A000127 -id:A000127 - cp's OEIS frontend

A117717 Maximal number of regions obtained by a straight line drawing of the complete bipartite graph K_{n,n}.

0, 2, 13, 45, 116, 250, 477, 833, 1360, 2106, 3125, 4477, 6228, 8450, 11221, 14625, 18752, 23698, 29565, 36461, 44500, 53802, 64493, 76705, 90576, 106250, 123877, 143613, 165620, 190066, 217125, 246977, 279808, 315810, 355181, 398125, 444852, 495578, 550525

Offset: 1

Patricia A. Carey and Anant Godbole, Apr 13 2006

This sequence is in the same spirit as A000127 where a formula is given for the maximal number of regions obtained by a straight line drawing of the complete graph K_n with the vertices located on the perimeter of a circle. This yields the often quoted sequence A000127.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000127.

[(n-1)^2*(n^2+4)/4: n in [1..40]]; // Vincenzo Librandi, Sep 09 2011

A117717 := proc(n)
    (n-1)^2*(n^2+4)/4 ;
end proc:
seq(A117717(n),n=1..10) ; # R. J. Mathar, Sep 15 2013

Table[n^2-2n+Binomial[n,2]^2+1,{n,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,2,13,45,116},40] (* Harvey P. Dale, Oct 16 2012 *)

a(n) = n^2 - 2n + C(n,2)^2 + 1
a(n) = (n-1)^2*(n^2+4)/4. - Vincenzo Librandi, Sep 09 2011
G.f.: x^2*(2+3*x+x^3)/(1-x)^5. - Colin Barker, Feb 15 2012
a(n)=5*a(n-1)-10*a(n-2)+ 10*a(n-3)- 5*a(n-4)+a(n-5), n>5. - Harvey P. Dale, Oct 16 2012

More terms from Harvey P. Dale, Oct 16 2012

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

Mokhtar Mohamed, Nov 23 2012

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.

			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.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531.

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 *)

a(n)=sum(k=1,12,binomial(n,k)) \\ Charles R Greathouse IV, Nov 27 2012

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

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

Mokhtar Mohamed, Nov 24 2012

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.

			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.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531.

f:= n -> add(binomial(n,k),k=0..13):
map(f, [$0..100]); # Robert Israel, Mar 14 2018

Table[Sum[Binomial[n, k], {k, 0, 13}], {n, 0, 40}] (* T. D. Noe, Nov 26 2012 *)

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.

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

A058923 a(n) = binomial(n,0) - binomial(n,2) + binomial(n,4).

Original entry on oeis.org

1, 1, 0, -2, -4, -4, 1, 15, 43, 91, 166, 276, 430, 638, 911, 1261, 1701, 2245, 2908, 3706, 4656, 5776, 7085, 8603, 10351, 12351, 14626, 17200, 20098, 23346, 26971, 31001, 35465, 40393, 45816, 51766, 58276, 65380, 73113, 81511, 90611, 100451, 111070, 122508

Offset: 0

N. J. A. Sloane, Jan 12 2001

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000127.

CoefficientList[Series[-((z - 1)*z*((z - 1)*z + 4) + 1)/(z - 1)^5, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 16 2011 *)
Table[Binomial[n,0]-Binomial[n,2]+Binomial[n,4],{n,0,50}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,1,0,-2,-4},50] (* Harvey P. Dale, Mar 02 2015 *)

a(n) = 1 - binomial(n,2) + binomial(n,4) \\ Harry J. Smith, Jun 24 2009

Vec(-(1-4*x-2*x^3+x^4+5*x^2)/(x-1)^5 + O(x^60)) \\ Michel Marcus, Jan 03 2016

From R. J. Mathar, Mar 17 2009: (Start)
a(n) = 5a(n-1)-10a(n-2)+10a(n-3)-5a(n-4)+a(n-5).
G.f.: -(1-4*x-2*x^3+x^4+5*x^2)/(x-1)^5. (End)

A209257 A musically inspired Titius-Bode-like sequence based on the geometric division of 4- and 5-dimensional space: Z_(n+1) = 3 * (C(n-1, 0) + C(n-1, 1) + C(n-1, 2) + C(n-1, 3) + C(n-1, 4) + C(n-1, 5)*A059620(n+6)) + 4.

Original entry on oeis.org

4, 7, 10, 16, 28, 52, 97, 193, 301, 493, 1150, 1162, 3076, 2386, 3283, 10423, 5827, 20659, 9646, 37852, 15112, 18592, 83692, 27331, 133660, 38857, 45832, 251050, 62566, 367318, 83527, 523315, 109375, 124351, 852826, 158872, 1152508, 200140, 223561, 1754809

Offset: 0

Raphie Frank, Jan 14 2013

The classical Titius-Bode version of this sequence is given in A003461.
C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) + C(n, 4) = A000127(n) = A059173(n+1)/2.
C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) + C(n, 4) + C(n, 5) = A006261(n) = A059174(n+1)/2.
Where planetary and dwarf-planetary distances from the Sun at semi-major axis are expressed in astronomical units/10, then compare the following (noting that the running correlation coefficient, r, trends upwards as the population size increases):
n = 0, Mercury @ semi-major =   3.8710 vs.   4.0 -->  96.78%.
n = 1, Venus   @ semi-major =   7.2333 vs.   7.0 --> 103.33%.
n = 2, Earth   @ semi-major =  10.0000 vs.  10.0 --> 100.00%,  r = 0.998430.
n = 3, Mars    @ semi-major =  15.2368 vs.  16.0 -->  95.23%,  r = 0.998356.
n = 4, Ceres   @ semi-major =  27.654  vs.  28.0 -->  98.76%,  r = 0.999412.
n = 5, Jupiter @ semi-major =  52.0427 vs.  52.0 --> 100.08%,  r = 0.999809.
n = 6, Saturn  @ semi-major =  95.8202 vs.  97.0 -->  98.78%,  r = 0.999937.
n = 7, Uranus  @ semi-major = 192.2941 vs. 193.0 -->  99.63%,  r = 0.999981.
n = 8, Neptune @ semi-major = 301.0366 vs. 301.0 --> 100.01%,  r = 0.999990.
The correspondence between this sequence and planetary distances breaks down subsequent to Neptune unless one adopts the conceit of considering the outer four dwarf planets -- Pluto, Haumea, MakeMake and Eris -- as one unit occupying one "planetary band" (note that Eris @ perihelion is inside the Kuiper Belt). Then:
n = 9, Pluto/Haumea/MakeMake/Eris @ semi-major ~ 490.492 average vs. 493.0 --> 99.49%, r = 0.999994.
Empirical source: Wikipedia planet pages as of Jan 14 2013.
This sequence originated as part of an attempt to compare and contrast the "good" numerology of Johann Balmer to the "bad" numerology of Titius-Bode. Coincidentally, (Totient(C(31, 0) + C(31, 1) + C(31, 2) + C(31, 3) + C(31, 4)))/10^11 equals 3.6456*10^-7, in meters, the Balmer constant as given by Johann Balmer in 1885.

			Z_1 = 3*((1 - 1 +  1 -  1 +  1) + (-1 * 1)) + 4 =   4,
Z_2 = 3*((1 + 0 +  0 +  0 +  0) +  (0 * 0)) + 4 =   7,
Z_3 = 3*((1 + 1 +  0 +  0 +  0) +  (0 * 0)) + 4 =  10,
Z_4 = 3*((1 + 2 +  1 +  0 +  0) +  (0 * 1)) + 4 =  16,
Z_5 = 3*((1 + 3 +  3 +  1 +  0) +  (0 * 0)) + 4 =  28,
Z_6 = 3*((1 + 4 +  6 +  4 +  1) +  (0 * 1)) + 4 =  52,
Z_7 = 3*((1 + 5 + 10 + 10 +  5) +  (1 * 0)) + 4 =  97,
Z_8 = 3*((1 + 6 + 15 + 20 + 15) +  (6 * 1)) + 4 = 193,
Z_9 = 3*((1 + 7 + 21 + 35 + 35) + (21 * 0)) + 4 = 301.

G. C. Greubel, Table of n, a(n) for n = 0..1000
J. Baez, This Week's Finds in Mathematical Physics (Week 234)
H. Chang, Titius-Bode’s Relation and Distribution of Exoplanets
D. M. F. Chapman, The Titius-Bode Rule, Part 2: Science or Numerology?
Chemteam, The Balmer Formula
A. L. Kholodenko, Role of general relativity and quantum mechanics in dynamics of Solar System
Wikipedia, Dwarf planet
Wikipedia, Planet
Wikipedia, Titius-Bode law

Cf. A003461, A059620, A000127, A059173, A006261, A059174.

[3*(Binomial(n-1,0) + Binomial(n-1,1) + Binomial(n-1,2) + Binomial(n-1,3) + Binomial(n-1,4) + Binomial(n-1,5)*(Floor((5*(n+6) + 7)/12) - Floor((5*(n+6)+2)/12))) + 4: n in [0..30]]; // G. C. Greubel, Jan 07 2018

Z[n_]:= 3*(Binomial[n - 1, 0] + Binomial[n - 1, 1] + Binomial[n - 1, 2] + Binomial[n - 1, 3] + Binomial[n - 1, 4] + Binomial[n - 1, 5]*(Floor[(5 (n + 6) + 7)/12] - Floor[(5 (n + 6) + 2)/12])) + 4; Table[Z[n], {n, 0, 50}] (* G. C. Greubel, Jan 07 2018 *)

{z(n) = 3*(binomial(n-1,0) + binomial(n-1,1) + binomial(n-1,2) + binomial(n-1,3) + binomial(n-1,4) + binomial(n-1,5)*(floor((5*(n+6) + 7)/12) - floor((5*(n+6)+2)/12))) + 4};
for(n=0,30, print1(z(n), ", ")) \\ G. C. Greubel, Jan 07 2018

Z_(n+1) = 3 * (C(n-1, 0) + C(n-1, 1) + C(n-1, 2) + C(n-1, 3) + C(n-1, 4) + C(n-1, 5)*(floor((5*(n+6)+7)/12) - floor((5*(n+6)+2)/12))) + 4.

a(18) corrected by G. C. Greubel, Jan 07 2018

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

Mokhtar Mohamed, Dec 03 2012

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.

			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.

Indranil Ghosh, Table of n, a(n) for n = 0..1000

Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531, A219615, A219676.

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 *)

a(n)=sum(k=0,14,binomial(n,k)) \\ Indranil Ghosh, Feb 23 2017

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.

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

Stefano Spezia, May 19 2024

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.

			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, ...
  ...

Noga Alon, Zhihan Jin, and Benny Sudakov, The Helly number of Hamming balls and related problems, arXiv:2405.10275 [math.CO], 2024. See p. 3.
S. L. Bezrukov, Specification of all maximal subsets of the unit cube with respect to given diameter. Problemy Peredachi Informatsii, pages 106-109, 1987. On ResearchGate.
G. Katona, Intersection theorems for systems of finite sets, Acta Mathematica Academiae Scientiarum Hungaricae 15, 329-337 (1964).
Daniel J. Kleitman, On a combinatorial conjecture of Erdös, Journal of Combinatorial Theory, Series A 1, 209-214, (1966).

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).

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

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.

A144841 a(n) = 280C(n,9)+10C(n,6)+6*C(n,4)+C(n,3)+1 where C = binomial.

Original entry on oeis.org

1, 1, 2, 11, 41, 121, 316, 757, 1961, 6281, 22166, 74031, 221937, 596961, 1460096, 3294761, 6945521, 13818417, 26161386, 47446611, 82880281, 140069161, 229877572, 367512861, 573882201, 877268601, 1315379326, 1937825527, 2809097761, 4012108241, 5652377096, 7862946641, 10810114657, 14700084961, 19786641106

Offset: 1

Sep, Jorik & Noud Thijssen (sep.thijssen(AT)gmail.com), Sep 22 2008

Conjectured values for maximal number of regions obtained by joining each triple of n points on a 3 dimensional sphere by a plane.

This is the first step in the generalization of the maximal number of regions obtained by joining n points around a circle by straight lines. See A000127.

Cf. A000127.

function a(n) return 280*Binomial(n,9)+10*Binomial(n,6)+6*Binomial(n,4)+Binomial(n,3)+1; end function;

G.f.: x*(1-9*x+37*x^2-84*x^3+111*x^4-76*x^5+9*x^6+18*x^7+274*x^8-x^9)/(1-x)^10. - Colin Barker, Feb 15 2012

More terms and new name using formula given from Joerg Arndt, Aug 21 2025

cp's OEIS Frontend

A117717 Maximal number of regions obtained by a straight line drawing of the complete bipartite graph K_{n,n}.

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

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

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A058923 a(n) = binomial(n,0) - binomial(n,2) + binomial(n,4).

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A209257 A musically inspired Titius-Bode-like sequence based on the geometric division of 4- and 5-dimensional space: Z_(n+1) = 3 * (C(n-1, 0) + C(n-1, 1) + C(n-1, 2) + C(n-1, 3) + C(n-1, 4) + C(n-1, 5)*A059620(n+6)) + 4.

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

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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.

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A144841 a(n) = 280C(n,9)+10C(n,6)+6*C(n,4)+C(n,3)+1 where C = binomial.