1/a(n) = 3320303/2601 + 1400*Pi^2/17 + (8960/17)*sqrt(2/17)*Pi*cot(sqrt(17/2)*Pi). - Amiram Eldar, Jan 26 2022

A254870 Seventh partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 23, 221, 1355, 6239, 23465, 75803, 217373, 566150, 1361802, 3063502, 6508450, 13159666, 25481470, 47493274, 85567222, 149553199, 254336185, 421956275, 684451365, 1087616985, 1695917535, 2598828765, 3918943275, 5822229660, 8530902276, 12339433068
Offset: 1

Views

Author

Luciano Ancora, Feb 17 2015

Keywords

Examples

			Second differences:   2, 14,  50,  110,  194,   302, ...   A120328(2k+1)
First differences:    1, 15,  65,  175,  369,   671, ...   A005917
--------------------------------------------------------------------------
The fourth powers:    1, 16,  81,  256,  625,  1296, ...   A000583
--------------------------------------------------------------------------
First partial sums:   1, 17,  98,  354,  979,  2275, ...   A000538
Second partial sums:  1, 18, 116,  470, 1449,  3724, ...   A101089
Third partial sums:   1, 19, 135,  605, 2054,  5778, ...   A101090
Fourth partial sums:  1, 20, 155,  760, 2814,  8592, ...   A101091
Fifth partial sums:   1, 21, 176,  936, 3750, 12342, ...   A254681
Sixth partial sums:   1, 22, 198, 1134, 4884, 17226, ...   A254470
Seventh partial sums: 1, 23, 221, 1355, 6239, 23465, ...   (this sequence)
		

Crossrefs

Programs

  • Magma
    [n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)*(7+n)*(7+2*n)*(7 +42*n+6*n^2)/19958400: n in [1..30]]; // Vincenzo Librandi, Feb 19 2015
  • Mathematica
    Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (7 + 2 n)((7 + 42 n + 6 n^2)/19958400), {n, 24}] (* or *)
    CoefficientList[Series[(1 + 11 x + 11 x^2 + x^3)/(- 1 + x)^12, {x, 0, 23}], x]
  • PARI
    vector(50,n,n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 2*n)*(7 + 42*n + 6*n^2)/19958400) \\ Derek Orr, Feb 19 2015
    

Formula

G.f.: (x + 11*x^2 + 11*x^3 + x^4)/(- 1 + x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 2*n)*(7 + 42*n + 6*n^2)/19958400.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + n^4.
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