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 1-3 of 3 results.

A175114 First differences of A175113.

Original entry on oeis.org

1, 364, 7448, 51012, 206896, 620060, 1527624, 3281908, 6373472, 11454156, 19360120, 31134884, 48052368, 71639932, 103701416, 146340180, 201982144, 273398828, 363730392, 476508676, 615680240, 785629404, 991201288, 1237724852
Offset: 0

Views

Author

R. J. Mathar, Feb 13 2010

Keywords

Comments

Convolution of the finite sequence 1,358,5279,11764,5279,358,1 with A000389. Number of points in the standard root system of the D_6 lattice having L_infinity norm n.

Crossrefs

Programs

  • Magma
    I:=[1,364,7448,51012,206896,620060,1527624]; [n le 7 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Dec 20 2012
  • Mathematica
    CoefficientList[Series[(358 x + 5279 x^2 + 11764 x^3 + 5279 x^4 + 358 x^5 + 1+x^6)/(x - 1)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 20 2012 *)

Formula

a(n)= 6*a(n-1) -15*a(n-2) +20*a(n-3) -15*a(n-4) +6*a(n-5) -a(n-6), n>6.
a(n) = ((2*n+1)^6-(2*n-1)^6)/2 = 4*n*(12*n^2+1)*(4*n^2+3), n>0. - Bruno Berselli, Dec 27 2010
G.f.: (358*x+5279*x^2+11764*x^3+5279*x^4+358*x^5+1+x^6)/(x-1)^6. - R. J. Mathar, Jan 03 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

A359844 a(n) = ((2*n+1)^8 + 1)/2.

Original entry on oeis.org

1, 3281, 195313, 2882401, 21523361, 107179441, 407865361, 1281445313, 3487878721, 8491781521, 18911429681, 39155492641, 76293945313, 141214768241, 250123206481, 426445518721, 703204309121, 1125937695313, 1756239726961, 2676004630241, 3992462614561, 5844100138801
Offset: 0

Views

Author

Jianing Song, Jan 15 2023

Keywords

Crossrefs

Cf. {((2*n+1)^k + 1)/2}: A000012 (k=0), A001477 (k=1), A219086 (k=2), A050492 (k=3), A175110 (k=4), A175113 (k=6), this sequence (k=8).

Programs

  • Mathematica
    ((2*Range[0, 25] + 1)^8 + 1)/2 (* Paolo Xausa, Jan 23 2025 *)
  • PARI
    a(n) = ((2*n+1)^8 + 1)/2
    
  • Python
    def A359844(n): return ((n<<1)+1)**8+1>>1 # Chai Wah Wu, Jan 15 2023

Formula

a(n) = A359499(n)/A359498(n) = 16 * A359498(n) + 1.
Showing 1-3 of 3 results.