A145882 -id:A145882 - cp's OEIS frontend

A037270 a(n) = n^2*(n^2 + 1)/2.

0, 1, 10, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10440, 14365, 19306, 25425, 32896, 41905, 52650, 65341, 80200, 97461, 117370, 140185, 166176, 195625, 228826, 266085, 307720, 354061, 405450, 462241, 524800, 593505, 668746, 750925, 840456, 937765

Offset: 0

Aaron Gulliver (gulliver(AT)elec.canterbury.ac.nz)

Sum of first n^2 positive integers.
Start from xanthene and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002; Amarnath Murthy, Aug 01 2002
Sum of the next n multiples of n. - Amarnath Murthy, Aug 01 2002
The sum of the terms in an n X n spiral. These are also triangular numbers. - William A. Tedeschi, Feb 27 2008
Hypotenuse of Pythagorean triangles with smallest side a cube: A000578(n)^2 + A083374(n)^2 = a(n)^2. - Martin Renner, Nov 12 2011
For n>1, triangular numbers that can be represented as a sum of a square and a triangular number. For example, a(2)=10=4+6=9+1. - Ivan N. Ianakiev, Apr 24 2012
A037270 can be constructed in the following manner: Take A000217 and for every n not in A000290 delete the corresponding A000217(n). - Ivan N. Ianakiev, Apr 26 2012
Starting at a(1)=1 simply take 1*1=1, a(2)= 2*(2+3)=10, a(3)= 3*(4+5+6)=45, a(4)=4*(7+8+9+10) and so on. - J. M. Bergot, May 01 2015
Observation: The digital roots of the terms repeat in the sequence 1, 1, 9; e.g., the digital roots of 1, 10, 45, 136, 325, and 666 are 1, 1, 9, 1, 1, and 9. Verified for the first 10000 terms. - Rob Barton, Mar 28 2018
The above observation is easily explained and proved given that the digital root of a positive number equals the number modulo 9, and a(n + 9k) == a(n) (mod 9). - M. F. Hasler, Apr 05 2018
Number of unoriented rows of length 4 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=10, there are 4 achiral (AAAA, ABBA, BAAB, BBBB) and 6 chiral pairs (AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, BABB-BBAB). - Robert A. Russell, Nov 14 2018
For n > 0, a(2n+1) is the number of non-isomorphic 6C_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k>=2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
Number of achiral colorings of the edges of a tetrahedron with n available colors. - Robert A. Russell, Sep 07 2019

C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 5.
C. Barrientos, Graceful labelings of cyclic snakes, Ars Combin., 60(2001), 85-96.
Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 55.
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.
R. A. Wilson, Cosmic Trigger, epilogue of S.-P. Sirag.

T. D. Noe, Table of n, a(n) for n = 0..1000
J. D. Bell, A translation of Leonhard Euler's "De Quadratis Magicis", E795, arXiv:math/0408230 [math.CO], 2004-2005.
N. G de Bruijn, Some classes of integer-valued functions, Nederl. Akad. Wetensch. Proc, Ser. A, 58 (1955), 363-367. See page 363.
Th. Gruner, A. Kerber, R. Laue, and M. Meringer, Mathematics for Combinatorial Chemistry, In: F. Keil, W. Mackens, H. Voß and J. Wether, Scientific Computing in Chemical Engineering II, Springer, 1999, 74-81.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A236770 (see crossrefs).
Row 4 of A277504.
Cf. A000583 (oriented), A083374 (chiral), A000290 (achiral).
Cf. A317617.
Row 3 of A327086 (achiral simplex edge colorings).

a:=List([0..30],n->n^2*(n^2+1)/2); # Muniru A Asiru, Mar 28 2018

[n^2*(n^2 + 1)/2: n in [0..30]] // Stefano Spezia, Jan 15 2019

seq(n^2*(n^2+1)/2,n=0..30); # Muniru A Asiru, Mar 28 2018

Table[ n^2*((n^2 + 1)/2), {n, 0, 30} ]
Table[(1/8) Round[N[Sinh[2 ArcSinh[n]]^2, 100]], {n, 0, 30}] (* Artur Jasinski, Feb 10 2010 *)
LinearRecurrence[{5,-10,10,-5,1},{0,1,10,45,136},30] (* Harvey P. Dale, Aug 03 2014 *)

a(n)=binomial(n^2+1,2) \\ Charles R Greathouse IV, Apr 25 2012

for n in range(0,30): print(n**2*(n**2+1)/2, end=', ') # Stefano Spezia, Jan 10 2019

a(n) = a(n-1) + n^3 + (n-1)^3.
a(n) = A000537(n)+A000537(n-1), i.e., square of sum of first n integers plus square of sum of first n-1 integers. - Henry Bottomley, Oct 15 2001
a(n) = Sum_{k=0..n^2} k. - William A. Tedeschi, Feb 27 2008
a(n) = (1/8)*sinh(2*arcsinh(n)). - Artur Jasinski, Feb 10 2010
G.f.: x*(1+x)*(1+4*x+x^2)/(1-x)^5. - Colin Barker, Mar 22 2012
a(n) = a(n-1) + A005898(n-1). - Ivan N. Ianakiev, May 13 2012
a(n) = 2 * A000217(n-1) * A000217(n) + A000290(n). - Ivan N. Ianakiev, May 26 2012
a(n) = A000217(n^2). - J. M. Bergot, Jun 07 2012
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5) n>4, a(0)=0, a(1)=1, a(2)=10, a(3)=45, a(4)=136. - Yosu Yurramendi, Sep 02 2013
For n>0, a(n) = A000217(n)^2 + A000217(n-1)^2. - Richard R. Forberg, Dec 25 2013
a(n) = T(T(n)) + T(T(n-1)) + T(T(n)-1) + T(T(n-1)-1), where T(n) = A000217(n). - Charlie Marion, Sep 10 2016
a(n) = t(n-3)*t(n)+t(n-1)*t(n+2), with t(n)=A000217(n). - J. M. Bergot, Apr 07 2018
From Robert A. Russell, Nov 14 2018: (Start)
a(n) = (A000583(n) + A000290(n)) / 2 = (n^4 + n^2) / 2.
a(n) = A000583(n) - A083374(n) = A083374(n) + A000290(n).
G.f.: (Sum_{j=1..4} S2(4,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..2} S2(2,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: Sum_{k=1..4} A145882(4,k) * x^k / (1-x)^5.
E.g.f.: (Sum_{k=1..4} S2(4,k)*x^k + Sum_{k=1..2} S2(2,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>4, a(n) = Sum_{j=1..5} -binomial(j-6,j) * a(n-j). (End)
a(n) = n*A006003(n). - Kritsada Moomuang, Dec 16 2018
For n > 0, a(n) = Sum_{k=1..n} A317617(n,k). - Stefano Spezia, Jan 10 2019
Sum_{n>=1} 1/a(n) = 1 + Pi^2/3 - Pi*coth(Pi) = 1.13652003875929052467672874379... - Vaclav Kotesovec, Jan 21 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi*csch(Pi) + Pi^2/6 - 1. - Amiram Eldar, Nov 02 2021

A277504 Array read by descending antidiagonals: T(n,k) is the number of unoriented strings with n beads of k or fewer colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 6, 1, 0, 1, 5, 10, 18, 10, 1, 0, 1, 6, 15, 40, 45, 20, 1, 0, 1, 7, 21, 75, 136, 135, 36, 1, 0, 1, 8, 28, 126, 325, 544, 378, 72, 1, 0, 1, 9, 36, 196, 666, 1625, 2080, 1134, 136, 1, 0, 1, 10, 45, 288, 1225, 3996, 7875, 8320, 3321, 272, 1, 0

Offset: 0

Jean-François Alcover, Oct 18 2016

From Petros Hadjicostas, Jul 07 2018: (Start)
Column k of this array is the "BIK" (reversible, indistinct, unlabeled) transform of k,0,0,0,....
Consider the input sequence (c_k(n): n >= 1) with g.f. C_k(x) = Sum_{n>=1} c_k(n)*x^n. Let a_k(n) = BIK(c_k(n): n >= 1) be the output sequence under Bower's BIK transform. It can proved that the g.f. of BIK(c_k(n): n >= 1) is A_k(x) = (1/2)*(C_k(x)/(1-C_k(x)) + (C_k(x^2) + C_k(x))/(1-C_k(x^2))). (See the comments for sequence A001224.)
For column k of this two-dimensional array, the input sequence is defined by c_k(1) = k and c_k(n) = 0 for n >= 1. Thus, C_k(x) = k*x, and hence the g.f. of column k is (1/2)*(C_k(x)/(1-C_k(x)) + (C_k(x^2) + C_k(x))/(1-C_k(x^2))) = (1/2)*(k*x/(1-k*x) + (k*x^2 + k*x)/(1-k*x^2)) = (2 + (1-k)*x - 2*k*x^2)*k*x/(2*(1-k*x^2)*(1-k*x)).
Using the first form the g.f. above and the expansion 1/(1-y) = 1 + y + y^2 + ..., we can easily prove J.-F. Alcover's formula T(n,k) = (k^n + k^((n + mod(n,2))/2))/2.
(End)

			Array begins with T(0,0):
1 1   1     1      1       1        1         1         1          1 ...
0 1   2     3      4       5        6         7         8          9 ...
0 1   3     6     10      15       21        28        36         45 ...
0 1   6    18     40      75      126       196       288        405 ...
0 1  10    45    136     325      666      1225      2080       3321 ...
0 1  20   135    544    1625     3996      8575     16640      29889 ...
0 1  36   378   2080    7875    23436     58996    131328     266085 ...
0 1  72  1134   8320   39375   140616    412972   1050624    2394765 ...
0 1 136  3321  32896  195625   840456   2883601   8390656   21526641 ...
0 1 272  9963 131584  978125  5042736  20185207  67125248  193739769 ...
0 1 528 29646 524800 4884375 30236976 141246028 536887296 1743421725 ...
...

See A005418.

Robert A. Russell, Antidiagonals n=0..52, flattened (antidiagonals 1..50 from Andrew Howroyd)
C. G. Bower, Transforms (2)

Columns 0-6 are A000007, A000012, A005418(n+1), A032120, A032121, A032122, A056308.
Rows 0-20 are A000012, A001477, A000217 (triangular numbers), A002411 (pentagonal pyramidal numbers), A037270, A168178, A071232, A168194, A071231, A168372, A071236, A168627, A071235, A168663, A168664, A170779, A170780, A170790, A170791, A170801, A170802.
Main diagonal is A275549.
Transpose is A284979.
Cf. A284871, A284949.
Cf. A003992 (oriented), A293500 (chiral), A321391 (achiral).

[[n le 0 select 1 else ((n-k)^k + (n-k)^Ceiling(k/2))/2: k in [0..n]]: n in [0..15]]; // G. C. Greubel, Nov 15 2018

Table[If[n>0, ((n-k)^k + (n-k)^Ceiling[k/2])/2, 1], {n, 0, 15}, {k, 0, n}] // Flatten (* updated Jul 10 2018 *) (* Adapted to T(0,k)=1 by Robert A. Russell, Nov 13 2018 *)

for(n=0,15, for(k=0,n, print1(if(n==0,1, ((n-k)^k + (n-k)^ceil(k/2))/2), ", "))) \\ G. C. Greubel, Nov 15 2018

T(n,k) = {(k^n + k^ceil(n/2)) / 2} \\ Andrew Howroyd, Sep 13 2019

T(n,k) = [n==0] + [n>0] * (k^n + k^ceiling(n/2)) / 2. [Adapted to T(0,k)=1 by Robert A. Russell, Nov 13 2018]
G.f. for column k: (1 - binomial(k+1,2)*x^2) / ((1-k*x)*(1-k*x^2)). - Petros Hadjicostas, Jul 07 2018 [Adapted to T(0,k)=1 by Robert A. Russell, Nov 13 2018]
From Robert A. Russell, Nov 13 2018: (Start)
T(n,k) = (A003992(k,n) + A321391(n,k)) / 2.
T(n,k) = A003992(k,n) - A293500(n,k) = A293500(n,k) + A321391(n,k).
G.f. for row n: (Sum_{j=0..n} S2(n,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=0..ceiling(n/2)} S2(ceiling(n/2),j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f. for row n>0: x*Sum_{k=0..n-1} A145882(n,k) * x^k / (1-x)^(n+1).
E.g.f. for row n: (Sum_{k=0..n} S2(n,k)*x^k + Sum_{k=0..ceiling(n/2)} S2(ceiling(n/2),k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
T(0,k) = 1; T(1,k) = k; T(2,k) = binomial(k+1,2); for n>2, T(n,k) = k*(T(n-3,k)+T(n-2,k)-k*T(n-1,k)).
For k>n, T(n,k) = Sum_{j=1..n+1} -binomial(j-n-2,j) * T(n,k-j). (End)

Array transposed for greater consistency by Andrew Howroyd, Apr 04 2017

Origin changed to T(0,0) by Robert A. Russell, Nov 13 2018

A170801 a(n) = n^10*(n^9 + 1)/2.

Original entry on oeis.org

0, 1, 262656, 581160258, 137439477760, 9536748046875, 304679900238336, 5699447733924196, 72057594574798848, 675425860579888245, 5000000005000000000, 30579545237175985446, 159739999716270145536

Offset: 0

N. J. A. Sloane, Dec 11 2009

Number of unoriented rows of length 19 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=262656, there are 2^19=524288 oriented arrangements of two colors. Of these, 2^10=1024 are achiral. That leaves (524288-1024)/2=261632 chiral pairs. Adding achiral and chiral, we get 262656. - Robert A. Russell, Nov 13 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (20, -190, 1140, -4845, 15504, -38760, 77520, -125970, 167960, -184756, 167960, -125970, 77520, -38760, 15504, -4845, 1140, -190, 20, -1).

Row 19 of A277504.
Cf. A010807 (oriented), A008454 (achiral).
Sequences of the form n^10*(n^m + 1)/2: A170793 (m=1), A170794 (m=2), A170795 (m=3), A170796 (m=4), A170797 (m=5), A170798 (m=6), A170799 (m=7), A170800 (m=8), this sequence (m=9), A170802 (m=10).

List([0..30], n -> n^10*(n^9+1)/2); # G. C. Greubel, Nov 15 2018

[n^10*(n^9+1)/2: n in [0..20]]; // Vincenzo Librandi, Aug 27 2011

seq(n^10*(n^9 +1)/2, n=0..20); # G. C. Greubel, Oct 11 2019

Table[(n^19 + n^10)/2, {n,0,30}] (* Robert A. Russell, Nov 13 2018 *)

vector(30, n, n--; n^10*(n^9+1)/2) \\ G. C. Greubel, Nov 15 2018

[n^10*(n^9+1)/2 for n in range(30)] # G. C. Greubel, Nov 15 2018

From Robert A. Russell, Nov 13 2018: (Start)
a(n) = (A010807(n) + A008454(n)) / 2 = (n^19 + n^10) / 2.
G.f.: (Sum_{j=1..19} S2(19,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..10} S2(10,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x*Sum_{k=0..18} A145882(19,k) * x^k / (1-x)^20.
E.g.f.: (Sum_{k=1..19} S2(19,k)*x^k + Sum_{k=1..10} S2(10,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>19, a(n) = Sum_{j=1..20} -binomial(j-21,j) * a(n-j). (End)

A170802 a(n) = n^10*(n^10 + 1)/2.

Original entry on oeis.org

0, 1, 524800, 1743421725, 549756338176, 47683720703125, 1828079250264576, 39896133290043625, 576460752840294400, 6078832731271856601, 50000000005000000000, 336374997479248716901, 1916879996254696243200

Offset: 0

N. J. A. Sloane, Dec 11 2009

By definition, all terms are triangular numbers. - Harvey P. Dale, Aug 12 2012

Number of unoriented rows of length 20 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=524800, there are 2^20=1048576 oriented arrangements of two colors. Of these, 2^10=1024 are achiral. That leaves (1048576-1024)/2=523776 chiral pairs. Adding achiral and chiral, we get 524800. - Robert A. Russell, Nov 13 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (21, -210, 1330, -5985, 20349, -54264, 116280, -203490, 293930, -352716, 352716, -293930, 203490, -116280, 54264, -20349, 5985, -1330, 210, -21, 1).

Row 20 of A277504.
Cf. A010808 (oriented), A008454 (achiral).
Sequences of the form n^10*(n^m + 1)/2: A170793 (m=1), A170794 (m=2), A170795 (m=3), A170896 (m=4), A170797 (m=5), A170798 (m=6), A170799 (m=7), A170800 (m=8), A170801 (m=9), this sequence (m=10).

List([0..30], n -> n^10*(n^10+1)/2); # G. C. Greubel, Nov 15 2018

[n^10*(n^10+1)/2: n in [0..20]]; // Vincenzo Librandi, Aug 27 2011

seq(n^10*(n^10 +1)/2, n=0..20); # G. C. Greubel, Oct 11 2019

n10[n_]:=Module[{c=n^10},(c(c+1))/2];Array[n10,15,0] (* Harvey P. Dale, Jul 17 2012 *)

vector(30, n, n--; n^10*(n^10+1)/2) \\ G. C. Greubel, Nov 15 2018

for n in range(0,20): print(int(n**10*(n**10 + 1)/2), end=', ') # Stefano Spezia, Nov 15 2018

[n^10*(n^10+1)/2 for n in range(30)] # G. C. Greubel, Nov 15 2018

From Robert A. Russell, Nov 13 2018: (Start)
a(n) = (A010808(n) + A008454(n)) / 2 = (n^20 + n^10) / 2.
G.f.: (Sum_{j=1..20} S2(20,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..10} S2(10,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x*Sum_{k=0..19} A145882(20,k) * x^k / (1-x)^21.
E.g.f.: (Sum_{k=1..20} S2(20,k)*x^k + Sum_{k=1..10} S2(10,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>20, a(n) = Sum_{j=1..21} -binomial(j-22,j) * a(n-j). (End)

A168178 a(n) = n^3*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 20, 135, 544, 1625, 3996, 8575, 16640, 29889, 50500, 81191, 125280, 186745, 270284, 381375, 526336, 712385, 947700, 1241479, 1604000, 2046681, 2582140, 3224255, 3988224, 4890625, 5949476, 7184295, 8616160, 10267769, 12163500

Offset: 0

N. J. A. Sloane, Dec 11 2009

Number of unoriented rows of length 5 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=20, there are 8 achiral (AAAAA, AABAA, ABABA, ABBBA, BAAAB, BABAB, BBABB, BBBBB) and 12 chiral pairs (AAAAB-BAAAA, AAABA-ABAAA, AAABB-BBAAA, AABAB-BABAA, AABBA-ABBAA, AABBB-BBBAA, ABAAB-BAABA, ABABB-BBABA, ABBAB-BABBA, ABBBB-BBBBA, BAABB-BBAAB, BABBB-BBBAB). - Robert A. Russell, Nov 14 2018

For n > 0, a(2n+1) is the number of non-isomorphic kC_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k>=2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 16 2019

C. Barrientos, Graceful labelings of cyclic snakes, Ars Combin., 60(2001), 85-96.

Vincenzo Librandi, Table of n, a(n) for n = 0..595
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A155977.
Row 5 of A277504.
Cf. A000584 (oriented), A000578 (achiral).

List([0..40], n -> n^3*(1 +n^2)/2); # G. C. Greubel, Nov 15 2018

[n^3*(n^2+1)/2: n in [0..50]]; // Vincenzo Librandi, Apr 25 2011

Table[(n^5+n^3)/2,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1},{0, 1, 20, 135, 544, 1625},40] (* Robert A. Russell, Nov 14 2018 *)

vector(40, n, n--; n^3*(1+n^2)/2) \\ G. C. Greubel, Nov 15 2018

[n^3*(1 + n^2)/2 for n in range(40)] # G. C. Greubel, Nov 15 2018

From Robert A. Russell, Nov 14 2018: (Start)
a(n) = (A000584(n) + A000578(n)) / 2 = (n^5 + n^3) / 2.
G.f.: (Sum_{j=1..5} S2(5,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..3} S2(3,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x*Sum_{k=0..4} A145882(5,k) * x^k / (1-x)^6.
E.g.f.: (Sum_{k=1..5} S2(5,k)*x^k + Sum_{k=1..3} S2(3,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>5, a(n) = Sum_{j=1..6} -binomial(j-7,j) * a(n-j). (End)
From G. C. Greubel, Nov 15 2018: (Start)
G.f.: x*(1 + 14*x + 30*x^2 + 14*x^3 + x^4)/(1-x)^6.
E.g.f.: x*(2 + 18*x + 26*x^2 + 10*x^3 + x^4)*exp(x)/2. (End)

A071231 a(n) = (n^8 + n^4)/2.

Original entry on oeis.org

0, 1, 136, 3321, 32896, 195625, 840456, 2883601, 8390656, 21526641, 50005000, 107186761, 215001216, 407879641, 737913736, 1281470625, 2147516416, 3487920481, 5510032776, 8491846681, 12800080000, 18911526921, 27438053896, 39155632561, 55037822976

Offset: 0

N. J. A. Sloane, Jun 11 2002

Number of unoriented rows of length 8 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=136, there are 2^8=256 oriented arrangements of two colors. Of these, 2^4=16 are achiral. That leaves (256-16)/2=120 chiral pairs. Adding achiral and chiral, we get 136. - Robert A. Russell, Nov 13 2018

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Row 8 of A277504.
Cf. A001016 (oriented), A000583 (achiral).
Cf. A000583, A001016.
Cf. A008277, A145882.
Subsequence of A000217 (triangular numbers).

List([0..50], n -> (n^4 + n^8)/2); # G. C. Greubel, Nov 15 2018

[(n^8 + n^4)/2: n in [0..50]]; // Vincenzo Librandi, Jun 14 2011

Table[(n^8+n^4)/2,{n,0,30}] (* or *) LinearRecurrence[{9,-36,84,-126,126, -84,36,-9,1},{0,1,136,3321,32896,195625,840456,2883601,8390656},30] (* Harvey P. Dale, Oct 11 2011 *)

vector(50, n, n--; (n^8 + n^4)/2) \\ G. C. Greubel, Nov 15 2018

[(n^4 + n^8)/2 for n in range(50)] # G. C. Greubel, Nov 15 2018

From Harvey P. Dale, Oct 11 2011: (Start)
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9); a(0)=0, a(1)=1, a(2)=136, a(3)=3321, a(4)=32896, a(5)=195625, a(6)=840456, a(7)=2883601, a(8)=8390656.
G.f.: -((x*(x+1)*(x*(x*(x*(x*(x*(x+126)+2007)+5812)+2007)+126)+1))/ (x-1)^9). (End)
From Robert A. Russell, Nov 13 2018: (Start)
a(n) = (A001016(n) + A000583(n)) / 2 = (n^8 + n^4) / 2.
G.f.: (Sum_{j=1..8} S2(8,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..4} S2(4,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x*Sum_{k=0..7} A145882(8,k) * x^k / (1-x)^9.
E.g.f.: (Sum_{k=1..8} S2(8,k)*x^k + Sum_{k=1..4} S2(4,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>8, a(n) = Sum_{j=1..9} -binomial(j-10,j) * a(n-j). (End)
E.g.f.: x*(2 + 134*x + 972*x^2 + 1702*x^3 + 1050*x^4 + 266*x^5 + 28*x^6 + x^7)*exp(x)/2. - G. C. Greubel, Nov 15 2018

A071232 a(n) = (n^6 + n^3)/2.

Original entry on oeis.org

0, 1, 36, 378, 2080, 7875, 23436, 58996, 131328, 266085, 500500, 886446, 1493856, 2414503, 3766140, 5697000, 8390656, 12071241, 17009028, 23526370, 32004000, 42887691, 56695276, 74024028, 95558400, 122078125, 154466676, 193720086, 240956128, 297423855, 364513500

Offset: 0

N. J. A. Sloane, Jun 11 2002

Number of unoriented rows of length 6 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=36, there are 2^6=64 oriented arrangements of two colors. Of these, 2^3=8 are achiral. That leaves (64-8)/2=28 chiral pairs. Adding achiral and chiral, we get 36. - Robert A. Russell, Nov 14 2018

For n > 0, a(2n+1) is the number of non-isomorphic 8C_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 16 2019

C. Barrientos, Graceful labelings of cyclic snakes, Ars Combin., 60 (2001), 85-96.
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Row 6 of A277504.

Cf. A001014 (oriented), A085744 (chiral), A000578 (achiral).

List([0..50], n -> (n^6 + n^3)/2); # G. C. Greubel, Nov 15 2018

[(n^6 + n^3)/2: n in [0..50]]; // Vincenzo Librandi, Jun 14 2011

Table[(n^6+n^3)/2,{n,0,40}] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,36,378,2080,7875,23436},40] (* Harvey P. Dale, Nov 06 2011 *)

vector(50, n, n--; (n^6 + n^3)/2) \\ G. C. Greubel, Nov 15 2018

[(n^6 + n^3)/2 for n in range(50)] # G. C. Greubel, Nov 15 2018

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); a(0)=0, a(1)=1, a(2)=36, a(3)=378, a(4)=2080, a(5)=7875, a(6)=23436. - Harvey P. Dale, Nov 06 2011
G.f.: x*(28*x^4 + 155*x^3 + 147*x^2 + 29*x + 1)/(1-x)^7. - Colin Barker, Oct 12 2012
From Robert A. Russell, Nov 14 2018: (Start)
a(n) = (A001014(n) + A000578(n)) / 2 = (n^6 + n^3) / 2.
a(n) = A001014(n) - A085744(n) = A085744(n) + A000578(n).
G.f.: (Sum_{j=1..6} S2(6,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..3} S2(3,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x*Sum_{k=0..5} A145882(6,k) * x^k / (1-x)^7.
E.g.f.: (Sum_{k=1..6} S2(6,k)*x^k + Sum_{k=1..3} S2(3,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>6, a(n) = Sum_{j=1..7} -binomial(j-8,j) * a(n-j). (End)
E.g.f.: x*(2 +34*x +91*x^2 +65*x^3 +15*x^4 +x^5)*exp(x)/2. - G. C. Greubel, Nov 15 2018
a(n) = A000217(n^3), sum of the integers up to the n'th cube. - R. J. Mathar, Mar 11 2025

A128612 Triangle T(n,k) read by rows: number of permutations in [n] with exactly k ascents that have an even number of inversions.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 1, 5, 5, 1, 1, 14, 30, 14, 1, 0, 28, 155, 147, 29, 1, 0, 56, 605, 1208, 586, 64, 1, 1, 127, 2133, 7819, 7819, 2133, 127, 1, 1, 262, 7288, 44074, 78190, 44074, 7288, 262, 1, 0, 496, 23947, 227623, 655039, 655315, 227569, 23893, 517, 1, 0, 992, 76305, 1102068, 4868556, 7862124, 4869558, 1101420, 76332, 1044, 1

Offset: 1

Ralf Stephan, May 08 2007

			Triangle starts:
  1;
  0,   1;
  0,   2,    1;
  1,   5,    5,    1;
  1,  14,   30,   14,    1;
  0,  28,  155,  147,   29,    1;
  0,  56,  605, 1208,  586,   64,   1;
  1, 127, 2133, 7819, 7819, 2133, 127, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Jason Fulman, Gene B. Kim, Sangchul Lee, and T. Kyle Petersen, On the joint distribution of descents and signs of permutations, arXiv:1910.04258 [math.CO], 2019.
S. Tanimoto, A new approach to signed Eulerian numbers, arXiv:math/0602263 [math.CO], 2006.

Cf. A008292, A049061, A128613.
Cf. A145882 (similar with rows reversed).
Row sums give A001710.
T(2n,n) gives A382309.

A008292 := proc(n,k) local j; add( (-1)^j*(k-j)^n*binomial(n+1,j),j=0..k) ; end: A049061 := proc(n,k) if k <= 0 or n <=0 or k > n then 0; elif n = 1 then 1 ; elif n mod 2 = 0 then A049061(n-1,k)-A049061(n-1,k-1) ; else k*A049061(n-1,k)+(n-k+1)*A049061(n-1,k-1) ; fi ; end: A128612 := proc(n,k) (A008292(n,n-k)+A049061(n,n-k))/2 ; end: for n from 1 to 11 do for k from 0 to n-1 do printf("%d,",A128612(n,k)) ; od: od: # R. J. Mathar, Nov 01 2007
# second Maple program:
b:= proc(u, o, i) option remember; expand(`if`(u+o=0, 1-i,
       add(b(u+j-1, o-j, irem(i+u+j-1, 2)), j=1..o)*x+
       add(b(u-j, o+j-1, irem(i+u-j, 2)), j=1..u)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n-1))(b(n, 0$2)):
seq(T(n), n=1..14);  # Alois P. Heinz, May 02 2017

b[u_, o_, i_] := b[u, o, i] = Expand[If[u + o == 0, 1 - i, Sum[b[u + j - 1, o - j, Mod[i + u + j - 1, 2]], {j, 1, o}]*x + Sum[b[u - j, o + j - 1, Mod[i + u - j, 2]], {j, 1, u}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n-1}]][b[n, 0,0]];
Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, Jul 25 2017, after Alois P. Heinz *)

T(n,k) = (1/2) * (A008292(n,n-k) + A049061(n,n-k)), n>=1, 0<=kR. J. Mathar, Nov 01 2007

More terms from R. J. Mathar, Nov 01 2007

A145883 Triangle read by rows: T(n,k) is the number of odd permutations of {1,2,...,n} having k descents. (n>=1, k>=1).

Original entry on oeis.org

0, 1, 2, 1, 6, 6, 12, 36, 12, 28, 155, 147, 29, 1, 56, 605, 1208, 586, 64, 1, 120, 2160, 7800, 7800, 2160, 120, 240, 7320, 44160, 78000, 44160, 7320, 240, 496, 23947, 227623, 655039, 655315, 227569, 23893, 517, 1, 992, 76305, 1102068, 4868556

Offset: 1

Emeric Deutsch, Nov 11 2008

Number of entries in row n is ceiling(binomial(n,2)/2) - ceiling(binomial(n-2,2)/2).

Sum of entries in row n is A001710(n) for n>=2.

			T(4,2) = 6 because we have 1432, 3142, 3214, 4312, 4231 and 3421.
Triangle begins with T(1,1):
    0
    1
    2     1
    6     6
   12    36      12
   28   155     147      29       1
   56   605    1208     586      64       1
  120  2160    7800    7800    2160     120
  240  7320   44160   78000   44160    7320     240
  496 23947  227623  655039  655315  227569   23893   517    1
  992 76305 1102068 4868556 7862124 4869558 1101420 76332 1044 1

Alois P. Heinz, Rows n = 1..143, flattened
J. Shareshian and M. L. Wachs, q-Eulerian polynomials: excedance number and major index, Electronic Research Announcements of the Amer. Math. Soc., 13 (2007), 33-45.
R. P. Stanley, Binomial posets, Möbius inversion and permutation enumeration, J. Combinat. Theory, A 20 (1976), 336-356.
S. Tanimoto, A study of Eulerian numbers for permutations in the alternating group, Integers, Electronic J. of Combinatorial Number Theory, 6 (2006), #A31.

Cf. A001710, A145882.

for n to 11 do qbr := proc (m) options operator, arrow; sum(q^i, i = 0 .. m-1) end proc; qfac := proc (m) options operator, arrow; product(qbr(j), j = 1 .. m) end proc; Exp := proc (z) options operator, arrow; sum(q^binomial(m, 2)*z^m/qfac(m), m = 0 .. 19) end proc; g := (1-t)/(Exp(z*(t-1))-t); gser := simplify(series(g, z = 0, 17)); a[n] := simplify(qfac(n)*coeff(gser, z, n)); b[n] := (a[n]-subs(q = -q, a[n]))*1/2; P[n] := sort(subs(q = 1, b[n])) end do; 0; for n to 11 do seq(coeff(P[n], t, j), j = 1 .. ceil((1/2)*binomial(n, 2))-ceil((1/2)*binomial(n-2, 2))) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(u, o, t) option remember; `if`(u+o=0, t, expand(
       add(b(u+j-1, o-j, irem(t+j-1+u, 2)), j=1..o)+
       add(b(u-j, o+j-1, irem(t+u-j, 2))*x, j=1..u)))
    end:
T:= n->`if`(n=1, 0, (p->seq(coeff(p, x, i), i=1..degree(p)))
       (add(b(j-1, n-j, irem(j+1, 2)), j=1..n))):
seq(T(n), n=1..12);  # Alois P. Heinz, Nov 19 2013

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, t, Expand[Sum[b[u+j-1, o-j, Mod[t+j-1+u, 2]], {j, 1, o}] + Sum[b[u-j, o+j-1, Mod[t+u-j, 2]]*x, {j, 1, u}]]]; T[n_] := If[n == 1, 0, Function[{p}, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][Sum[ b[j-1, n-j, Mod[j+1, 2]], {j, 1, n}]]]; Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)
Needs["Combinatorica`"];
Join[{0}, Table[(Eulerian[n, k] - Sum[Binomial[j-1-Floor[n/2], j] Eulerian[Ceiling[n/2], k-j], {j, Max[0, k+1-Ceiling[n/2]], Min[Floor[n/2], k]}])/2, {n, 2, 15}, {k, 1, n}] // Flatten // DeleteCases[0]] (* Robert A. Russell, Nov 16 2018 *)

In the Shareshian and Wachs reference (p. 35) a q-analog of the exponential g.f. of the Eulerian polynomials is given for the joint distribution of (inv, des) (see also the Stanley reference). The first Maple program given below makes use of this function by considering its odd part.

T(n,k) = (euler(n,k) - Sum_{j=max(0, k+1-ceiling(n/2))..min(floor(n/2), k)} binomial(j-1-floor(n/2), j) * euler(ceiling(n/2), k-j)) / 2, where euler(n,k) is the Eulerian number A173018 (not A008292, which has different indexing). - Robert A. Russell, Nov 16 2018

A071236 a(n) = (n^10 + n^5)/2.

Original entry on oeis.org

0, 1, 528, 29646, 524800, 4884375, 30236976, 141246028, 536887296, 1743421725, 5000050000, 12968792826, 30958806528, 68929431571, 144627596400, 288325575000, 549756338176, 1007997660153, 1785234558096, 3065534366950, 5120001600000, 8339942531151

Offset: 0

N. J. A. Sloane, Jun 12 2002

Subset of A000217. - Robert Israel, Nov 20 2014

Number of unoriented rows of length 10 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=528, there are 2^10=1024 oriented arrangements of two colors. Of these, 2^5=32 are achiral. That leaves (1024-32)/2=496 chiral pairs. Adding achiral and chiral, we get 528. - Robert A. Russell, Nov 13 2018

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1).

Cf. A000217.
Row 10 of A277504.
Cf. A008454 (oriented), A000584 (achiral).

List([0..40], n -> n^5*(1 + n^5)/2); # G. C. Greubel, Nov 15 2018

[n^5*(n+1)*(n^4-n^3+n^2-n+1)/2: n in [0..40]]; // Vincenzo Librandi, Jun 14 2011

seq((n^10 + n^5)/2, n=0..100); # Robert Israel, Nov 19 2014

Table[n^5(n+1)(n^4-n^3+n^2-n+1)/2,{n,0,30}] (* or *) LinearRecurrence[{11, -55, 165, -330,462,-462,330,-165,55,-11,1},{0, 1, 528,29646,524800,4884375, 30236976,141246028,536887296, 1743421725, 5000050000}, 30](* Harvey P. Dale, Jul 24 2012 *)

a(n)=binomial(n^5+1,2) \\ Charles R Greathouse IV, Nov 19 2014

[n^5*(1 + n^5)/2 for n in range(40)] # G. C. Greubel, Nov 15 2018

a(0)=0, a(1)=1, a(2)=528, a(3)=29646, a(4)=524800, a(5)=4884375, a(6)=30236976, a(7)=141246028, a(8)=536887296, a(9)=1743421725, a(10)=5000050000, a(n)= 11*a(n-1)- 55*a(n-2)+165*a(n-3)-330*a(n-4)+462*a(n-5)-462*a(n-6)+330*a(n-7)-165*a(n-8)+ 55*a(n-9)- 11*a(n-10)+a(n-11). - Harvey P. Dale, Jul 24 2012
From Robert Israel, Nov 19 2014: (Start)
a(n) = (A008454(n) + A000584(n))/2 = n^5*(n+1)*(n^4 -n^3 +n^2 -n +1)/2.
G.f.: x*(1 +517*x +23893*x^2 +227569*x^3 +655315*x^4 +655039*x^5 +227623*x^6 +23947*x^7 +496*x^8)/(1-x)^11. (End)
From Robert A. Russell, Nov 13 2018: (Start)
G.f.: (Sum_{j=1..10} S2(10,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..5} S2(5,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x*Sum_{k=0..9} A145882(10,k) * x^k / (1-x)^11.
E.g.f.: (Sum_{k=1..10} S2(10,k)*x^k + Sum_{k=1..5} S2(5,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n > 10, a(n) = Sum_{j=1..11} -binomial(j-12,j) * a(n-j). (End)
E.g.f.: x*(2 + 526*x + 9355*x^2 + 34115 x^3 + 42526*x^4 + 22827*x^5 + 5880*x^6 + 750*x^7 + 45*x^8 + x^9)*exp(x)/2. - G. C. Greubel, Nov 15 2018

Definition simplified by Robert Israel, Nov 19 2014

cp's OEIS Frontend

A037270 a(n) = n^2*(n^2 + 1)/2.

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A277504 Array read by descending antidiagonals: T(n,k) is the number of unoriented strings with n beads of k or fewer colors.

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A170801 a(n) = n^10*(n^9 + 1)/2.

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A170802 a(n) = n^10*(n^10 + 1)/2.

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A168178 a(n) = n^3*(n^2 + 1)/2.

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A071231 a(n) = (n^8 + n^4)/2.