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.

Previous Showing 11-17 of 17 results.

A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 2, 3, 1, 0, 1, 2, 4, 4, 1, 1, 1, 2, 4, 7, 5, 1, 0, 1, 2, 4, 8, 11, 6, 1, 1, 1, 2, 4, 8, 15, 16, 7, 1, 0, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 0, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 0
Offset: 0

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Author

Henry Bottomley, Nov 24 2000

Keywords

Comments

Changing the formula by replacing T(0,2n)=T(1,n) by T(0,2n)=T(m,n) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058394, A058395, A057884, (and effectively A007318).

Examples

			Rows are (1,0,1,0,1,0,1,...), (1,1,1,1,1,1,...), (1,2,2,2,2,2,...), (1,3,4,4,4,...) etc.
		

Crossrefs

Rows are A000035 (A000012 with zeros), A000012, A040000 etc. Columns are A000012, A001477, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863 etc. Diagonals include A000079, A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, etc. The triangles A008949, A054143 and A055248 also appear in the half of the array which is not powers of 2.

Formula

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(1, 1)=1, T(0, 2n)=T(1, n) and T(0, 2n+1)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2).

A219615 a(n) = Sum_{k=0..12} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16369, 32647, 64839, 127858, 249528, 480492, 910596, 1695222, 3096514, 5546382, 9740686, 16777216, 28354132, 47050564, 76717268, 123012781, 194129627, 301766029, 462411533, 699030226, 1043243132
Offset: 0

Views

Author

Mokhtar Mohamed, Nov 23 2012

Keywords

Comments

a(n) is the number of compositions (ordered partitions) of n+1 into thirteen or fewer parts.
a(n) is the sum of the first thirteen terms in the n-th row of Pascal's triangle.

Examples

			a(13)= 8191 because there are (2^13) -1 compositions of 14 into thirteen or fewer parts. When 1<= n <= 12, for n=5, a(5) = 2*a(4) = 2*16 = 32. For n=12, a(12) = 2*a(11)= 2*2048 = 4096. When n>12, for n=13, a(13) = 2*a(12) - binomial(12,12) = 2*4096 - 1 = 8191. For n = 15, a(15) = 2*a(14) - binomial(14,12) = 2*16369 - 91 = 32738 - 91 = 32647.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k], {k, 0, 12}], {n, 0, 40}] (* T. D. Noe, Nov 27 2012 *)
    LinearRecurrence[{13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1},{1,2,4,8,16,32,64,128,256,512,1024,2048,4096},40] (* Harvey P. Dale, Nov 29 2012 *)
  • PARI
    a(n)=sum(k=1,12,binomial(n,k)) \\ Charles R Greathouse IV, Nov 27 2012

Formula

a(n) = (n^12 - 54n^11 + 1397n^10 - 21450n^9 + 218823n^8 - 1508562n^7 + 7374191n^6 - 23551110n^5 + 58206676n^4 - 48306984n^3 + 173699712n^2 + 312888960n)/479001600. - Charles R Greathouse IV, Nov 27 2012
a(0)=1, a(1)=2, a(2)=4, a(3)=8, a(4)=16, a(5)=32, a(6)=64, a(7)=128, a(8)=256, a(9)=512, a(10)=1024, a(11)=2048, a(12)=4096, a(n)= 13*a(n-1)- 78*a(n-2)+286*a(n-3)-715*a(n-4)+1287*a(n-5)-1716*a(n-6)+ 1716*a(n-7)- 1287*a(n-8)+715*a(n-9)-286*a(n-10)+78*a(n-11)-13*a(n-12)+a(n-13). - Harvey P. Dale, Nov 29 2012

Extensions

Sequence corrected and extended by T. D. Noe, Nov 26 2012
Definition corrected by Harvey P. Dale, Nov 29 2012

A219676 a(n) = Sum_{k=0..13} binomial(n, k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32752, 65399, 130238, 258096, 507624, 988116, 1898712, 3593934, 6690448, 12236830, 21977516, 38754732, 67108864, 114159428, 190876696, 313889477, 508019104, 809785133, 1272196666
Offset: 0

Views

Author

Mokhtar Mohamed, Nov 24 2012

Keywords

Comments

a(n) is the number of compositions (ordered partitions) of n+1 into fourteen or fewer parts.
a(n) is the sum of the first fourteen terms in the n-th row of Pascal's triangle.

Examples

			a(14) = 16383 because there are 2^14 = 16384 compositions of 15 into any size parts but one of the compositions (1 + 1 + ... + 1 = 15) has more than fourteen parts.
When 1 <= n <= 13, a(7) = 2*a(6) = 2*64= 128, a(13) = 2*a(12) = 2*4096 = 8192.
When n > 13, a(14) = 2*a(13) - C(13, 13) = 2*8192 - 1 = 16383, a(15) = 2*a(14) - C(14, 13) = 2*16383 - 14 = 32766 - 14 = 32752.
		

Crossrefs

Programs

  • Maple
    f:= n -> add(binomial(n,k),k=0..13):
    map(f, [$0..100]); # Robert Israel, Mar 14 2018
  • Mathematica
    Table[Sum[Binomial[n, k], {k, 0, 13}], {n, 0, 40}] (* T. D. Noe, Nov 26 2012 *)

Formula

a(n) = Sum_{k=1..7} binomial(n+1, 2k-1).
a(n) = 1 +(n^13 -65*n^12 +2015*n^11 -37609*n^10 +470613*n^9 -4081935*n^8 +25378925*n^7 -110205667*n^6 +351042406*n^5 -657328100*n^4 +1303568760*n^3 +771653376*n^2 +4546558080*n)/13!. - corrected by Mokhtar Mohamed, Dec 01 2012
G.f.: (1 - 12*x + 67*x^2 - 230*x^3 + 541*x^4 - 920*x^5 + 1163*x^6 - 1106*x^7 + 791*x^8 - 420*x^9 + 161*x^10 - 42*x^11 + 7*x^12)/(1-x)^14.
a(n) = 2*a(n-1), for 1 <= n <= 13, with a(0) = 1, a(n) = 2*a(n-1) - C(n-1, 13), for n > 13.

Extensions

Corrected and extended by T. D. Noe, Nov 26 2012

A084638 Binomial transform of (1,0,1,0,1,0,1,0,2,0,2,0,2,....).

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 32, 64, 129, 265, 558, 1200, 2610, 5682, 12288, 26292, 55587, 116179, 240366, 493108, 1004780, 2036692, 4112144, 8278552, 16631717, 33364381, 66863358, 133903816, 268037862, 536371734, 1073120208, 2146715436, 4294024647, 8588785575
Offset: 0

Views

Author

Paul Barry, Jun 06 2003

Keywords

Comments

The sequence starting 1,2,4,... is the binomial transform of (1,1,1,1,1,1,1,2,2...) with a(n) = Sum_{k=0..6} C(n,k) + 2*Sum_{k=7..n} C(n,k) = 2^(n+1) - A008859(n). This gives the partial sums of A084637.

Crossrefs

Programs

  • Magma
    [2^n -4 -(n+1)*(n^5-16*n^4+131*n^3-536*n^2+1500*n-2160)/720 + 0^n: n in [0..50]]; // G. C. Greubel, Mar 20 2023
    
  • Mathematica
    Table[2^n -4 -(1/6!)*(n+1)*(n^5-16*n^4+131*n^3-536*n^2+1500*n-2160) + Boole[n==0], {n,0,50}] (* G. C. Greubel, Mar 20 2023 *)
  • PARI
    Vec((1-8*x+28*x^2-56*x^3+70*x^4-56*x^5+28*x^6-8*x^7+2*x^8)/((1-x)^7*(1-2*x)) + O(x^50)) \\ Colin Barker, Mar 17 2016
    
  • SageMath
    [2^n -4 -(n+1)*(n^5-16*n^4+131*n^3-536*n^2+1500*n-2160)/720 + 0^n for n in range(51)] # G. C. Greubel, Mar 20 2023

Formula

a(n) = Sum_{k=0..3, C(n, 2*k)} + 2*Sum_{k=4..floor(n/2), C(n, 2*k)}.
a(n) = (n^6-15*n^5+115*n^4-405*n^3+964*n^2-660*n+720)/720 + 2*Sum_{k=4..floor(n/2), C(n, 2k)}.
G.f.: (1-8*x+28*x^2-56*x^3+70*x^4-56*x^5+28*x^6-8*x^7+2*x^8) / ((1-x)^7*(1-2*x)). - Colin Barker, Mar 17 2016

A220051 Sum_{k=0..14} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65519, 130918, 261156, 519252, 1026876, 2014992, 3913704, 7507638, 14198086, 26434916, 48412432, 87167164, 154276028, 268435456, 459312152, 773201629, 1281220733, 2091005866
Offset: 0

Views

Author

Mokhtar Mohamed, Dec 03 2012

Keywords

Comments

a(n) is the number of compositions (ordered partitions) of n+1 into fifteen or fewer parts.
a(n) = sum(binomial(n+1,2k), for k = 0..7).
a(n) is the sum of the first fifteen terms in the n-th row of Pascal's triangle.

Examples

			a(15) = 32767 because there are 2^15 = 32768 compositions of 16 into any size parts but one of the compositions (1 + 1 + ... + 1 = 16) has more than fifteen parts.
When 1 <= n <= 14, for n=10, a(10) = 2*a(9) = 2*512 = 1024. For n=14, a(14) = 2*a(13) = 2*8192 = 16384.
When n > 14, for n = 15, a(15) = 2*a(14) -C(14,14) = 2*16384 -1 = 32767. For n=20, a(20) = 2*a(19) -C(19,14) = 2*519252 -11626 = 1038504 -11626 = 1026876.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k],{k,0,14}],{n,0,33}] (* Indranil Ghosh, Feb 22 2017 *)
    NestList[{#1 + 1, 2 #2 - Boole[#1 >= 14] Binomial[#1, 14]} & @@ # &, {0, 1}, 33][[All, -1]] (* Michael De Vlieger, Feb 22 2017 *)
  • PARI
    a(n)=sum(k=0,14,binomial(n,k)) \\ Indranil Ghosh, Feb 23 2017

Formula

a(n) = 1 + (n^14 - 77*n^13 + 2821*n^12 - 6288*n^11 + 947947*n^10 - 10081071*n^9 + 77889383*n^8 - 435638203*n^7 + 1793239448*n^6 - 5043110072*n^5 + 1111159696*n^4 - 8346754416*n^3 + 30605906304*n^2 + 57424792320*n)/14!.
G.f.: (1 - 13x + 79x^2 - 297x^3 + 771x^4 - 1461x^5 + 2083x^6 - 2269x^7 + 1897x^8 - 1211x^9 + 581x^10 - 203x^11 + 49x^12 - 7x^13 + x^14)/(1-x)^15.
a(n) = 2*a(n-1), for 1 <= n <= 14, with a(0) = 1, a(n) = 2*a(n-1) - C(n-1,14), for n> 14.

A362194 Number of Grassmannian permutations of size n that avoid a pattern, sigma, where sigma is a pattern of size 7 with exactly one descent.

Original entry on oeis.org

1, 1, 2, 5, 12, 27, 58, 120, 239, 457, 838, 1475, 2498, 4083, 6462, 9934, 14877, 21761, 31162, 43777, 60440, 82139, 110034, 145476, 190027, 245481, 313886, 397567, 499150, 621587, 768182, 942618, 1148985, 1391809, 1676082, 2007293, 2391460, 2835163, 3345578, 3930512
Offset: 0

Views

Author

Jessica A. Tomasko, Apr 20 2023

Keywords

Comments

A permutation is said to be Grassmannian if it has at most one descent. The definition for sigma is a pattern of size 7 with exactly one descent. For example, sigma can be chosen to be 1247356, 2413567, 3671245, 5712346, etc.

Crossrefs

Programs

  • PARI
    a(n) = 1 + sum(i=2, 6, binomial(n,i)) \\ Andrew Howroyd, Apr 20 2023

Formula

a(n) = 1 + Sum_{i=2..6} binomial(n, i).
a(n) = A008859(n) - n.
G.f.: (1-6*x+16*x^2-23*x^3+19*x^4-8*x^5+2*x^6)/(1-x)^7.
E.g.f.: exp(x)*(720 + 360*x^2 + 120*x^3 + 30*x^4 + 6*x^5 + x^6)/720. - Stefano Spezia, Apr 20 2023

A373005 Array read by ascending antidiagonals: A(n,k) is the maximum possible cardinality of a set of points of diameter at most k-1 in {0,1}^n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 0, 1, 2, 2, 0, 0, 1, 2, 3, 2, 1, 0, 1, 2, 4, 4, 2, 2, 0, 1, 2, 5, 6, 4, 2, 1, 0, 1, 2, 6, 8, 7, 4, 2, 0, 0, 1, 2, 7, 10, 11, 8, 4, 2, 1, 0, 1, 2, 8, 12, 16, 14, 8, 4, 2, 2, 0, 1, 2, 9, 14, 22, 22, 15, 8, 4, 2, 1, 0, 1, 2, 10, 16, 29, 32, 26, 16, 8, 4, 2, 0
Offset: 0

Views

Author

Stefano Spezia, May 19 2024

Keywords

Comments

A(n,k) is also the size of the Hamming ball in {0,1}^n of radius (k-1)/2 if k is odd and of the union of two Hamming balls in {0,1}^n of radius k/2-1 whose centers are of Hamming distance 1 if k is even.

Examples

			The array begins:
  1, 1, 2, 1,  0,  1,  2,  1, ...
  0, 1, 2, 2,  2,  2,  2,  2, ...
  0, 1, 2, 3,  4,  4,  4,  4, ...
  0, 1, 2, 4,  6,  7,  8,  8, ...
  0, 1, 2, 5,  8, 11, 14, 15, ...
  0, 1, 2, 6, 10, 16, 22, 26, ...
  0, 1, 2, 7, 12, 22, 32, 42, ...
  0, 1, 2, 8, 14, 29, 44, 64, ...
  ...
		

Crossrefs

Cf. A000007 (k=0), A000012 (k=1), A000124 (k=5), A000125 (k=7), A005843 (k=4), A006261 (k=11), A007395 (k=2), A008859 (k=13), A011782 (main diagonal), A014206, A046127 (k=8), A059173, A059174, A130130 (n=1), A158411 (n=2), A373006 (antidiagonal sums).

Programs

  • Mathematica
    A[n_,k_]:=If[OddQ[k],Sum[Binomial[n,i],{i,0,(k-1)/2}], Binomial[n-1,k/2-1]+Sum[Binomial[n,i],{i,0,k/2-1}]]; Table[A[n-k,k],{n,0,12},{k,0,n}]//Flatten

Formula

A(n,k) = Sum_{i=0..(k-1)/2} binomial(n,i) if k is odd;
A(n,k) = binomial(n-1,k/2-1) + Sum_{i=0..k/2-1} binomial(n,i) if k is even.
A(n,3) = n+1.
A(n,6) = A014206(n-1).
A(n,9) = A000127(n+1).
A(n,10) = A059173(n) for n > 0.
A(n,12) = A059174(n) for n > 0.
A(0,k) = A007877(k) for k > 0.
Previous Showing 11-17 of 17 results.