A079908 -id:A079908 - cp's OEIS frontend

A079919 Solution to the Dancing School Problem with 14 girls and n+14 boys: f(14,n).

1, 15, 4163, 158364, 3904260, 60560175, 671224467, 5697401802, 38983643908, 223245029176, 1100925116264, 4780871048064, 18612106195456, 65909241461760, 214868401724416, 650515953570304, 1842743223078144, 4916155345428736, 12422627638293760, 29881211844270336, 68721268507385344, 151698799246127104

Offset: 0

Jaap Spies, Jan 28 2003

f(g,h) = per(B), the permanent of the (0,1)-matrix B of size g X g+h with b(i,j)=1 if and only if i <= j <= i+h. See A079908 for more information.

For fixed g, f(g,n) is polynomial in n for n >= g-2. See reference.

Jaap Spies, Dancing School Problems, Nieuw Archief voor Wiskunde 5/7 nr. 4, Dec 2006, pp. 283-285.
Jaap Spies, Dancing School Problems, Permanent solutions of Problem 29.
Jaap Spies, Sage program "dance" for computing the polynomial a(n).

Cf. A079908-A079928.

a(n) = n^14 - 77*n^13 + 3094*n^12 - 83083*n^11 + 1637636*n^10 - 24785761*n^9 + 294696402*n^8 - 2779448529*n^7 + 20797459683*n^6 - 122389753486*n^5 + 555826054784*n^4 - 1883902028008*n^3 + 4494445040176*n^2 - 6742111050752*n + 4789534153984 for n >= 12. - Georg Fischer, Apr 27 2021 (polynomial computed by the program in links)

Corrected by Jaap Spies, Feb 01 2004

a(13)-a(21) from Georg Fischer, Apr 27 2021

A079920 Solution to the Dancing School Problem with 15 girls and n+15 boys: f(15,n).

Original entry on oeis.org

1, 16, 6746, 313464, 9479292, 174763208, 2262089361, 22088730348, 171764779170, 1106667645872, 6087616677864, 29267369636800, 125299076209408, 485013257865472, 1718947213795328, 5636819806209792, 17235204961273600, 49467590616190208, 134058587073795072, 344809293460572928, 845577589114049792, 1985060631106310400

Offset: 0

Jaap Spies, Jan 28 2003

f(g,h) = per(B), the permanent of the (0,1)-matrix B of size g X g+h with b(i,j)=1 if and only if i <= j <= i+h. See A079908 for more information.

For fixed g, f(g,n) is polynomial in n for n >= g-2. See reference.

Jaap Spies, Dancing School Problems, Nieuw Archief voor Wiskunde 5/7 nr. 4, Dec 2006, pp. 283-285.
Jaap Spies, Dancing School Problems, Permanent solutions of Problem 29.
Jaap Spies, Sage program "dance" for computing the polynomial a(n).

Cf. A079908-A079928.

a(n) = n^15 - 90*n^14 + 4200*n^13 - 131040*n^12 + 3011190*n^11 - 53441388*n^10 + 751250500*n^9 - 8470570680*n^8 + 76896261585*n^7 - 560015385930*n^6 + 3235452199980*n^5 - 14525684311320*n^4 + 48947506506080*n^3 - 116650912956480*n^2 + 175512302620800*n - 125495209214208 for n >= 13. - Georg Fischer, Apr 27 2021 (polynomial computed by the program in links)

Corrected by Jaap Spies, Feb 01 2004

a(12)-a(21) from Georg Fischer, Apr 27 2021

A169629 Array T(n,k) read by antidiagonals: T(n,k) = Sum_{v=1..n, v odd} binomial(n,v)*k^v.

Original entry on oeis.org

1, 2, 2, 4, 4, 3, 8, 14, 6, 4, 16, 40, 36, 8, 5, 32, 122, 120, 76, 10, 6, 64, 364, 528, 272, 140, 12, 7, 128, 1094, 2016, 1684, 520, 234, 14, 8, 256, 3280, 8256, 7448, 4400, 888, 364, 16, 9, 512, 9842, 32640, 40156, 21280, 9966, 1400, 536, 18, 10

Offset: 1

Roger L. Bagula, Mar 03 2010

Antidiagonal sums are: 1, 4, 11, 32, 105, 366, 1387, ...

			   1,    2,    3,     4,      5,      6,       7, ...
   2,    4,    6,     8,     10,     12,      14, ...
   4,   14,   36,    76,    140,    234,     364, ...
   8,   40,  120,   272,    520,    888,    1400, ...
  16,  122,  528,  1684,   4400,   9966,   20272, ...
  32,  364, 2016,  7448,  21280,  51012,  107744, ...
  64, 1094, 8256, 40156, 148160, 450834, 1188544, ...

Cf. A152011.

Cf. A005843 (2nd line), A079908 (3rd line), A105374 (4th line).

tabl(nn) = {for (n=1, nn, for (k=1, nn, print1(sum(v=1, n, (v%2)*binomial(n, v)*k^v), ", ");); print(););} \\ Michel Marcus, Jul 22 2015

A180359 n^9+9n.

Original entry on oeis.org

0, 10, 530, 19710, 262180, 1953170, 10077750, 40353670, 134217800, 387420570, 1000000090, 2357947790, 5159780460, 10604499490, 20661046910, 38443359510, 68719476880, 118587876650, 198359290530, 322687697950, 512000000180

Offset: 0

Odimar Fabeny, Aug 30 2010

Index entries for linear recurrences with constant coefficients, signature (10, -45, 120, -210, 252, -210, 120, -45, 10, -1).

Cf. A132411, A079908, A180354, A180355, A180356, A180357, A180358.

a(n)= +10*a(n-1) -45*a(n-2) +120*a(n-3) -210*a(n-4) +252*a(n-5) -210*a(n-6) +120*a(n-7) -45*a(n-8) +10*a(n-9) -a(n-10). G.f.: 10*x*(1+43*x+1486*x^2+8773*x^3+15682*x^4+8773*x^5+1486*x^6+43*x^7+x^8)/(x -1)^10. [From R. J. Mathar, Sep 19 2010]

a(0) corrected by R. J. Mathar, Sep 19 2010

A356976 Least positive integer m such that the numbers k^3 + 3*k (k = 1..n) are pairwise distinct modulo m.

Original entry on oeis.org

1, 3, 3, 7, 15, 15, 19, 27, 27, 39, 39, 39, 61, 61, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243

Offset: 1

Zhi-Wei Sun, Sep 07 2022

nonn

Conjecture 1: If n is at least 15, then a(n) is the least power of 3 not smaller than 3*n.
Conjecture 2: For each positive integer n, the least positive integer m such that those numbers 2*k^3 + k (k = 1..n) are pairwise distinct modulo m, is just the least power of 2 not smaller than n.
Conjecture 3: For any positive integer n, the least positive integer m such that those numbers 2*k^3 - 4*k (k = 1..n) are pairwise distinct modulo m, is just the least power of 3 not smaller than n.
Conjecture 4: For each positive integer n not equal to 4, the least positive integer m such that those numbers 16*k^3 - 8*k (k = 1..n) are pairwise distinct modulo m, is just the least power of 3 not smaller than n.
The author formulated Conjectures 1-4 on Nov. 16, 2021, and verified them for n up to 10^5.

			a(2) = 3, for, 1^3 + 3*1 = 4 and 2^3 + 3*2 = 14 are incongruent modulo 3, but congruent modulo 1 and 2.

Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.
Quan-Hui Yang and Lilu Zhao, On a conjecture of Sun involving powers of three, arXiv:2111.02746 [math.NT], 2021.

Cf. A000578, A079908, A349459, A349530, A349537.

f[k_]:=f[k]=k^3+3*k;
U[m_, n_]:=U[m, n]=Length[Union[Table[Mod[f[k], m], {k, 1, n}]]]
tab={}; s=1; Do[m=s; Label[bb]; If[U[m, n]==n, s=m; tab=Append[tab, s]; Goto[aa]];
m=m+1; Goto[bb]; Label[aa], {n, 1, 80}]; Print[tab]

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A079919 Solution to the Dancing School Problem with 14 girls and n+14 boys: f(14,n).

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A079920 Solution to the Dancing School Problem with 15 girls and n+15 boys: f(15,n).

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A169629 Array T(n,k) read by antidiagonals: T(n,k) = Sum_{v=1..n, v odd} binomial(n,v)*k^v.

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PARI

A180359 n^9+9n.

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A356976 Least positive integer m such that the numbers k^3 + 3*k (k = 1..n) are pairwise distinct modulo m.

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Mathematica