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 1-8 of 8 results.

A062392 a(n) = n^4 - (n-1)^4 + (n-2)^4 - ... 0^4.

Original entry on oeis.org

0, 1, 15, 66, 190, 435, 861, 1540, 2556, 4005, 5995, 8646, 12090, 16471, 21945, 28680, 36856, 46665, 58311, 72010, 87990, 106491, 127765, 152076, 179700, 210925, 246051, 285390, 329266, 378015, 431985, 491536, 557040, 628881, 707455, 793170, 886446, 987715
Offset: 0

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Author

Henry Bottomley, Jun 21 2001

Keywords

Comments

Number of edges in the join of two complete graphs of order n^2 and n, K_n^2 * K_n. - Roberto E. Martinez II, Jan 07 2002
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-5)(P(4,1)-(-1)^k P(4,2k+1))|. - Peter Luschny, Jul 12 2009
Define an infinite symmetric array by T(n,m) = n*(n-1) + m for 0 <= m <= n and T(n,m) = T(m,n), n >= 0. Then a(n) is the sum of terms in the top left (n+1) X (n+1) subarray: a(n) = Sum_{r=0..n} Sum_{c=0..n} T(r,c). - J. M. Bergot, Jul 05 2013
a(n) is the sum of all positive numbers less than A002378(n). - J. M. Bergot, Aug 30 2013
Except the first term, these are triangular numbers that remain triangular when divided by their index, e.g., 66 divided by 11 gives 6. - Waldemar Puszkarz, Sep 14 2017
a(n) is the semiperimeter of the unique primitive Pythagorean triple such that (a-b+c)/2 = T(n) = A000217(n). Its long leg and hypotenuse are consecutive natural numbers and the triple is (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1). - Miguel-Ángel Pérez García-Ortega, May 27 2025

Examples

			From _Bruno Berselli_, Oct 30 2017: (Start)
After 0:
1   =                 -(1) + (2);
15  =             -(1 + 2) + (3 + 4 + 5 + 2*3);
66  =         -(1 + 2 + 3) + (4 + 5 + 6 + 7 + ... + 11 + 3*4);
190 =     -(1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + ... + 19 + 4*5);
435 = -(1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + ... + 29 + 5*6), etc. (End)
		

References

  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Crossrefs

Cf. A000538, A000583. A062393 provides the result for 5th powers, A011934 for cubes, A000217 for squares, A001057 (unsigned) for nonnegative integers, A000035 (offset) for 0th powers.
Cf. A236770 (see crossrefs).

Programs

  • Maple
    a := n -> (2*n^2+n^3-1)*n/2; # Peter Luschny, Jul 12 2009
  • Mathematica
    Table[n (n + 1) (n^2 + n - 1)/2, {n, 0, 40}] (* Harvey P. Dale, Oct 19 2011 *)
  • PARI
    { a=0; for (n=0, 1000, write("b062392.txt", n, " ", a=n^4 - a) ) } \\ Harry J. Smith, Aug 07 2009

Formula

a(n) = n*(n+1)*(n^2 + n - 1)/2 = n^4 - a(n-1) = A000583(n) - a(n-1) = A000217(A028387(n-1)) = A000217(n)*A028387(n-1).
a(n) = Sum_{i=0..n} A007588(i) for n > 0. - Jonathan Vos Post, Mar 15 2006
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Harvey P. Dale, Oct 19 2011
G.f.: x*(x*(x + 10) + 1)/(1 - x)^5. - Harvey P. Dale, Oct 19 2011
a(n) = A000384(A000217(n)). - Bruno Berselli, Jan 31 2014
a(n) = A110450(n) - A002378(n). - Gionata Neri, May 13 2015
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*2*Pi/sqrt(5). - Amiram Eldar, Jan 22 2024
a(n) = sqrt(144*A288876(n-2) + 72*A006542(n+2) + A000537(n)). - Yasser Arath Chavez Reyes, Jul 22 2024
E.g.f.: exp(x)*x*(2 + 13*x + 8*x^2 + x^3)/2. - Stefano Spezia, Apr 27 2025
a(n) = A000217(n)*(2*A000217(n)-1). - Miguel-Ángel Pérez García-Ortega, May 27 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

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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

A108679 a(n) = (n+1)*(n+2)^2*(n+3)^2*(n+4)^2*(n+5)^2*(n+6)/86400.

Original entry on oeis.org

1, 21, 196, 1176, 5292, 19404, 60984, 169884, 429429, 1002001, 2186184, 4504864, 8836464, 16604784, 30046752, 52581816, 89311761, 147685461, 238369516, 376372920, 582481900, 885069900, 1322357400, 1945206900, 2820550005, 4035556161, 5702666256, 7965629056
Offset: 0

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Author

Emeric Deutsch, Jun 17 2005

Keywords

Comments

Kekulé numbers for certain benzenoids.
6th column of the table of Narayana numbers A001263. - Zerinvary Lajos, Jun 18 2007
Sequence provided by binomial(n-1,m)*binomial(n,m)/(m+1) for m=5 and n>5 (these numbers are also called Runyon numbers, see T. Koshy in References). - Vincenzo Librandi, Sep 04 2014

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 232, # 1).
  • T. Koshy, Catalan Numbers with Applications, Oxford University Press, 2009, p. 7.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; Prop. 8.4, case n=7. - N. J. A. Sloane, Aug 28 2010

Crossrefs

Cf. A001263, A002378, A005585, A006542, A006857 (sequences having a similar structure), A288876.

Programs

  • Magma
    [Binomial(n-1,5)*Binomial(n,5)/6: n in [6..45]]; // Vincenzo Librandi, Sep 04 2014
    
  • Maple
    a:=n->(n+1)*(n+2)^2*(n+3)^2*(n+4)^2*(n+5)^2*(n+6)/86400: seq(a(n),n=0..40);
  • Mathematica
    Table[(n + 1) (n + 2)^2 (n + 3)^2 (n + 4)^2 (n + 5)^2 (n + 6)/86400,{n, 0, 50}]  (* Harvey P. Dale, Mar 13 2011 *)
  • PARI
    Vec((1+10*x+20*x^2+10*x^3+x^4)/(1-x)^11 + O(x^99)) \\ Altug Alkan, Sep 02 2016
    
  • SageMath
    def A108679(n): return binomial(n+5,5)*binomial(n+6,5)//6
    print([A108679(n) for n in range(41)]) # G. C. Greubel, Mar 12 2025

Formula

a(n) = binomial(n+5, 5)*binomial(n+6, 5)/6 = binomial(n+6, 6)*binomial(n+6, 5)/(n+6).
a(n) = A001263(n+6,6).
G.f.: (1 + 10*x + 20*x^2 + 10*x^3 + x^4)/(1 - x)^11. Numerator polynomial is the fifth row polynomial of the Narayana triangle.
a(n) = binomial(n+5,5)^2 - binomial(n+5,4)*binomial(n+5,6). - Gary Detlefs, Dec 05 2011
a(n) = Product_{i=1..5} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 27637/2 - 1400*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 2560*log(2) - 3547/2. (End)
a(n) = (A005585(n+2)^2 - A288876(n+1))/24. - Yasser Arath Chavez Reyes, Aug 19 2024

A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

Original entry on oeis.org

1, 2, 1, 8, 8, 1, 48, 72, 18, 1, 384, 768, 288, 32, 1, 3840, 9600, 4800, 800, 50, 1, 46080, 138240, 86400, 19200, 1800, 72, 1, 645120, 2257920, 1693440, 470400, 58800, 3528, 98, 1, 10321920, 41287680, 36126720, 12042240, 1881600, 150528, 6272, 128, 1, 185794560, 836075520, 836075520, 325140480, 60963840, 6096384, 338688, 10368, 162, 1, 3715891200, 18579456000, 20901888000, 9289728000, 2032128000, 243855360, 16934400, 691200, 16200, 200, 1
Offset: 0

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Author

Wolfdieter Lang, Jun 16 2017

Keywords

Comments

These generalized unsigned Lah numbers are the instance L[2,1] of the Sheffer triangles called L[d,a], with integers d >= 1 and integers 0 <= a < d with gcd(d,a) = 1. The standard unsigned Lah numbers are L[1,0] = A271703.
The Sheffer structure of L[d,a] is ((1 - d*t)^(-2*a/d), t/(1 - d*t)). This follows from the defining property
risefac[d,a](x, n) = Sum_{m=0..n} L[d,a](n, m)*fallfac[d,a](x, m), where risefac[d,a](x, n):= Product_{0..n-1} (x + (a+d*j)) for n >= 1 and risefac[d,a](x, 0) := 1, and fallfac[d,a](x, n):= Product_{0..n-1} (x - (a+d*j)) = for n >= 1 and fallfac[d,a](x, 0) := 1. Such rising and falling factorials arise in the generalization of Stirling numbers of both kinds S2[d,a] and S1[d,a]. See the Peter Bala link under A143395 for these falling factorials called there [t;a,b,c]_n with t=x, a=d, b=0, c=a.
In matrix notation: L[d,a] = S1phat[d,a].S2hat[d,a] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations with Sheffer structures S1phat[d,a] = ((1 - d*t)^(-a/d), -(1/d)*(log(1 - d*t))) and S2hat[d,a] = (exp(a*t), (1/d)*(exp(d*t) - 1). See, e.g., S1phat[2,1] = A028338 and S2hat[2,1] = A039755.
The a- and z-sequences for these Sheffer matrices have e.g.f.s 1 + d*t and ((1 + d*t)/t)*(1 - (1 + d*t)^(-2*a/d)), respectively. See a W. Lang link under A006232 for these types of sequences.
E.g.f. of row polynomials R[d,a](n, x) := Sum_{m=0..n} L[d,a](n, m)*x^m
(1 - d*x)^(-2*a/d)*exp(t*x/(1 - d*x)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - d*t)^(-2*a/d)*(t/(1 - d*t))^m/m, m >= 0.
Meixner type identity for (monic) row polynomials: (D_x/(1 + d*D_x)) * R[d,a](n, x) = n*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1. The series in the differentiations D_x = d/dx terminates.
General Sheffer recurrence for row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R[d,a](n, x) = [(2*a+x)*1 + 2*d*(a + x)*D_x + d^2*x*(D_x)^2]*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1.
The inverse matrix L^(-1)[d,a] is Sheffer (g[d,a](-t), -f[d,a](-t)) with L[d,a] Sheffer (g[d,a](t), f[d,a](t)) from above. This means (see the column e.g.f. of Sheffer matrices) that L^(-1)[d,a](n, m) = (-1)^(n-m)*L[d,a](n, m). Therefore, the recurrence relations can easily be rewritten for L^(-1)[d,a] by replacing a -> -a and d -> -d.
fallfac[d,a](x, n) = Sum_{m=0..n} L^(-1)[d,a](n, m)*risefac[d,a](x, m), n >= 0.
From Wolfdieter Lang, Aug 12 2017: (Start)
The Sheffer row polynomials R[d,a](n, x) belong to the Boas-Buck class and satisfy therefore the Boas-Buck identity (see the reference, and we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function) (E_x - n*1)*R[d,a](n, x) = - n*(2*a*1 + d*E_x) * Sum_{k=0..n-1} d^k*R(d,a;n-1-k,x)/(n-1-k)!, with E_x = x*d/dx (Euler operator).
This implies a recurrence for the sequence of column m: L[d,a](n, m) = (n!*(2*a + d*m)/(n-m))*Sum_{p=0..n-1-m} d^p*L[d,a](n-1-p, m)/(n-1-p)!, for n > m>=0, and input L[d,a](m, m) = 1. For the present [d,a] = [2,1] instance see the formula and example sections. (End)
From Wolfdieter Lang, Sep 14 2017: (Start)
The diagonal sequences are 2^D*D!*(binomial(m+D, m))^2, m >= 0, for D >= 0 (main diagonal D = 0). From the o.g.f.s obtained via Lagrange's theorem. See the second W. Lang link below for the general Sheffer case.
The o.g.f. of the diagonal D sequence is 2^D*D!*Sum_{m=0..D} A008459(D, m)*x^m /(1- x)^(2*D + 1), D >= 0. (End)
It appears that this is also the matrix square of unsigned triangle of coefficients of Laguerre polynomials n!*L_n(x), abs(A021009(n, k)). - Ali Pourzand, Mar 10 2025 [This observation is correct. - Peter Luschny, Mar 10 2025]

Examples

			The triangle T(n, m) begins:
  n\m        0         1         2         3        4       5      6     7   8 9
  0:         1
  1:         2         1
  2:         8         8         1
  3:        48        72        18         1
  4:       384       768       288        32        1
  5:      3840      9600      4800       800       50       1
  6:     46080    138240     86400     19200     1800      72      1
  7:    645120   2257920   1693440    470400    58800    3528     98     1
  8:  10321920  41287680  36126720  12042240  1881600  150528   6272   128   1
  9: 185794560 836075520 836075520 325140480 60963840 6096384 338688 10368 162 1
  ...
From _Wolfdieter Lang_, Aug 12 2017: (Start)
Recurrence for column elements with m >= 1, and input column m = 0: T(3, 2) = (3/2)*T(2, 1) + 2*3*T(2, 2) = (3/2)*8 + 6 = 18.
Four term recurrence: T(3, 2) = T(2, 1) + 2*5*T(2, 2) - 4*2^2*T(1, 2) = 8 + 10 + 0 = 18.
Meixner type identity, n=2: 2*R(1, x) = (D_x - 2*(D_x)^2)*R(2, x), 2*(2 + x) = (8 + 2*x) - 2*2.
Sheffer recurrence: R(2, x) = (2 + x)*(2 + x) + 4*(1 + x)*1 + 0 = 8 + 8*x + x^2.
Boas-Buck recurrence for column m = 2 and n = 4: T(4, 2) = (2*4!*3/2)*(1*T(3, 2)/3! + 2*T(2, 2)/2!) = 4!*3*(18/3! + 1) = 288. (End)
Diagonal sequence D = 1: o.g.f. 2*1!*(1 + 1*x)/(1- x)^3 generating
{2*(binomial(m+1, m))^2}_{m >= 0} = {2, 8, 18, 32, ...}. - _Wolfdieter Lang_, Sep 14 2017
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
  • Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Crossrefs

Column sequences (no leading zeros): A000165, A014479, A286725.
Diagonal sequences: A000012, 2*A000290(m+1), 8*A000537(n+1), 48*A001249, 384*A288876. - Wolfdieter Lang, Sep 14 2017
Row sums are A025167. - Michael Somos, Sep 27 2017

Programs

  • Maple
    T := (n, k) -> ifelse(n < k, 0, ifelse(k = 0, n!*2^n, (n/k)*T(n-1, k-1) + 2*n*T(n-1, k))): seq(seq(T(n, k), k = 0..n), n = 0..10);  # Peter Luschny, Mar 10 2025
  • Mathematica
    T[ n_, k_] := Coefficient[ Integrate[ Exp[-x^2 - y x] HermiteH[n, x]^2, {x, -Infinity, Infinity}] / (Sqrt[Pi] Exp[y^2 / 4]), y, 2 k]; (* Michael Somos, Sep 27 2017 *)
  • SageMath
    # Using the function A021009_triangle, displays as a matrix. Following the observation of Ali Pourzand.
    print(A021009_triangle(9)^2)  # Peter Luschny, Mar 10 2025

Formula

T(n, m) = L[2,1](n, m) = Sum_{k=m..n} A028338(n, k)*A039755(k, m).
Three term recurrence for column elements with m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 2*n*T(n-1, m) with T(n, m) = 0 for n < m and the column m = 0 is T(n, 0) = (2*n)!! = n*2^n = A000165(n). (From the a- and z-sequences {1, 2, repeat(0)} and {2, repeat(0)}, respectively.)
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(2*n-1)*T(n-1, m) - 4*(n-1)^2*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
E.g.f. of row polynomials R(n, x) = R[2,1](n, x) (i.e., e.g.f. of the triangle): (1/(1-2*t))*exp(x*t/(1-2*t)).
E.g.f. of column m sequences: (t^m/(1-2*t)^(m+1))/m!, m >= 0.
Meixner type identity: Sum_{k=0..n-1} (-1)^k*2^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx.
Sheffer recurrence: R(n, x) = [(2 + x)*1 + 4*(1 + x)*D_x + 4*x*(D_x)^2]*R(n-1, x), n >= 1, and R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment above): T(n, m) = (2*n!*(1 + m)/(n-1))*Sum_{p=0..n-1-m} 2^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, and input T(m, m) = 1. - Wolfdieter Lang, Aug 12 2017
Explicit form (from the diagonal sequences with the o.g.f.s given as a comment above): T(n, m) = 2^(n-m)*(n-m)!*(binomial(n, n-m))^2 for n >= m >= 0. - Wolfdieter Lang, Sep 23 2017
Let R(n,x) denote the n-th row polynomial. Then x^n*R(n,x) = x^n o x^n, where o denotes the deformed Hadamard product of power series defined in Bala, Section 3.1. - Peter Bala, Jan 18 2018

A123095 Sum of first n 11th powers.

Original entry on oeis.org

0, 1, 2049, 179196, 4373500, 53201625, 415998681, 2393325424, 10983260016, 42364319625, 142364319625, 427675990236, 1170684360924, 2962844754961, 7012409924625, 15662165784000, 33254351828416, 67526248136049, 131794658215281, 248284917113500, 453084917113500
Offset: 0

Views

Author

Zerinvary Lajos, Sep 27 2006

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), A023002 (m=10), this sequence (m=11), A123094 (m=12), A181134 (m=13).

Programs

  • Magma
    [(&+[j^11: j in [0..n]]): n in [0..30]]; // G. C. Greubel, Jul 21 2021
    
  • Maple
    [seq(add(i^11, i=1..n), n=0..20)];
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^11 od: seq(a[n], n=0..13); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    Table[Sum[k^11, {k, n}], {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
    Accumulate[Range[0,20]^11] (* Harvey P. Dale, Sep 17 2021 *)
  • Python
    A123095_list, m = [0], [39916800, -199584000, 419126400, -479001600, 322494480, -129230640, 29607600, -3498000, 171006, -2046, 1, 0 , 0]
    for _ in range(10**2):
        for i in range(12):
            m[i+1]+= m[i]
        A123095_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Sage
    [(bernoulli_polynomial(n+1, 12) - bernoulli(12))/12  for n in (0..30)] # G. C. Greubel, Jul 21 2021

Formula

a(n) = n*A023002(n) - Sum_{i=0..n-1} A023002(i). - Bruno Berselli, Apr 27 2010
a(n) = n^2*(n+1)^2*(2*n^8 +8*n^7 +4*n^6 -16*n^5 -5*n^4 +26*n^3 -3*n^2 -20*n +10)/24. - Bruno Berselli, Oct 03 2010
G.f.: x*(x^10 +2036*x^9 +152637*x^8 +2203488*x^7 +9738114*x^6 +15724248*x^5 +9738114*x^4 +2203488*x^3 +152637*x^2 +2036*x +1)/(1-x)^13. - Colin Barker, May 27 2012
a(n) = (-1)*Sum_{j=1..11} j*Stirling1(n+1,n+1-j)*Stirling2(n+11-j,n). - Mircea Merca, Jan 25 2014
a(n) = 1728*A006542(n+2)^2 + 216*A288876(n-2) + 96*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, May 25 2024

A331575 a(n) is the number of subsets of {1..n} that contain 4 even and 4 odd numbers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 5, 25, 75, 225, 525, 1225, 2450, 4900, 8820, 15876, 26460, 44100, 69300, 108900, 163350, 245025, 353925, 511225, 715715, 1002001, 1366365, 1863225, 2484300, 3312400, 4331600, 5664400, 7282800, 9363600, 11860560, 15023376, 18779220, 23474025, 28997325
Offset: 0

Views

Author

Enrique Navarrete, Jan 20 2020

Keywords

Examples

			a(9)=5 and the 5 subsets are {1,2,3,4,5,6,7,8}, {1,2,3,4,5,6,8,9}, {1,2,3,4,6,7,8,9}, {1,2,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}.
		

Crossrefs

Cf. A288876 (even bisection, shifted).

Programs

  • Magma
    [IsEven(n) select Binomial((n div 2),4)^2 else Binomial((n-1) div 2,4)*Binomial((n+1) div 2,4): n in [0..41]]; // Marius A. Burtea, Jan 21 2020
  • Maple
    a:= n-> ((b, q)-> b(q, 4)*b(n-q, 4))(binomial, iquo(n, 2)):
    seq(a(n), n=0..50);  # Alois P. Heinz, Jan 30 2020
  • Mathematica
    a[n_] := If[OddQ[n], Binomial[(n - 1)/2, 4]*Binomial[(n + 1)/2, 4], Binomial[n/2, 4]^2]; Array[a, 42, 0] (* Amiram Eldar, Jan 21 2020 *)
    LinearRecurrence[{2,6,-14,-14,42,14,-70,0,70,-14,-42,14,14,-6,-2,1},{0,0,0,0,0,0,0,0,1,5,25,75,225,525,1225,2450},50] (* Harvey P. Dale, Jul 20 2025 *)
  • PARI
    a(n) = if (n%2, binomial((n-1)/2,4)*binomial((n+1)/2,4), binomial(n/2,4)^2); \\ Michel Marcus, Jan 21 2020
    
  • PARI
    concat([0,0,0,0,0,0,0,0], Vec(x^8*(1 + 3*x + 9*x^2 + 9*x^3 + 9*x^4 + 3*x^5 + x^6) / ((1 - x)^9*(1 + x)^7) + O(x^40))) \\ Colin Barker, Jan 21 2020
    

Formula

a(n) = binomial(n/2,4)^2, n even;
a(n) = binomial((n-1)/2,4)*binomial((n+1)/2,4), n odd.
From Colin Barker, Jan 21 2020: (Start)
G.f.: x^8*(1 + 3*x + 9*x^2 + 9*x^3 + 9*x^4 + 3*x^5 + x^6) / ((1 - x)^9*(1 + x)^7).
a(n) = 2*a(n-1) + 6*a(n-2) - 14*a(n-3) - 14*a(n-4) + 42*a(n-5) + 14*a(n-6) - 70*a(n-7) + 70*a(n-9) - 14*a(n-10) - 42*a(n-11) + 14*a(n-12) + 14*a(n-13) - 6*a(n-14) - 2*a(n-15) + a(n-16) for n>15.
(End)
E.g.f.: (cosh(x)-sinh(x))*(1575+1350*x+630*x^2+204*x^3+54*x^4+12*x^5+4*x^6+(-1575+1800*x-1080*x^2+456*x^3-156*x^4+48*x^5-16*x^6+8*x^7+2*x^8)*(cosh(2*x)+sinh(2*x)))/294912. - Stefano Spezia, Jan 27 2020

A303987 Triangle read by rows: T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0, k = 0..n.

Original entry on oeis.org

1, 1, 4, 1, 36, 36, 1, 144, 900, 400, 1, 400, 8100, 19600, 4900, 1, 900, 44100, 313600, 396900, 63504, 1, 1764, 176400, 2822400, 9922500, 7683984, 853776, 1, 3136, 571536, 17640000, 133402500, 276623424, 144288144, 11778624, 1, 5184, 1587600, 85377600, 1200622500, 5194373184, 7070119056, 2650190400, 165636900
Offset: 0

Views

Author

Wolfdieter Lang, May 14 2018

Keywords

Comments

The row sums of this triangle are b(n) = A005259(n), for n >= 0. This sequence b was used in R. Apéry's 1979 proof of the irrationality of Zeta(3). See A005259 for references and links.
Row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], x), hence b(n) = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], 1) (see the formula in A005259 given by K. A. Penson. This is the solution to Exercise 2.14 of the Koepf reference given there, p. 29).

Examples

			The triangle T begins:
n\k  0    1       2        3          4          5          6          7 ...
0:   1
1:   1    4
2:   1   36      36
3:   1  144     900      400
4:   1  400    8100    19600       4900
5:   1  900   44100   313600     396900      63504
6:   1 1764  176400  2822400    9922500    7683984     853776
7:   1 3136  571536 17640000  133402500  276623424  144288144   11778624
----------------------------------------------------------------------------
row n = 8:   1 5184 1587600 85377600 1200622500 5194373184 7070119056 2650190400 165636900,
row n = 9: 1 8100 3920400 341510400 8116208100 63631071504 176752976400 169612185600 47869064100 2363904400,
row n = 10: 1 12100 8820900 1177862400 44188244100 572679643536 2828047622400 5446435737600 3877394192100 853369488400 34134779536.
...
		

Crossrefs

The column sequences (without zeros) are A000012, A035287(n+1) = 4*A000217(n)^2, 36*A288876, 400*A000579(n+6)^2, 4900*A000581(n+8)^2, 63504*A001287(n+10)^2, ...

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->(Binomial(n,k)*Binomial(n+k,k))^2))); # Muniru A Asiru, May 15 2018
  • Mathematica
    T[n_, k_] := (Gamma[k + n + 1]/(Gamma[k + 1]^2*Gamma[-k + n + 1]))^2;
    Flatten[Table[T[n, k], {n, 0, 8}, {k, 0, n}]] (* Peter Luschny, May 14 2018 *)

Formula

T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0 and k = 0..n.
T(n, k) = (binomial(n+k, 2*k)*cbi(k))^2, with cbi(k) = A000984(k) = binomial(2*k, k), and cbi(k)^2 = A002894(k).
G.f. for column sequences (without leading zeros):
cbi(k)^2*P2(2*k, x)/(1 - x)^(4*k+1), with the row polynomials of A008459 (Pascal entries squared) P2(2*k, x) = Sum_{j=0..2*k} A008459(2*k, j)*x^j. For a proof see the general comment in A288876 on the diagonals and columns of A008459.

A290310 Irregular triangle read by rows. Row n gives the coefficients of the polynomial multiplying the exponential function in the e.g.f. of the (n+1)-th diagonal sequences of triangle A008459 (Pascal squares). T(n,k) for n >= 0 and k = 0..2*n.

Original entry on oeis.org

1, 1, 3, 2, 1, 8, 19, 18, 6, 1, 15, 69, 147, 162, 90, 20, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 63, 1169, 10703, 58821, 214123, 545629, 1004307, 1356194, 1347318, 974862, 500346, 172788, 36036, 3432
Offset: 0

Views

Author

Wolfdieter Lang, Jul 27 2017

Keywords

Comments

The length of row n of this irregular triangle is 2*n+1.
A008459 gives the squares of the entries of Pascal's triangle A007318.
The e.g.f. of the (n+1)-th diagonal sequence of the squares of Pascal's triangle (A008459) is EDP2(n, t) = Sum_{m=0..n} binomial(n+m, m)^2*t^m/m!, for n >= 0. It turns out to be EDP2(n, t) = exp(t)*Sum_{k=0..2*n} T(n, k)*t^k/k!.
This has been computed from the corresponding o.g.f.: GDP2(n, x) = Sum_{m=0..n} binomial(n+m, m)^2*x^m, which is GDP2(n, x) = Sum_{m=0..n} binomial(n,m)^2*x^m / (1 - x)^(2*n+1) (see the triangle A008459, and comments in A288876 on how to compute these o.g.f.s). To obtain the e.g.f.s from the o.g.f.s the formulas (23) - (25) of the W. Lang link given in A060187 have been used.

Examples

			The irregular triangle T(n, k) begins:
  n\k 0  1   2    3     4     5      6      7      8     9    10   11  12 ..,
  0:  1
  1:  1  3   2
  2:  1  8  19   18     6
  3:  1 15  69  147   162    90     20
  4:  1 24 176  624  1251  1500   1070    420     70
  5:  1 35 370 1920  5835 11253  14240  11830   6230  1890   252
  6:  1 48 687 4850 20385 55908 104959 137886 127050 80640 33642 8316 924
  ...
  7: 1 63 1169 10703 58821 214123 545629 1004307 1356194 1347318 974862 500346 172788 36036 3432
  ...
n = 3: The e.g.f. of the fourth diagonal sequence of A008459 is A001249 = [1, 16, 100, ...] is exp(t)*(1 + 15*t + 69*t^2/2! + 147*t^3/3! + 162*t^4/4! + 90*t^5/5! + 20*t^6/6!). The corresponding o.g.f. from which the e.g.f. has been computed is (1 + x)*(1 + 8*x + x^2)/(1 - x)^7 = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7.
		

Crossrefs

T(n, n) = A005258(n). The squares of the first diagonals are in A000012, A000290(n+1), A000537(n+1), A001249, A288876 (for d = 0..4).

Programs

  • Maple
    T := (n,k) -> binomial(2*n, k)*hypergeom([-k, -n, -n], [1, -2*n], 1):
    seq(seq(simplify(T(n,k)),k=0..2*n),n=0..7); # Peter Luschny, Feb 10 2018
  • Mathematica
    Table[Sum[Binomial[2 n - i, k - i] Binomial[n, i]^2, {i, 0, k}], {n, 0, 7}, {k, 0, 2 n}] // Flatten (* Michael De Vlieger, Jul 30 2017 *)

Formula

T(n, k) = Sum_{i = 0..k} binomial(2*n - i, k-i)*binomial(n, i)^2.
From Peter Bala, Feb 06 2018: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(n+i,i)^2.
T(n,k) = Sum_{i = 0..k} binomial(n,i)*binomial(n,k-i)*binomial(n+k-i,k-i).
T(n,2*n) = binomial(2*n,n) = A000984(n); T(n+1,2*n+1) = 3*(2*n+1)!/n!^2 = 3*A002457(n).
Recurrence: (2*n-k)*(2*n-k-1)T(n,k) = (5*n^2+2*n*k+1-4*n-k)*T(n-1,k) - (n-1)^2*T(n-2,k).
n-th row polynomial R(n,x) = (1 + x)^n * P(n,2*x + 1) = (1 + x)^n * the n-th row polynomial of A063007, where P(n,x) is the n-th Legendre polynomial.
R(n,x) = Sum_{k >= 0} binomial(n+k,k)^2 * x^k/(1 + x)^(k+1).
(x - 1)^(2*n)/x^n * R(n,1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)^2*x^k, the n-th row polynomial of A008459.
R(n,x) = (1 + x)^n o (1 + x)^n where o denotes the black diamond product of power series as defined in Dukes and White.
R(n,x) = coefficient of u^n*v^n in the expansion of the rational function 1/((1+x)*(1+u)(1+v) - x). (End)
T(n,k) = binomial(2*n, k)*hypergeom([-k, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018
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