A006261 -id:A006261 - cp's OEIS frontend

A161702 a(n) = (-n^3 + 9n^2 - 5n + 3)/3.

1, 2, 7, 14, 21, 26, 27, 22, 9, -14, -49, -98, -163, -246, -349, -474, -623, -798, -1001, -1234, -1499, -1798, -2133, -2506, -2919, -3374, -3873, -4418, -5011, -5654, -6349, -7098, -7903, -8766, -9689, -10674, -11723, -12838, -14021, -15274

Offset: 0

Reinhard Zumkeller, Jun 17 2009

{a(k): 0 <= k < 4} = divisors of 14:

a(n) = A027750(A006218(13) + k + 1), 0 <= k < A000005(14).

			Differences of divisors of 14 to compute the coefficients of their interpolating polynomial, see formula:
  1     2     7    14
     1     5     7
        4     2
          -2

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. Zumkeller, Enumerations of Divisors
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000124, A000125, A000127, A002522, A005408, A006261, A016813, A058331, A080856, A086514, A161701, A161703, A161704, A161706-A161708, A161710, A161711-A161713, A161715.

[(-n^3 + 9*n^2 - 5*n + 3)/3: n in [0..50]]; // Vincenzo Librandi, Dec 27 2010

A161702:=n->(-n^3 + 9*n^2 - 5*n + 3)/3: seq(A161702(n), n=0..60); # Wesley Ivan Hurt, Jul 16 2017

Table[(-n^3+9n^2-5n+3)/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,2,7,14},40] (* Harvey P. Dale, Jun 15 2013 *)

a(n)=(-n^3+9*n^2-5*n+3)/3 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = C(n,0) + C(n,1) + 4*C(n,2) - 2*C(n,3).
G.f.: (1-2*x+5*x^2-6*x^3)/(1-x)^4. - Colin Barker, Jan 08 2012
a(0)=1, a(1)=2, a(2)=7, a(3)=14, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Jun 15 2013

A161703 a(n) = (4n^3 - 12n^2 + 14*n + 3)/3.

Original entry on oeis.org

1, 3, 5, 15, 41, 91, 173, 295, 465, 691, 981, 1343, 1785, 2315, 2941, 3671, 4513, 5475, 6565, 7791, 9161, 10683, 12365, 14215, 16241, 18451, 20853, 23455, 26265, 29291, 32541, 36023, 39745, 43715, 47941, 52431, 57193, 62235, 67565, 73191, 79121

Offset: 0

Reinhard Zumkeller, Jun 17 2009

{a(k): 0 <= k < 4} = divisors of 15:

a(n) = A027750(A006218(14) + k + 1), 0 <= k < A000005(15).

			Differences of divisors of 15 to compute the coefficients of their interpolating polynomial, see formula:
  1     3     5    15
     2     2    10
        0     8
           8

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. Zumkeller, Enumerations of Divisors
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000124, A000125, A000127, A002522, A005408, A006261, A016813, A058331, A080856, A086514, A161701, A161702, A161704, A161706-A161708, A161710, A161711-A161713, A161715.

[(4*n^3 - 12*n^2 + 14*n + 3)/3: n in [0..50]]; // Vincenzo Librandi, Dec 27 2010

A161703:=n->(4*n^3 - 12*n^2 + 14*n + 3)/3: seq(A161703(n), n=0..100); # Wesley Ivan Hurt, Jul 16 2017

CoefficientList[Series[(1 - x - x^2 + 9*x^3)/(1 - x)^4, {x, 0, 50}], x] (* G. C. Greubel, Jul 16 2017 *)
LinearRecurrence[{4,-6,4,-1},{1,3,5,15},50] (* Harvey P. Dale, Oct 04 2024 *)

a(n)=n*(4*n^2-12*n+14)/3+1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = C(n,0) + 2*C(n,1) + 8*C(n,3).

G.f.: (1-x-x^2+9*x^3)/(1-x)^4. - Colin Barker, Jan 08 2012

A161711 a(n) = (-4n^3 + 27n^2 - 20*n + 3)/3.

Original entry on oeis.org

1, 2, 13, 26, 33, 26, -3, -62, -159, -302, -499, -758, -1087, -1494, -1987, -2574, -3263, -4062, -4979, -6022, -7199, -8518, -9987, -11614, -13407, -15374, -17523, -19862, -22399, -25142, -28099, -31278, -34687, -38334, -42227, -46374, -50783

Offset: 0

Reinhard Zumkeller, Jun 17 2009

{a(k): 0 <= k < 4} = divisors of 26:

a(n) = A027750(A006218(25) + k + 1), 0 <= k < A000005(26).

			Differences of divisors of 26 to compute the coefficients of their interpolating polynomial, see formula:
  1     2    13    26
     1    11    13
       10     2
          -8

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
R. Zumkeller, Enumerations of Divisors
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A005408, A000124, A016813, A086514, A000125, A058331, A002522, A161701, A161702, A161703, A000127, A161704, A161706, A161707, A161708, A161710, A080856, A161712, A161713, A161715, A006261.

[(-4*n^3 + 27*n^2 - 20*n + 3)/3: n in [0..40]]; // Vincenzo Librandi, Jul 17 2011

LinearRecurrence[{4,-6,4,-1},{1,2,13,26},40] (* Harvey P. Dale, Jul 02 2017 *)

x='x+O('x^50); Vec((1-2*x+11*x^2-18*x^3)/(1-x)^4) \\ G. C. Greubel, Jul 16 2017

a(n) = C(n,0) + C(n,1) + 10*C(n,2) - 8*C(n,3).

G.f.: (1-2*x+11*x^2-18*x^3)/(1-x)^4. - Bruno Berselli, Jul 17 2011

A008860 a(n) = Sum_{k=0..7} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 255, 502, 968, 1816, 3302, 5812, 9908, 16384, 26333, 41226, 63004, 94184, 137980, 198440, 280600, 390656, 536155, 726206, 971712, 1285624, 1683218, 2182396, 2804012, 3572224, 4514873, 5663890, 7055732

Offset: 0

N. J. A. Sloane and R. K. Guy

This is a general comment about sequences: A000012, A000027, A000124, A000125, A000127, A006261, A008859, this sequence, A008861, A008862, A008863. Let j in {1, 2, ..., 11} index these 11 sequences respective to their order above. Then a(n) in each sequence is the number of compositions of (n+1) into j or fewer parts. From this we see that the ordinary generating function for each sequence is Sum_{i=0..j-1} x^i/(1-x)^(i+1). - Geoffrey Critzer, Jan 19 2009

a(n) is the maximal number of regions in 7-space formed by n-1 6-dimensional hypercubes. Also the number of binary words of length n matching the regular expression 1*0*1*0*1*0*1*0*. A000124, A000125, A000127, A006261, A008859 count binary words of the form 0*1*0*, 1*0*1*0*, 0*1*0*1*0*, 1*0*1*0*1*0*, and 0*1*0*1*0*1*0* respectively. - Manfred Scheucher, Jun 22 2023

			a(8)=255 because there are 255 compositions of 9 into eight or fewer parts. - _Geoffrey Critzer_, Jan 23 2009

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

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

Cf. A007318, A219531.

List([0..40], n-> Sum([0..7], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019

a008860 = sum . take 8 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012

[&+[Binomial(n, k): k in [0..7]]: n in [0..55]]; // Vincenzo Librandi, May 20 2019

seq(sum(binomial(n,j), j=0..7), n=0..40); # G. C. Greubel, Sep 13 2019

CoefficientList[Series[(1-6x+16x^2-24x^3+22x^4-12x^5+4x^6)/(1-x)^8, {x, 0, 34}], x] (* Georg Fischer, May 19 2019 *)
Sum[Binomial[Range[41]-1, j-1], {j,8}] (* G. C. Greubel, Sep 13 2019 *)

a(n)=(n+1)*(n^6-15*n^5+127*n^4-477*n^3+1576*n^2-1212*n+5040)/5040 \\ Charles R Greathouse IV, Dec 07 2011

[binomial(n,1)+binomial(n,3)+binomial(n,5)+binomial(n,7) for n in range(1, 36)] # Zerinvary Lajos, May 17 2009

[sum(binomial(n,k) for k in (0..7)) for n in (0..40)] # G. C. Greubel, Sep 13 2019

a(n) = Sum_{k=1..4} binomial(n+1, 2k-1) = (n^6 - 14*n^5 + 112*n^4 - 350*n^3 + 1099*n^2 + 364*n + 3828)*n/5040 + 1. [Len Smiley's formula for A006261, copied by Frank Ellermann]

G.f.: (1 - 6*x + 16*x^2 - 24*x^3 + 22*x^4 - 12*x^5 + 4*x^6)/(1-x)^8. - Geoffrey Critzer, Jan 19 2009 [Corrected by Georg Fischer, May 19 2019]

A008863 a(n) = Sum_{k=0..10} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4083, 8100, 15914, 30827, 58651, 109294, 199140, 354522, 616666, 1048576, 1744436, 2842226, 4540386, 7119516, 10970272, 16628809, 24821333, 36519556, 53009102, 75973189, 107594213, 150676186, 208791332

Offset: 0

N. J. A. Sloane, R. K. Guy

a(n) is the number of compositions (ordered partitions) of n+1 into eleven or fewer parts. - Geoffrey Critzer, Jan 24 2009

a(n) is the maximal number of regions in 10-space formed by n-1 9-dimensional hypercubes. Also the number of binary words of length n matching the regular expression 0*1*0*1*0*1*0*1*0*1*0*. A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862 count binary words of the form 0*1*0*, 1*0*1*0*, 0*1*0*1*0*, 1*0*1*0*1*0*, 0*1*0*1*0*1*0*, 1*0*1*0*1*0*1*0*, 0*1*0*1*0*1*0*1*0* and 1*0*1*0*1*0*1*0*1*0* respectively. - Manfred Scheucher, Jun 23 2023

			a(11) = 2047 because there are 2^11=2048 compositions of 12 into any size parts but one of the compositions (1+1+...+1=12) has more than eleven parts. - _Geoffrey Critzer_, Jan 24 2009

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.

T. D. Noe, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165, 55,-11,1).

Cf. A008859, A008860, A008861, A008862, A006261, A000127, A007318, A219531.

List([0..40], n-> Sum([0..10], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019

a008863 = sum . take 11 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012

[(&+[Binomial(n,k): k in [0..10]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019

A008863:=n->add(binomial(n,k), k=0..10): seq(A008863(n), n=0..40); # Wesley Ivan Hurt, Apr 28 2017

Table[Sum[Binomial[n, i], {i, 0, 10}], {n, 0, 40}] (* T. D. Noe, Mar 27 2012 *)
LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,2,4,8, 16,32,64,128,256,512,1024}, 40] (* Harvey P. Dale, Apr 25 2012 *)

a(n)=sum(k=0,10,binomial(n,k)) \\ Charles R Greathouse IV, Apr 07 2016

A008863_list, m = [], [1, -8, 29, -62, 86, -80, 50, -20, 5, 0, 1]
for _ in range(10**2):
    A008863_list.append(m[-1])
    for i in range(10):
        m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

[sum(binomial(n,k) for k in (0..10)) for n in (0..40)] # G. C. Greubel, Sep 13 2019

a(n) = Sum_{k=0..5} binomial(n+1, 2k), compare A008859.
From Geoffrey Critzer, Jan 24 2009: (Start)
G.f.: (1 - 9*x + 37*x^2 - 91*x^3 + 148*x^4 - 166*x^5 + 130*x^6 - 70*x^7 + 25*x^8 - 5*x^9 + x^10)/(1-x)^11.
a(n) = (n^10 - 35*n^9 + 600*n^8 - 5790*n^7 + 36813*n^6 - 140595*n^5 + 408050*n^4 - 382060*n^3 + 1368936*n^2 + 2342880*n + 3628800)/10!. (End)
a(n) = 11*a(n-1) - 55*a(n-2) + 165*a(n-3) - 330*a(n-4) + 462*a(n-5) - 462*a(n-6) + 330*a(n-7) - 165*a(n-8) + 55*a(n-9) - 11*a(n-10) + a(n-11); 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. - Harvey P. Dale, Apr 25 2012

A008861 a(n) = Sum_{k=0..8} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1013, 1981, 3797, 7099, 12911, 22819, 39203, 65536, 106762, 169766, 263950, 401930, 600370, 880970, 1271626, 1807781, 2533987, 3505699, 4791323, 6474541, 8656937, 11460949, 15033173, 19548046

Offset: 0

N. J. A. Sloane, R. K. Guy

a(n) is the number of compositions (ordered partitions) of n+1 into nine or fewer parts. - Geoffrey Critzer, Jan 24 2009

a(n) is the maximal number of regions in 8-space formed by n-1 7-dimensional hypercubes. Also the number of binary words of length n matching the regular expression 0*1*0*1*0*1*0*1*0*. A000124, A000125, A000127, A006261, A008859, A008860 count binary words of the form 0*1*0*, 1*0*1*0*, 0*1*0*1*0*, 1*0*1*0*1*0*, 0*1*0*1*0*1*0*, and 1*0*1*0*1*0*1*0* respectively. - Manfred Scheucher, Jun 22 2023

			a(9)=511 because all but one (namely 1+1+1+...+1=10) of the 2^9 compositions of 10 are in nine or fewer parts. - _Geoffrey Critzer_, Jan 24 2009

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

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

Cf. A007318, A219531.

List([0..40], n-> Sum([0..8], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019

a008861 = sum . take 9 . a007318_row -- Reinhard Zumkeller, Nov 24 2012

[(&+[Binomial(n,k): k in [0..8]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019

seq(sum(binomial(n,j), j=0..8), n=0..40); # G. C. Greubel, Sep 13 2019

Sum[Binomial[Range[41]-1, j-1], {j,9}] (* G. C. Greubel, Sep 13 2019 *)

vector(40, n, sum(j=0,8, binomial(n-1,j))) \\ G. C. Greubel, Sep 13 2019

[sum(binomial(n,k) for k in (0..8)) for n in (0..40)] # G. C. Greubel, Sep 13 2019

a(n) = Sum_{k=0..4} binomial(n+1, 2*k), compare A008859.
From Geoffrey Critzer, Jan 24 2009: (Start)
G.f.: (1 - 7*x + 22*x^2 - 40*x^3 + 46*x^4 - 34*x^5 + 16*x^6 - 4*x^7 + x^8)/(1-x)^9.
a(n) = (n^8 - 20*n^7 + 210*n^6 - 1064*n^5 + 3969*n^4 - 4340*n^3 + 15980*n^2 + 25584*n + 40320)/8!. (End)

A008862 a(n) = Sum_{k=0..9} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2036, 4017, 7814, 14913, 27824, 50643, 89846, 155382, 262144, 431910, 695860, 1097790, 1698160, 2579130, 3850756, 5658537, 8192524, 11698223, 16489546, 22964087, 31621024, 43081973, 58115146, 77663192, 102875128

Offset: 0

N. J. A. Sloane, R. K. Guy

a(n) is the number of compositions (ordered partitions) of n+1 into ten or fewer parts. - Geoffrey Critzer, Jan 24 2009

a(n) is the maximal number of regions in 9-space formed by n-1 8-dimensional hypercubes. Also the number of binary words of length n matching the regular expression 1*0*1*0*1*0*1*0*1*0*. A000124, A000125, A000127, A006261, A008859, A008860, A008861 count binary words of the form 0*1*0*, 1*0*1*0*, 0*1*0*1*0*, 1*0*1*0*1*0*, 0*1*0*1*0*1*0*, 1*0*1*0*1*0*1*0* and 0*1*0*1*0*1*0*1*0* respectively. - Manfred Scheucher, Jun 23 2023

			a(10)=1023 because there are (2^10)-1 compositions of 11 into ten or fewer parts. - _Geoffrey Critzer_, Jan 24 2009

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

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

Cf. A007318, A219531.

List([0..40], n-> Sum([0..9], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019

a008862 = sum . take 10 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012

[(&+[Binomial(n,k): k in [0..9]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019

seq(add(binomial(n,j), j=0..9), n=0..40); # G. C. Greubel, Sep 13 2019

Table[Sum[Binomial[n,k],{k,0,9}],{n,0,40}] (* or *) LinearRecurrence[ {10,-45,120,-210,252,-210,120,-45,10,-1}, {1,2,4,8,16,32,64,128,256, 512}, 40] (* Harvey P. Dale, Mar 18 2012 *)

vector(40, n, sum(j=0,9, binomial(n-1,j))) \\ G. C. Greubel, Sep 13 2019

[sum(binomial(n,k) for k in (0..9)) for n in (0..40)] # G. C. Greubel, Sep 13 2019

a(n) = Sum_{k=1..5} binomial(n+1, 2*k-1), compare A008860.
From Geoffrey Critzer, Jan 24 2009: (Start)
G.f.: (1 - 8*x + 29*x^2 - 62*x^3 + 86*x^4 - 80*x^5 + 50*x^6 - 20*x^7 + 5*x^8)/(1-x)^10.
a(n) = (n^9 - 27*n^8 + 366*n^7 - 2646*n^6 + 12873*n^5 - 31563*n^4 + 79064*n^3 + 34236*n^2 + 270576*n + 362880)/9!. (End)
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); 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. - Harvey P. Dale, Mar 18 2012

A219531 a(n) = Sum_{k=0..11} C(n, k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095, 8178, 16278, 32192, 63019, 121670, 230964, 430104, 784626, 1401292, 2449868, 4194304, 7036530, 11576916, 18696432, 29666704, 46295513, 71116846, 107636402, 160645504, 236618693, 344212906, 494889092

Offset: 0

Mokhtar Mohamed, Nov 21 2012

a(n) is the number of compositions (ordered partitions) of n+1 into twelve or fewer parts. a(n) = sum(binomial(n + 1, 2k - 1), for k = 1 .. 6). a(n) is the sum of the first twelve terms in the n-th row of Pascal's triangle.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

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

Cf. A007318.

List([0..40], n-> Sum([0..11], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019

a219531 = sum . take 12 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012

[(&+[Binomial(n,k): k in [0..11]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019

seq(sum(binomial(n,j), j=0..11), n=0..40); # G. C. Greubel, Sep 13 2019

Table[Sum[Binomial[n, k], {k, 0, 11}], {n, 0, 40}] (* T. D. Noe, Nov 23 2012 *)
LinearRecurrence[{12,-66,220,-495,792,-924,792,-495,220,-66,12,-1},{1,2,4,8,16,32,64,128,256,512,1024,2048},40] (* Harvey P. Dale, Sep 19 2019 *)

vector(40, n, sum(j=0,11, binomial(n-1,j))) \\ G. C. Greubel, Sep 13 2019

A219531_list, m = [], [1, -9, 37, -91, 148, -166, 130, -70, 25, -5, 1, 1]
for _ in range(10**2):
    A219531_list.append(m[-1])
    for i in range(11):
        m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

[sum(binomial(n,k) for k in (0..11)) for n in (0..40)] # G. C. Greubel, Sep 13 2019

a(n) = 1 + (n^11 - 44*n^10 + 935*n^9 - 11550*n^8 + 94083*n^7 - 497112*n^6 +1870385*n^5 -3920950*n^4 +8550916*n^3 +4429656*n^2 +29400480*n)/11!. a(n) = 2*a(n - 1), for 1 <= n <= 11 with a(0) = 1, a(n) = 2*a(n - 1) - C(n - 1, 11), for n > 11. - Mohamed

G.f.: (1 - 10*x + 46*x^2 - 128*x^3 + 239*x^4 - 314*x^5 + 296*x^6 - 200*x^7 + 95*x^8 - 30*x^9 + 6*x^10)/(1-x)^12. - Mokhtar Mohamed, Nov 23 2012

A256816 T(n,k) = Number of length n+k 0..1 arrays with at most two downsteps in every k consecutive neighbor pairs.

Original entry on oeis.org

4, 8, 8, 16, 16, 16, 32, 32, 32, 32, 63, 64, 64, 64, 64, 120, 124, 128, 128, 128, 128, 219, 229, 245, 256, 256, 256, 256, 382, 402, 442, 484, 512, 512, 512, 512, 638, 673, 753, 856, 956, 1024, 1024, 1024, 1024, 1024, 1080, 1220, 1424, 1656, 1888, 2048, 2048, 2048

Offset: 1

R. H. Hardin, Apr 10 2015

Table starts
....4....8...16....32....63...120...219...382....638...1024...1586...2380
....8...16...32....64...124...229...402...673...1080...1670...2500...3638
...16...32...64...128...245...442...753..1220...1894...2836...4118...5824
...32...64..128...256...484...856..1424..2249...3402...4965...7032...9710
...64..128..256...512...956..1656..2693..4158...6153...8792..12202..16524
..128..256..512..1024..1888..3204..5088..7677..11120..15579..21230..28264
..256..512.1024..2048..3728..6192..9613.14168..20075..27566..36888..48304
..512.1024.2048..4096..7362.11955.18104.26117..36218..48738..64024..82440
.1024.2048.4096..8192.14539.23088.34013.47858..65130..86008.110976.140536
.2048.4096.8192.16384.28712.44617.63928.87338.116104.150906.191620.238932

			Some solutions for n=4, k=4
..1....1....0....0....0....0....1....0....0....0....0....0....1....0....0....1
..0....0....1....1....0....1....0....1....1....0....0....0....1....0....0....1
..1....1....0....1....0....0....1....0....1....1....1....1....0....1....0....1
..0....1....1....1....0....1....1....1....1....1....0....0....1....1....0....0
..0....1....0....0....1....1....1....1....0....0....1....1....1....1....0....0
..0....1....1....0....1....1....0....0....0....1....0....0....0....1....0....1
..0....0....1....0....1....1....1....0....0....1....1....1....0....0....0....0
..0....1....0....1....1....1....0....1....0....1....0....1....0....1....0....1

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A000079(n+1).
Column 2 is A000079(n+2).
Column 3 is A000079(n+3).
Column 4 is A000079(n+4).
Row 1 is A006261(n+1).

Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 2*a(n-1)
k=3: a(n) = 2*a(n-1)
k=4: a(n) = 2*a(n-1)
k=5: a(n) = 2*a(n-1) -a(n-2) +2*a(n-3) -a(n-4) +2*a(n-5) -a(n-6)
k=6: a(n) = 2*a(n-1) -a(n-2) +2*a(n-3) -a(n-4) +3*a(n-6) -2*a(n-7) -6*a(n-9) +4*a(n-10)
k=7: [order 15]
Empirical for row n:
n=1: a(n) = (1/120)*n^5 + (1/8)*n^3 + (1/2)*n^2 + (41/30)*n + 2
n=2: a(n) = (1/120)*n^5 + (1/24)*n^4 + (3/8)*n^3 - (1/24)*n^2 + (277/60)*n + 3
n=3: a(n) = (1/120)*n^5 + (1/12)*n^4 + (31/24)*n^3 - (31/12)*n^2 + (66/5)*n + 4
n=4: [polynomial of degree 5] for n>2
n=5: [polynomial of degree 5] for n>3
n=6: [polynomial of degree 5] for n>4
n=7: [polynomial of degree 5] for n>5

A059174 Maximal number of regions into which 5-space can be divided by n hyperspheres.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 126, 240, 438, 764, 1276, 2048, 3172, 4760, 6946, 9888, 13770, 18804, 25232, 33328, 43400, 55792, 70886, 89104, 110910, 136812, 167364, 203168, 244876, 293192, 348874, 412736, 485650, 568548, 662424, 768336, 887408, 1020832, 1169870

Offset: 0

N. J. A. Sloane, Feb 15 2001

n hyperspheres divide R^k into at most binomial(n-1, k) + Sum_{i=0..k} binomial(n, i) regions.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.

Muniru A Asiru, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), this sequence (dim 5).
Fifth row (k=5) of A059250.
Cf. A006261.

Concatenation([1], List([1..40], n-> Binomial(n-1,5) + Sum([0..5], i-> Binomial(n,i)))); # Muniru A Asiru, Dec 18 2018

[1] cat [(n^5-5*n^4+25*n^3+5*n^2+94*n+120)/60: n in [0..40]]; // Vincenzo Librandi, Dec 21 2018

seq(coeff(series((x^6+3*x^4-6*x^3+7*x^2-4*x+1)/(1-x)^6,x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Dec 18 2018

Join[{1}, Table[((n^5 - 5 n^4 + 25 n^3 + 5 n^2 + 94 n + 120) / 60), {n, 0, 50}]] (* Vincenzo Librandi, Dec 21 2018 *)

a(n) = binomial(n-1, 5) + sum(i=0, 5, binomial(n, i)); \\ Michel Marcus, Jan 29 2016

a(n) = binomial(n-1, 5) + Sum_{i=0..5} binomial(n, i).
G.f.: (x^6 + 3*x^4 - 6*x^3 + 7*x^2 - 4*x + 1)/(x - 1)^6. - Colin Barker, Oct 06 2012
a(n) = 2*A006261(n-1), for n > 0. - Günter Rote, Dec 18 2018, by elementary manipulations.
E.g.f.: 1 + (1/60)*(120*x + 20*x^3 + x^5)*exp(x). - Franck Maminirina Ramaharo, Dec 21 2018

cp's OEIS Frontend

A161702 a(n) = (-n^3 + 9n^2 - 5n + 3)/3.

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A161703 a(n) = (4*n^3 - 12*n^2 + 14*n + 3)/3.

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A161711 a(n) = (-4*n^3 + 27*n^2 - 20*n + 3)/3.

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A008860 a(n) = Sum_{k=0..7} binomial(n,k).

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A008863 a(n) = Sum_{k=0..10} binomial(n,k).

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A161703 a(n) = (4n^3 - 12n^2 + 14*n + 3)/3.

A161711 a(n) = (-4n^3 + 27n^2 - 20*n + 3)/3.