A271870 -id:A271870 - cp's OEIS frontend

A005995 Alkane (or paraffin) numbers l(8,n).

1, 3, 12, 28, 66, 126, 236, 396, 651, 1001, 1512, 2184, 3108, 4284, 5832, 7752, 10197, 13167, 16852, 21252, 26598, 32890, 40404, 49140, 59423, 71253, 85008, 100688, 118728, 139128, 162384, 188496, 218025, 250971, 287964, 329004, 374794, 425334, 481404, 543004

Offset: 0

N. J. A. Sloane, Winston C. Yang (yang(AT)math.wisc.edu)

From M. F. Hasler, May 01 2009: (Start)
Also, number of 5-element subsets of {1,...,n+5} whose elements sum to an odd integer, i.e. column 5 of A159916.
A linear recurrent sequence with constant coefficients and characteristic polynomial x^9 - 3*x^8 + 8*x^6 - 6*x^5 - 6*x^4 + 8*x^3 - 3*x + 1. (End)
Equals (1/2)*((1, 6, 21, 56, 126, 252, ...) + (1, 0, 3, 0, 6, 0, 10, ...)), see A000389 and A000217.
Equals row sums of triangle A160770.
F(1,5,n) is the number of bracelets with 1 blue, 5 identical red and n identical black beads. If F(1,5,1) = 3 and F(1,5,2) = 12 taken as a base, F(1,5,n) = n(n+1)(n+2)(n+3)/24 + F(1,3,n) + F(1,5,n-2). [F(1,3,n) is the number of bracelets with 1 blue, 3 identical red and n identical black beads. If F(1,3,1) = 2 and F(1,3,2) = 6 taken as a base F(1,3,n) = n(n+1)/2 + [|n/2|] + 1 + F(1,3,n-2)], where [|x|]: if a is an integer and a<=x

S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Winston C. Yang (paper in preparation).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
N. J. A. Sloane, Classic Sequences
L. Smith, Polynomial invariants of finite groups. A survey of recent developments. Bull. Amer. Math. Soc. (N.S.) 34 (1997), no. 3, 211-250.
Ata A. Uslu and Hamdi G. Ozmenekse, F(1,3,n)
Ata A. Uslu and Hamdi G. Ozmenekse, F(1,5,n)
Index entries for linear recurrences with constant coefficients, signature (3, 0, -8, 6, 6, -8, 0, 3, -1).

Cf. A160770, A053132 (bisection), A271870 (bisection), A018210 (partial sums).

a:= n-> (Matrix([[1, 0$6, -3, -9]]). Matrix(9, (i,j)-> if (i=j-1) then 1 elif j=1 then [3, 0, -8, 6, 6, -8, 0, 3, -1][i] else 0 fi)^n)[1, 1]: seq(a(n), n=0..40); # Alois P. Heinz, Jul 31 2008

a[n_?OddQ] := 1/240*(n+1)*(n+2)*(n+3)*(n+4)*(n+5); a[n_?EvenQ] := 1/240*(n+2)*(n+4)*(n+6)*(n^2+3*n+5); Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Mar 17 2014, after M. F. Hasler *)
LinearRecurrence[{3, 0, -8, 6, 6, -8, 0, 3, -1},{1, 3, 12, 28, 66, 126, 236, 396, 651},40] (* Ray Chandler, Sep 23 2015 *)

a(k)= if(k%2,(k+1)*(k+3)*(k+5),(k+6)*(k^2+3*k+5))*(k+2)*(k+4)/240 \\ M. F. Hasler, May 01 2009

G.f.: (1+3*x^2)/((1-x)^3*(1-x^2)^3).
a(2n-1) = n(n+1)(n+2)(2n+1)(2n+3)/30, a(2n) = (n+1)(n+2)(n+3)(4n^2+6n+5)/ 30. [M. F. Hasler, May 01 2009]
a(n) = (1/240)*(n+1)*(n+2)*(n+3)*(n+4)*(n+5) + (1/16)*(n+2)*(n+4)*(1/2)*(1+(-1)^n). [Yosu Yurramendi, Jun 20 2013]
a(n) = A038163(n)+3*A038163(n-2). - R. J. Mathar, Mar 29 2018

A271662 Convolution of nonzero pentagonal numbers (A000326) with themselves.

Original entry on oeis.org

1, 10, 49, 164, 434, 980, 1974, 3648, 6303, 10318, 16159, 24388, 35672, 50792, 70652, 96288, 128877, 169746, 220381, 282436, 357742, 448316, 556370, 684320, 834795, 1010646, 1214955, 1451044, 1722484, 2033104, 2387000, 2788544, 3242393, 3753498, 4327113, 4968804, 5684458

Offset: 0

Ilya Gutkovskiy, Apr 12 2016

More generally, the ordinary generating function for the convolution of nonzero k-gonal numbers with themselves is (1 + (k - 3)*x)^2/(1 - x)^6.

G. C. Greubel, Table of n, a(n) for n = 0..1000
OEIS Wiki, Figurate numbers
Eric Weisstein's World of Mathematics, Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Cf. A000326.

Cf. similar sequences of the convolution of k-gonal numbers with themselves: A000389 (k=3, without zeros), A033455 (k=4), this sequence (k=5), A271870 (k=6).

/* From definition: */ P:=func; /*, where P(n,k) is the n-th k-gonal number, */ [&+[P(n+1-i,5)*P(i,5): i in [1..n]]: n in [1..40]]; // Bruno Berselli, Apr 13 2016

[(n+1)*(n+2)*(n+3)*(9*n^2+21*n+20)/120: n in [0..40]]; // Bruno Berselli, Apr 13 2016

LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 10, 49, 164, 434, 980}, 40]
Table[(n + 1) (n + 2) (n + 3) (9 n^2 + 21 n + 20)/120, {n, 0, 40}]
With[{nmax = 50}, CoefficientList[Series[(120 + 1080*x + 1800*x^2 + 920*x^3 + 165*x^4 + 9*x^5)*Exp[x]/120, {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Jun 07 2017 *)

vector(40, n, n--; (n+1)*(n+2)*(n+3)*(9*n^2+21*n+20)/120) \\ Altug Alkan, Apr 12 2016

O.g.f.: (1 + 2*x)^2/(1 - x)^6.
E.g.f.: (120 + 1080*x + 1800*x^2 + 920*x^3 + 165*x^4 + 9*x^5)*exp(x)/120.
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) = (n + 1)*(n + 2)*(n + 3)*(9*n^2 + 21*n + 20)/120.
Sum_{n>=0} 1/a(n) = 1.13108002...

Edited by Bruno Berselli, Apr 13 2016

A213840 a(n) = n(1 + n)(3 - 4n + 4n^2)/6.

Original entry on oeis.org

1, 11, 54, 170, 415, 861, 1596, 2724, 4365, 6655, 9746, 13806, 19019, 25585, 33720, 43656, 55641, 69939, 86830, 106610, 129591, 156101, 186484, 221100, 260325, 304551, 354186, 409654, 471395, 539865, 615536, 698896, 790449, 890715, 1000230, 1119546, 1249231

Offset: 1

Clark Kimberling, Jul 05 2012

Antidiagonal sums of the convolution array A213838.
The sequence is the binomial transform of (1, 10, 33, 40, 16, 0, 0, 0, ...). - Gary W. Adamson, Jul 31 2015
From Mircea Dan Rus, Jul 11 2020: (Start)
a(n) is also the number of rectangles in a square biscuit of order n, which is obtained by stacking 2n-1 rows with their centers vertically aligned which consist successively of 1, 3, ..., 2n-3, 2n-1, 2n-3, ..., 3, 1 consecutive unit lattice squares. The order 2 and 3 square biscuits are shown below which contain 11 and 54 rectangles respectively.
                        
                   |__|
     |__|     |__||__|
    ||__||   ||__||__||
       ||         ||__||
                       ||
(End)

Clark Kimberling, Table of n, a(n) for n = 1..200
Teofil Bogdan and Mircea Rus, Numărând dreptunghiuri pe foaia de matematică (in Romanian). Gazeta Matematică, seria B, 2020 (6-7-8), pp. 281-288.
Teofil Bogdan and Mircea Dan Rus, Counting the lattice rectangles inside Aztec diamonds and square biscuits, arXiv:2007.13472 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000332, A002417, A213838.

First differences of A271870. - J. M. Bergot, Aug 29 2016

[n*(1+n)*(3-4*n+4*n^2)/6: n in [1..60]]; // Vincenzo Librandi, Aug 01 2015

A213840:=n->n*(1 + n)*(3 - 4*n + 4*n^2)/6: seq(A213840(n), n=1..50); # Wesley Ivan Hurt, Sep 16 2017

Table[n (1 + n) (3 - 4 n + 4 n^2)/6, {n, 50}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 11, 54, 170, 415}, 40] (* Vincenzo Librandi, Aug 01 2015 *)

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: x*(1 + 3*x)^2/(1 - x)^5.
From Mircea Dan Rus, Aug 26 2020: (Start)
a(n) = A000332(n+3) + 6*A000332(n+2) + 9*A000332(n+1).
a(n) = A002417(n) + 3*A002417(n-1). (End)

Edited (with simpler definition) by N. J. A. Sloane, Sep 19 2017

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A005995 Alkane (or paraffin) numbers l(8,n).

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A271662 Convolution of nonzero pentagonal numbers (A000326) with themselves.

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A213840 a(n) = n*(1 + n)*(3 - 4*n + 4*n^2)/6.

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A213840 a(n) = n(1 + n)(3 - 4n + 4n^2)/6.