A071231 -id:A071231 - cp's OEIS frontend

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

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

A078876 a(n) = n^4*(n^4-1)/240.

Original entry on oeis.org

0, 0, 1, 27, 272, 1625, 6993, 24010, 69888, 179334, 416625, 893101, 1791504, 3398759, 6148961, 10678500, 17895424, 29065308, 45916065, 70764303, 106666000, 157594437, 228648497, 326294606, 458645760, 635781250, 870110865, 1176787521, 1574172432, 2084357107

Offset: 0

Benoit Cloitre, Jan 11 2003

Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005

For n>=2, the triple (n^6, 120*a(n), (n^8 + n^4)/2) form a Pythagorean triple whose short leg is a square and the other sides are triangular numbers. - Michel Marcus, Mar 15 2021

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, #14).

Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A002415.

Cf. A001014 (n^6), A071231 ((n^8 + n^4)/2).

Table[n^4*(n^4 - 1)/240, {n, 0, 30}] (* Amiram Eldar, May 31 2022 *)

G.f.: -x^2*(x+1)*(x^4+17*x^3+48*x^2+17*x+1) / (x-1)^9. - Colin Barker, Jun 18 2013
From Amiram Eldar, May 31 2022: (Start)
Sum_{n>=2} 1/a(n) = 450 - 8*Pi^4/3 - 60*Pi*coth(Pi).
Sum_{n>=2} (-1)^n/a(n) = 7*Pi^4/3 - 60*Pi*cosech(Pi) - 210. (End)

A342858 a(n) is the least integer h such that there exists a Pythagorean triple (x, y, h) that satisfies f(x)+f(y)+f(h)=n where f(m)=A176774(m) is the smallest polygonality of m; a(n) = 0 if no such h exists.

Original entry on oeis.org

13530, 136, 35, 5, 4510, 10, 100, 45, 51, 1404

Offset: 9

Michel Marcus, Mar 26 2021

a(19) > 10^9 if it exists.

It appears that the triples whose sum is 10 (as in the 2nd example below) have legs n^6 = A001014(n), (n^8 - n^4)/2 = A218131(n+1)/2 and (n^8 + n^4)/2 = A071231(n) for n >= 2; they consist of 2 triangular numbers and 1 square number. - Michel Marcus, Apr 12 2021

			a(9)  = 13530 with A176774([8778, 10296, 13530]) = [3,3,3].
a(10) = 136   with A176774([64, 120, 136])       = [4,3,3].
a(11) = 35    with A176774([21, 28, 35])         = [3,3,5].
a(12) = 5     with A176774([3, 4, 5])            = [3,4,5].
a(13) = 4510  with A176774([2926, 3432, 4510])   = [3,5,5].
a(14) = 10    with A176774([6, 8, 10])           = [3,8,3].
a(15) = 100   with A176774([28, 96, 100])        = [3,8,4].
a(16) = 45    with A176774([27, 36, 45])         = [10,3,3].
a(17) = 51    with A176774([45, 24, 51])         = [3,9,5].
a(18) = 1404  with A176774([540, 1296, 1404])    = [7,4,7].

Michel Marcus, Table of n, a(n) for n = 9..10000, with 0 for a(19)

Cf. A009000, A046083, A046084, A176774.
Cf. A213188 (see 2nd comment).
Cf. A245646, A245647, A245648, A342491.
Cf. A001014, A071231, A218131.

tp(n) = if (n<3, [n], my(v=List()); fordiv(2*n, k, if(k<2, next); if(k==n, break); my(s=(2*n/k-4+2*k)/(k-1)); if(denominator(s)==1, listput(v, s))); v = Vec(v); v[#v]); \\ A176774
vsum(v) = vecsum(apply(tp, v));
lista(limp, lim) = {my(vr = vector(limp)); for(u = 2, sqrtint(lim), for(v = 1, u, if (u*u+v*v > lim, break); if ((gcd(u,v) == 1) && (0 != (u-v)%2), for (i = 1, lim, if (i*(u*u+v*v) > lim, break); my(w = [i*(u*u - v*v), i*2*u*v, i*(u*u+v*v)]); my(h = i*(u*u+v*v)); my(sw = vsum(w)); if (sw <= limp, if (vr[sw] == 0, vr[sw] = h, if (h < vr[sw], vr[sw] = h))););););); vector(#vr - 8, k, vr[k+8]);}
lista(80, 15000) \\ Michel Marcus, Apr 16 2021

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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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A078876 a(n) = n^4*(n^4-1)/240.

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A342858 a(n) is the least integer h such that there exists a Pythagorean triple (x, y, h) that satisfies f(x)+f(y)+f(h)=n where f(m)=A176774(m) is the smallest polygonality of m; a(n) = 0 if no such h exists.

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