A272378 -id:A272378 - cp's OEIS frontend

A015237 a(n) = (2n - 1)n^2.

0, 1, 12, 45, 112, 225, 396, 637, 960, 1377, 1900, 2541, 3312, 4225, 5292, 6525, 7936, 9537, 11340, 13357, 15600, 18081, 20812, 23805, 27072, 30625, 34476, 38637, 43120, 47937, 53100, 58621, 64512

Offset: 0

N. J. A. Sloane

Structured hexagonal prism numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of divisors of 60^(n-1) for n>0. - J. Lowell, Aug 30 2008
The sum of the 2*n+1 numbers between n*(n+1) and (n+1)*(n+2) gives a(n+1). - J. M. Bergot, Apr 17 2013
Partial sums of A080859. - J. M. Bergot, Jul 03 2013
a(n) = number of 2 X 2 matrices having all elements in {0..n} with determinant = permanent. - Indranil Ghosh, Dec 26 2016
Number of additions and multiplications needed to multiply two n X n matrices naively. - Charles R Greathouse IV, Jan 19 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
M. Janjic and B. Petkovic, A counting function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A100177 (structured prisms); A100145 (more on structured numbers).
Cf. A000578, A045991, A000384, A080859 (first diffs), A103220 (partial sums).
Cf. similar sequences, with the formula (k*n-k+2)*n^2/2, listed in A262000.
Cf. A036970, A272378.

List([0..40],n->(2*n-1)*n^2); # Muniru A Asiru, Feb 05 2019

[(2*n-1)*n^2: n in [0..40]]; // Vincenzo Librandi, Jul 19 2011

[(2*n-1)*n^2$n=0..40]; # Muniru A Asiru, Feb 05 2019

RecurrenceTable[{a[0]==0, a[1]==1, a[2]==12, a[n]== 3*a[n-1] - 3*a[n-2] + a[n-3] + 12}, a, {n, 30}] (* G. C. Greubel, Jul 31 2015 *)
Table[(2 n - 1) n^2, {n, 0, 40}] (* Bruno Berselli, Sep 08 2015 *)

a(n)=(2*n-1)*n^2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A000578(n) + A045991(n). - Zerinvary Lajos, Jun 11 2008
a(n) = A199771(2*n-1) for n > 0. - Reinhard Zumkeller, Nov 23 2011
G.f.: x*(1+8*x+3*x^2)/(1-x)^4. - Colin Barker, Jun 08 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 12, a(0)=1, a(1)=1, a(2)=12. - G. C. Greubel, Jul 31 2015
E.g.f.: x*(2*x^2 + 5*x + 1)*exp(x). - G. C. Greubel, Jul 31 2015
a(n) = Sum_{i=0..n-1} n*(4*i+1) for n>0. - Bruno Berselli, Sep 08 2015
Sum_{n>=1} 1/a(n) = 4*log(2) - Pi^2/6. - Vaclav Kotesovec, Oct 04 2016
a(n) = Sum_{i=n^2-n+1..n^2+n-1} i. - Wesley Ivan Hurt, Dec 27 2016
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have (2*x - 1)*x^2 = Sum_{n >= 0} ((n+1)^5 + n^5)*a(n,x) and (2*x - 1)*x = Sum_{n >= 0} ((n+1)^4 - n^4)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 2. See the Bala link in A036970. Cf. A272378. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi - Pi^2/12 - 2*log(2). - Amiram Eldar, Jul 12 2020

A036970 Triangle of coefficients of Gandhi polynomials.

Original entry on oeis.org

1, 1, 2, 3, 8, 6, 17, 54, 60, 24, 155, 556, 762, 480, 120, 2073, 8146, 12840, 10248, 4200, 720, 38227, 161424, 282078, 263040, 139440, 40320, 5040, 929569, 4163438, 7886580, 8240952, 5170800, 1965600, 423360, 40320, 28820619, 135634292

Offset: 1

N. J. A. Sloane

Another version of triangle T(n,k), 0 <= k <= n, read by rows; given by [0, 1, 2, 4, 6, 9, 12, 16, 20, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, ...] = 1; 0, 1; 0, 1, 2; 0, 3, 8, 6; 0, 17, 54, 60, 24; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 07 2004

			Triangle begins:
    1;
    1,   2;
    3,   8,   6;
   17,  54,  60,  24;
  155, 556, 762, 480, 120;
  ...

Robert Israel, Table of n, a(n) for n = 1..10011 (rows 1 to 141, flattened)
Peter Bala, The Gandhi polynomials as hypergeometric series
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
W. D. Cairns, Certain properties of binomial coefficients, Bull. Amer. Math. Soc. 26 (1920), 160-164. See p. 163 for a signed version.
Dominique Dumont, Sur une conjecture de Gandhi concernant les nombres de Genocchi, Discrete Mathematics 1 (1972) 321-327.
Dominique Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318. (Annotated scanned copy)
Marc Joye, Pascal Paillier and Berry Schoenmakers, On Second-Order Differential Power Analysis, in Cryptographic Hardware and Embedded Systems-CHES 2005, editors: Josyula R. Rao and Berk Sunar, Lecture Notes in Computer Science 3659 (2005) 293-308, Springer-Verlag.
Arthur Randrianarivony and Jiang Zeng, Une famille de polynomes qui interpole plusieurs suites classiques de nombres, Adv. Appl. Math. 17 (1996), 1-26.
Hans J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, 40(2):175-180, May 2002.

First 2 columns are Genocchi numbers A001469, A005440, row sums are also A001469.

Cf. A083061, A272378, A272379, A272380.

B[1]:= X -> X^2:
for n from 2 to 12 do B[n]:= unapply(expand(X^2*(B[n-1](X+1)-B[n-1](X))),X) od:
seq(seq(coeff(B[i](X),X,1+j),j=1..i),i=1..12); # Robert Israel, Apr 21 2016

B[1][X_] = X^2;
B[n_][X_] := B[n][X] = X^2*(B[n-1][X+1] - B[n-1][X]) // Simplify;
Table[Coefficient[B[i][X], X, j+1], {i, 1, 12}, {j, 1, i}] // Flatten (* Jean-François Alcover, Sep 19 2018, from Maple *)

Let B(X, n) = X^2 (B(X+1, n-1) - B(X, n-1)), B(X, 1) = X^2; then the (i, j)-th entry in the table is the coefficient of X^(1+j) in B(X, i). - Mike Domaratzki (mdomaratzki(AT)alumni.uwaterloo.ca), Nov 17 2001
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row = top row of M^(n-1), M = an infinite square matrix in which the first "1" and right border of 1's of Pascal's triangle are deleted, as follows:
  1,  2,  0,  0,  0,  0, ...
  1,  3,  3,  0,  0,  0, ...
  1,  4,  6,  4,  0,  0, ...
  1,  5, 10, 10,  5,  0, ...
  1,  6, 15, 20, 15,  6, ...
  ...
(End)
Let G(n,x) = (-1)^(n+1)*B(-x,n). Then G(n,x) = (2*x/(x+1))*( 1 + 2^(2*n+1)*(x-1)/(x+2) + 3^(2*n+1)*(x-1)*(x-2)/((x+2)*(x+3)) + ... ). Cf. A083061. - Peter Bala, Feb 04 2019

More terms from David W. Wilson, Jan 12 2001

A272379 a(n) = n(24n^3 - 60n^2 + 54n - 17).

Original entry on oeis.org

0, 1, 86, 759, 3100, 8765, 19986, 39571, 70904, 117945, 185230, 277871, 401556, 562549, 767690, 1024395, 1340656, 1725041, 2186694, 2735335, 3381260, 4135341, 5009026, 6014339, 7163880, 8470825, 9948926, 11612511, 13476484, 15556325, 17868090, 20428411

Offset: 0

Vincenzo Librandi, Apr 29 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1)

Cf. A272378, A272380.

[n*(24*n^3 - 60*n^2 + 54*n - 17): n in [0..50]];

Table[n (24 n^3 - 60 n^2 + 54 n - 17), {n, 0, 50}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,86,759,3100},40] (* Harvey P. Dale, Mar 24 2021 *)

vector(100, n, n--; n*(24*n^3 - 60*n^2 + 54*n - 17)) \\ Altug Alkan, Apr 29 2016

G.f.: x*(1 + 81*x + 339*x^2 + 155*x^3)/(1 - x)^5.
E.g.f.: x*(1 + 42*x + 84*x^2 + 24*x^3)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
See page 7 in Brent's paper:
a(n) = n^2*A272378(n) - n*(n-1)*A272378(n-1),
A272380(n) = n^2*a(n) - n*(n-1)*a(n-1).
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(24*x^3 - 60*x^2 + 54*x - 17) = Sum_{n >= 0} ((n+1)^9 + n^9)*a(n,x) and x*(24*x^3 - 60*x^2 + 54*x - 17)  = Sum_{n >= 0} ((n+1)^8 - n^8)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 9/2. See the Bala link in A036970. Cf. A272378 and A272380. (End)

A272380 a(n) = n(120n^4 - 480n^3 + 762n^2 - 556*n + 155).

Original entry on oeis.org

0, 1, 342, 6315, 40492, 157125, 456546, 1099567, 2321880, 4448457, 7907950, 13247091, 21145092, 32428045, 48083322, 69273975, 97353136, 133878417, 180626310, 239606587, 313076700, 403556181, 513841042, 647018175, 806479752, 995937625, 1219437726, 1481374467

Offset: 0

Vincenzo Librandi, Apr 29 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A272378, A272379.

[n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155): n in [0..50]];

Table[n (120 n^4 - 480 n^3 + 762 n^2 - 556 n + 155), {n, 0, 50}]
LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,342,6315,40492,157125},40] (* Harvey P. Dale, Mar 15 2018 *)

vector(100, n, n--; n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155)) \\ Altug Alkan, Apr 29 2016

O.g.f.: x*(1 + 336*x + 4278*x^2 + 7712*x^3 + 2073*x^4)/(1-x)^6.
E.g.f.: x*(1 + 170*x + 882*x^2 + 720*x^3 + 120*x^4)*exp(x).
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), for n>5.
a(n) = n^2*A272379(n) - n*(n-1)*A272379(n-1), see page 7 in Brent's paper.
From Peter Bala, Jan 30 2019: (Start)
Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(120*x^4 - 480*x^3 + 762*x^2 - 556*x + 155) = Sum_{n >= 0} ((n+1)^11 + n^11)*a(n,x) and x*(120*x^4 - 480*x^3 + 762*x^2 - 556*x + 155)  = Sum_{n >= 0} ((n+1)^10 - n^10)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 11/2. See the Bala link in A036970. Cf. A272378 and A272379. (End)

cp's OEIS Frontend

A015237 a(n) = (2*n - 1)*n^2.

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A036970 Triangle of coefficients of Gandhi polynomials.

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A272379 a(n) = n*(24*n^3 - 60*n^2 + 54*n - 17).

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Magma

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A272380 a(n) = n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155).

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Magma

Mathematica

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Formula

A015237 a(n) = (2n - 1)n^2.

A272379 a(n) = n(24n^3 - 60n^2 + 54n - 17).

A272380 a(n) = n(120n^4 - 480n^3 + 762n^2 - 556*n + 155).