A019583 -id:A019583 - cp's OEIS frontend

A295925 Number of bilaterally asymmetric 8-hoops with n symbols.

6, 336, 3795, 23520, 102795, 355656, 1039626, 2674440, 6223140, 13354440, 26807781, 50885016, 92095185, 159981360, 268161060, 435614256, 688255506, 1060829280, 1599170055, 2362871280, 3428409831, 4892775096, 6877654350

Offset: 2

Petros Hadjicostas, Nov 30 2017

This is a corrected version of the entries in sequence A210769, which is copied from the third row of Table 2 (p. 381) in Williamson (1972). Apparently, in that row, there are probable typographical errors for the values of a(4) and a(7). The formula for a(n) can be obtained by letting z_1=z_2=...=z_8=n in equation (24) on p. 377 in Williamson (1972). In any case, to be certain, we provide a sketch of an independent derivation of the formula.
Bilaterally symmetric bracelets are also known as circular palindromes. This kind of necklaces was first studied by Sommerville (1909) in the context of circular compositions.
Consider sequence A081720, which contains the numbers T(n,k) that are the number of bracelets (turn over necklaces) with n beads each of which is colored with one of k colors. The g.f. for column k of that triangle is (1/2)*((k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)). The part of the g.f. that is the number of bilaterally symmetric bracelets with n beads of k colors is (k*x+k*(k+1)*x^2/2)/(1-k*x^2). Thus, for fixed k (= number of colors for sequence A081720), the g.f. of the number of bilaterally asymmetric bracelets with n beads of k colors is the difference between the two g.f.'s, that is, (1/2)*(-(k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)). The coefficient of x^8 in the Taylor expansion w.r.t. x (around x=0) for the latter g.f. gives the number of bilaterally asymmetric 8-hoops obtained using (up to) k symbols. Taking the 8th derivative w.r.t. x of the last g.f., evaluating at x=0, dividing by 8!, and replacing k with n, we get the formulae given below.

			From the A060560(2) = 30 8-hoops (i.e., from the total number of ways of coloring the vertices of an octagon using up to n=2 colors, allowing for rotations and reflections), there are A019583(2+1) = 24 that are circular palindromes (i.e., bilaterally symmetric bracelets). Hence, there are 30-24=6 bilaterally asymmetric 8-hoops using up to 2 colors. They are the following: 01001111, 01000111, 01000011, 00101011, 00110111 and 11001000. (To view these 6 asymmetric bracelets, the 0's and 1's must be placed on the vertices of a regular octagon inscribed in a circle as it is done in Fig. 4 on p. 379 in Williamson (1972), where 0 is replaced by a and 1 by b.)

D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
S. G. Williamson, The combinatorial analysis of patterns and the principle of inclusion-exclusion, Discrete Math. 1 (1972), 357-388.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A019583, A054622, A060560, A081720, A119963, A210699, A210766, A210768, A295925, A292200.

Drop[#, 2] &@ CoefficientList[Series[3 (7 x^4 + 82 x^3 + 237 x^2 + 92 x + 2) (x + 1) x^2/(1 - x)^9, {x, 0, 24}], x] (* Michael De Vlieger, Dec 02 2017 *)

a(n) = (1/16)*(n^3-n^2-2)*(n^2+n+2)*(n+1)*(n-1)*n = (n^8-4*n^5-3*n^4+2*n^2+4*n)/16.
a(n) = A060560(n) - A019583(n+1) = (A054622(n) - A019583(n+1))/2. (Notice that the offsets of the sequences in these formulae are not necessarily the same as the offset of the current sequence.)
G.f.: 3*(7*x^4 + 82*x^3 + 237*x^2 + 92*x + 2)*(x + 1)*x^2/(1-x)^9.
Recurrence: (1-Delta)^9 a(n) = 0, where Delta^m a(n) = a(n-m). Hence, 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).
E.g.f.: exp(x)*x^2*(48 + 848*x + 1658*x^2 + 1046*x^3 + 266*x^4 + 28*x^5 + x^6)/16. - Stefano Spezia, Feb 18 2024

A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 9, 10, 5, 1, 1, 6, 10, 16, 15, 6, 1, 1, 5, 18, 20, 25, 21, 7, 1, 1, 8, 15, 40, 35, 36, 28, 8, 1, 1, 9, 27, 35, 75, 56, 49, 36, 9, 1

Offset: 1

Alford Arnold, Mar 29 2007

The array can be constructed by beginning with A007318 (Pascal's triangle) placing each diagonal on a prime row. The other rows are filled in by mapping the prime factorization of the row number to the known sequences on the prime rows and multiplying term by term.

			Row six begins 1 6 18 40 75 126 ... because rows two and three are
1 2 3 4 5 6 ...
1 3 6 10 15 21 ...
The array begins
1 1 1 1 1 1 1 1 1 A000012
1 2 3 4 5 6 7 8 9 A000027
1 3 6 10 15 21 28 36 45 A000217
1 4 9 16 25 36 49 64 81 A000290
1 4 10 20 35 56 84 120 165 A000292
1 6 18 40 75 126 196 288 405 A002411
1 5 15 35 70 126 210 330 495 A000332
1 8 27 64 125 216 343 512 729 A000578
1 9 36 100 225 441 784 1296 2025 A000537
1 8 30 80 175 336 588 960 1485 A002417
1 6 21 56 126 252 462 792 1287 A000389
1 12 54 160 375 756 1372 2304 3645 A019582
1 7 28 84 210 462 924 1716 3003 A000579
1 10 45 140 350 756 1470 2640 4455 A027800
1 12 60 200 525 1176 2352 4320 7425 A004302
1 16 81 256 625 1296 2401 4096 6561 A000583
1 8 36 120 330 792 1716 3432 6435 A000580
1 18 108 400 1125 2646 5488 10368 18225 A019584
1 9 45 165 495 1287 3003 6435 12870 A000581
1 16 90 320 875 2016 4116 7680 13365 A119771
1 15 90 350 1050 2646 5880 11880 22275 A001297
1 12 63 224 630 1512 3234 6336 11583 A027810
1 10 55 220 715 2002 5005 11440 24310 A000582
1 24 162 640 1875 4536 9604 18432 32805 A019583
1 16 100 400 1225 3136 7056 14400 27225 A001249
1 14 84 336 1050 2772 6468 13728 27027 A027818
1 27 216 1000 3375 9261 21952 46656 91125 A059827
1 20 135 560 1750 4536 10290 21120 40095 A085284

Cf. A000040 A007318.

Cf. A064553 (second diagonal), A080688 (second diagonal resorted).

A128629 := proc(n,m) if n = 1 then 1; elif isprime(n) then p := numtheory[pi](n) ; binomial(p+m-1,p) ; else a := 1 ; for p in ifactors(n)[2] do a := a* procname(op(1,p),m)^ op(2,p) ; od: fi; end: # R. J. Mathar, Sep 09 2009

A-number added to each row of the examples by R. J. Mathar, Sep 09 2009

A101362 a(n) = (n+1)*n^4.

Original entry on oeis.org

0, 2, 48, 324, 1280, 3750, 9072, 19208, 36864, 65610, 110000, 175692, 269568, 399854, 576240, 810000, 1114112, 1503378, 1994544, 2606420, 3360000, 4278582, 5387888, 6716184, 8294400, 10156250, 12338352, 14880348, 17825024, 21218430, 25110000, 29552672

Offset: 0

Jonathan Vos Post, Dec 25 2004

For n>=4, a(n-1) is equal to the number of functions f:{1,2,3,4,5}->{1,2,...,n} such that for fixed, different x_1, x_2, x_3, x_4 in {1,2,3,4,5} and fixed y_1, y_2, y_3, y_ 4 in {1,2,...n} we have f(x_i)<>y_i, (i=1,2,3,4). - Milan Janjic, May 13 2007

Pierce expansion of the constant 1 - Sum_{k >= 1} (-1)^(k+1)*k^4/k!^5 = 0.48961 54584 28443 62043 ... = 1/2 - 1/(2*48) + 1/(2*48*324) - .... - Peter Bala, Feb 01 2015

			a(5) = (5+1)*5^4 = 3750 = 2 * 3 * 5^4, the sum of the divisors of which is 30008.
a(7) = 8*7^4 = 19208 = 2^3 * 7^4 = 98^2 + 98^2.
a(8) = 9*8^4 = 36864 = 2^12*3^2 = 192^2.
a(9) = 10*9^4 = 65610 = 2*3^8*5 = 243^2 + 81^2.
a(10) = 11*10^4 = 110000 = 2^4*5^4*11 = 300^2 + 100^2 + 100^2.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Katherine Kanim, Proof without Words: The Sum of Cubes: An Extension of Archimedes' Sum of Squares, Mathematics Magazine, Vol. 77, No. 4 (2004), pp. 298-299.
Eric Weisstein's World of Mathematics, Pierce Expansion.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A019583.

[n^4+n^5: n in [0..40]]; // Vincenzo Librandi, Aug 15 2016

Maple
```
a:= n-> (n+1)*n^4: seq(a(n), n=0..35);
```

Table[(n + 1)*n^4, {n, 0, 30}]
LinearRecurrence[{6,-15,20,-15,6,-1},{0,2,48,324,1280,3750},40] (* Harvey P. Dale, Jun 10 2019 *)

a(n) + 6*Sum_{i=1..n} i^3 + 4*Sum_{i=1..n} i^2 + Sum_{i=1..n} i = 5*Sum_{i=1..n} i^4.
G.f.: 2*x*(8*x^3+33*x^2+18*x+1) / (x-1)^6. - Colin Barker, May 06 2013
Sum_{n>=1} 1/a(n) = 0.5252003... = Pi^2/6+Pi^4/90-Zeta(3)-1. - R. J. Mathar, Oct 18 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - 2*log(2) + Pi^2/12 - 3*zeta(3)/4 + 7*Pi^4/720. - Amiram Eldar, Nov 05 2020

Corrected and extended by Ray Chandler, Dec 26 2004

cp's OEIS Frontend

A295925 Number of bilaterally asymmetric 8-hoops with n symbols.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Extensions

A101362 a(n) = (n+1)*n^4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

Extensions