cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A191494 Number of compositions of even natural numbers in 7 parts <= n.

Original entry on oeis.org

1, 64, 1094, 8192, 39063, 139968, 411772, 1048576, 2391485, 5000000, 9743586, 17915904, 31374259, 52706752, 85429688, 134217728, 205169337, 306110016, 446935870, 640000000, 900544271, 1247178944, 1702412724, 2293235712, 3051757813, 4015905088
Offset: 0

Views

Author

Adi Dani, Jun 03 2011

Keywords

Comments

Number of ways of placing an even number of indistinguishable objects in 7 distinguishable boxes with the condition that in each box can be at most n objects.

Examples

			a(1)=64 and compositions of even natural numbers into 7 parts no greater than 1 are
:(0,0,0,0,0,0,0) --> 7!/7!0! =  1
:(0,0,0,0,0,1,1) --> 7!/5!2! = 21
:(0,0,0,1,1,1,1) --> 7!/3!4! = 35
:(0,1,1,1,1,1,1) --> 7!/1!6! =  7
		

Crossrefs

Cf. A036486 (3 parts), A171714 (4 parts), A191484 (5 parts), A191489 (6 parts), A191495 (8 parts).

Programs

  • Magma
    [((n + 1)^7 + (1+(-1)^n)/2 )/2: n in [0..40]]; // Vincenzo Librandi, Jun 16 2011
  • Mathematica
    Table[1/2*((n + 1)^7 + (1 + (-1)^n)*1/2), {n, 0, 25}]

Formula

a(n) = ((n + 1)^7 + (1 + (-1)^n)/2 )/2.
G.f.: ( 1 + 57*x + 666*x^2 + 1786*x^3 + 1821*x^4 + 645*x^5 + 64*x^6 ) / ( (1+x)*(x-1)^8 ). - R. J. Mathar, Jun 08 2011

A191489 Number of compositions of even natural numbers into 6 parts <= n.

Original entry on oeis.org

1, 32, 365, 2048, 7813, 23328, 58825, 131072, 265721, 500000, 885781, 1492992, 2413405, 3764768, 5695313, 8388608, 12068785, 17006112, 23522941, 32000000, 42883061, 56689952, 74017945, 95551488, 122070313, 154457888
Offset: 0

Views

Author

Adi Dani, Jun 03 2011

Keywords

Comments

Number of ways of placing of an even number of indistinguishable objects in 6 distinguishable boxes with condition that in each box can be at most n objects.

Examples

			a(1)=32 compositions of even natural numbers in 6 parts <= 1 are
:(0,0,0,0,0,0)--> 6!/(6!0!) =  1
:(0,0,0,0,1,1)--> 6!/(4!2!) = 15
:(0,0,1,1,1,1)--> 6!/(2!4!) = 15
:(1,1,1,1,1,1)--> 6!/(0!6!) =  1
a(2)=365 compositions of even natural numbers in 6 parts <= 2 are
:(0,0,0,0,0,0)--> 6!/(6!0!0!) =  1
:(0,0,0,0,1,1)--> 6!/(4!2!0!) = 15
:(0,0,0,0,0,2)--> 6!/(5!0!1!) =  6
:(0,0,1,1,1,1)--> 6!/(2!4!0!) = 15
:(0,0,0,1,1,2)--> 6!/(3!2!1!) = 60
:(0,0,0,0,2,2)--> 6!/(4!0!2!) = 15
:(0,1,1,1,1,2)--> 6!/(1!4!1!) = 30
:(0,0,0,2,2,2)--> 6!/(3!0!3!) = 20
:(0,0,1,1,2,2)--> 6!/(2!2!2!) = 90
:(1,1,1,1,1,1)--> 6!/(0!6!0!) =  1
:(0,1,1,2,2,2)--> 6!/(1!2!3!) = 60
:(0,0,2,2,2,2)--> 6!/(2!0!4!) = 15
:(1,1,1,1,2,2)--> 6!/(0!4!2!) = 15
:(0,2,2,2,2,2)--> 6!/(1!0!5!) =  6
:(1,1,2,2,2,2)--> 6!/(0!2!4!) = 15
:(2,2,2,2,2,2)--> 6!/(0!0!6!) =  1
		

Crossrefs

Cf. A036486 (3 parts), A171714 (4 parts), A191484 (5 parts), A191494 (7 parts), A191495 (8 parts).

Programs

  • Magma
    [((n + 1)^6 + (1+(-1)^n)/2 )/2: n in [0..40]]; // Vincenzo Librandi, Jun 16 2011
  • Mathematica
    Table[1/2*((n + 1)^6 + (1 + (-1)^n)*1/2), {n, 0, 25}]

Formula

a(n) = ((n + 1)^6 + (1+(-1)^n)/2 )/2.
G.f.: (x^2 + 10*x + 1)*(x^4 + 16*x^3 + 26*x^2 + 16*x + 1) / ( (1+x)*(1-x)^7 ). - R. J. Mathar, Jun 06 2011
a(2n) = A175113(n). - R. J. Mathar, Jun 07 2011

A191495 Number of compositions of even natural numbers into 8 parts <= n.

Original entry on oeis.org

1, 128, 3281, 32768, 195313, 839808, 2882401, 8388608, 21523361, 50000000, 107179441, 214990848, 407865361, 737894528, 1281445313, 2147483648, 3487878721, 5509980288, 8491781521, 12800000000, 18911429681, 27437936768, 39155492641, 55037657088, 76293945313, 104413532288
Offset: 0

Views

Author

Adi Dani, Jun 03 2011

Keywords

Comments

Number of ways of placing an even number of indistinguishable objects in 8 distinguishable boxes with the condition that in each box can be at most n objects.

Examples

			a(1)=128 compositions of even natural numbers into 8 parts no greater than 1 are
  (0,0,0,0,0,0,0,0) --> 8!/(8!0!) =  1
  (0,0,0,0,0,0,1,1) --> 8!/(6!2!) = 28
  (0,0,0,0,1,1,1,1) --> 8!/(4!4!) = 70
  (0,0,1,1,1,1,1,1) --> 8!/(2!6!) = 28
  (1,1,1,1,1,1,1,1) --> 8!/(0!8!) =  1
		

Crossrefs

Cf. A036486 (3 parts), A171714 (4 parts), A191484 (5 parts), A191489 (6 parts), A191494 (7 parts).

Programs

  • Magma
    [ 1/2*((n + 1)^8 + (1 + (-1)^n)*1/2): n in [0..35]]; // Vincenzo Librandi, Jun 06 2011
    
  • Mathematica
    Table[1/2*((n + 1)^8 + (1 + (-1)^n)*1/2), {n, 0, 25}]
    LinearRecurrence[{8,-27,48,-42,0,42,-48,27,-8,1},{1,128,3281,32768,195313,839808,2882401,8388608,21523361,50000000},30] (* Harvey P. Dale, Mar 13 2013 *)
  • PARI
    a(n)=((n+1)^8+1)>>1 \\ Charles R Greathouse IV, Jun 06 2011

Formula

a(n) = ( (n+1)^8 + (1 + (-1)^n)/2 )/2.
G.f.: ( -1 - 120*x - 2284*x^2 - 9928*x^3 - 15654*x^4 - 9928*x^5 - 2284*x^6 - 120*x^7 - x^8 ) / ( (1+x)*(x-1)^9 ). - R. J. Mathar, Jun 06 2011
a(n) = 8*a(n-1) - 27*a(n-2) + 48*a(n-3) - 42*a(n-4) + 42*a(n-6) - 48*a(n-7) + 27*a(n-8) - 8*a(n-9) + a(n-10); a(0)=1, a(1)=128, a(2)=3281, a(3)=32768, a(4)=195313, a(5)=839808, a(6)=2882401, a(7)=8388608, a(8)=21523361, a(9)=50000000. - Harvey P. Dale, Mar 13 2013

A191902 Number of compositions of odd positive integers into 5 parts <= n.

Original entry on oeis.org

0, 16, 121, 512, 1562, 3888, 8403, 16384, 29524, 50000, 80525, 124416, 185646, 268912, 379687, 524288, 709928, 944784, 1238049, 1600000, 2042050, 2576816, 3218171, 3981312, 4882812, 5940688, 7174453, 8605184, 10255574, 12150000, 14314575
Offset: 0

Views

Author

Adi Dani, Jun 19 2011

Keywords

Examples

			a(1)=16: the 16 compositions of odd numbers into 5 parts <= 1 are
1: (0,0,0,0,1) --> 5!/(4!1!) =  5;
3: (0,0,1,1,1) --> 5!/(2!3!) = 10;
5: (1,1,1,1,1) --> 5!/(0!5!) =  1.
		

Crossrefs

Cf. A191484.

Programs

  • Magma
    [((n + 1)^5 - (1 + (-1)^n)/2)/2: n in [0..50]]; // Vincenzo Librandi, Jul 04 2011
    
  • Mathematica
    Table[Floor[1/2*((n + 1)^5 - (1 + (-1)^n)/2)], {n, 0, 30}]
  • PARI
    a(n)=((n+1)^5-(1+(-1)^n)/2)/2 \\ Charles R Greathouse IV, Jul 06 2017

Formula

a(n) = ((n + 1)^5 - (1 + (-1)^n)/2)/2.
From R. J. Mathar, Jun 22 2011: (Start)
a(2n+1) = A191484(2n+1); a(2n) = A191484(2n) - 1.
G.f.: x*(16 + 41*x + 51*x^2 + 11*x^3 + x^4) / ( (1+x)*(x-1)^6 ). (End)

A191687 Table T(n,k) = ceiling((1/2)*((k+1)^n+(1+(-1)^k)/2)) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 1, 4, 5, 2, 1, 1, 8, 14, 8, 3, 1, 1, 16, 41, 32, 13, 3, 1, 1, 32, 122, 128, 63, 18, 4, 1, 1, 64, 365, 512, 313, 108, 25, 4, 1, 1, 128, 1094, 2048, 1563, 648, 172, 32, 5, 1
Offset: 1

Views

Author

Adi Dani, Jun 11 2011

Keywords

Comments

T(n,k) is the number of compositions of even natural numbers into n parts <= k.

Examples

			Top left corner:
  1, 1, 1,  1,  1,...
  1, 1, 2,  2,  3,...
  1, 2, 5,  8, 13,...
  1, 4,14, 32, 63,...
  1, 8,41,128,313,...
T(2,4)=13: there are 13 compositions of even natural numbers into 2 parts <=4
0: (0,0);
2: (0,2), (2,0), (1,1);
4: (0,4), (4,0), (1,3), (3,1), (2,2);
6: (2,4), (4,2), (3,3);
8: (4,4).
		

Crossrefs

Programs

  • Mathematica
    Table[Table[Ceiling[1/2*((k+1)^n+(1+(-1)^k)/2)],{n,0,9},{k,0,9}]]
Showing 1-5 of 5 results.