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 1-10 of 10 results.

A000127 Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes.

Original entry on oeis.org

1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196, 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457, 41449, 46938, 52956, 59536, 66712, 74519, 82993, 92171, 102091, 112792, 124314, 136698
Offset: 1

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Author

Keywords

Comments

a(n) is the sum of the first five terms in the n-th row of Pascal's triangle. - Geoffrey Critzer, Jan 18 2009
{a(k): 1 <= k <= 5} = divisors of 16. - Reinhard Zumkeller, Jun 17 2009
Equals binomial transform of [1, 1, 1, 1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 02 2010
From Bernard Schott, Apr 05 2021: (Start)
As a(n) = 2^(n-1) for n = 1..5, it is misleading to believe that a(n) = 2^(n-1) for n > 5 (see Patrick Popescu-Pampu link); other curiosities: a(6) = 2^5 - 1 and a(10) = 2^8.
The sequence of the first differences is A000125, the sequence of the second differences is A000124, the sequence of the third differences is A000027 and the sequence of the fourth differences is the all 1's sequence A000012 (see J. H. Conway and R. K. Guy reference, p. 80). (End)
a(n) is the number of binary words of length n matching the regular expression 0*1*0*1*0*. A000124 and A000125 count binary words of the form 0*1*0* and 1*0*1*0*, respectively. - Manfred Scheucher, Jun 22 2023

Examples

			a(7)=99 because the first five terms in the 7th row of Pascal's triangle are 1 + 7 + 21 + 35 + 35 = 99. - _Geoffrey Critzer_, Jan 18 2009
G.f. = x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 31*x^6 + 57*x^7 + 99*x^8 + 163*x^9 + ...
		

References

  • R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 28.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, Chap. 3.
  • J. H. Conway and R. K. Guy, Le Livre des Nombres, Eyrolles, 1998, p. 80.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 33 pp. 18; 128 Ellipses Paris 2004.
  • A. Deledicq and D. Missenard, A La Recherche des Régions Perdues, Math. & Malices, No. 22 Summer 1995 issue pp. 22-3 ACL-Editions Paris.
  • M. Gardner, Mathematical Circus, pp. 177; 180-1 Alfred A. Knopf NY 1979.
  • M. Gardner, The Colossal Book of Mathematics, 2001, p. 561.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • M. de Guzman, Aventures Mathématiques, Prob. B pp. 115-120 PPUR Lausanne 1990.
  • Ross Honsberger; Mathematical Gems I, Chap. 9.
  • Ross Honsberger; Mathematical Morsels, Chap. 3.
  • Jeux Mathématiques et Logiques, Vol. 3 pp. 12; 51 Prob. 14 FFJM-SERMAP Paris 1988.
  • J. N. Kapur, Reflections of a Mathematician, Chap.36, pp. 337-343, Arya Book Depot, New Delhi 1996.
  • C. D. Miller, V. E. Heeren, J. Hornsby, M. L. Morrow and J. Van Newenhizen, Mathematical Ideas, Tenth Edition, Pearson, Addison-Wesley, Boston, 2003, Cptr 1, 'The Art of Problem Solving, page 6.
  • I. Niven, Mathematics of Choice, pp. 158; 195 Prob. 40 NML 15 MAA 1965.
  • C. S. Ogilvy, Tomorrow's Math, pp. 144-6 OUP 1972.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 252-255.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, NY, 2007, page 81-87.
  • A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a000127 = sum . take 5 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(n^4-6*n^3+23*n^2-18*n+24)/24: n in [1..50]]; // Vincenzo Librandi, Feb 16 2015
    
  • Maple
    A000127 := n->(n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24;
    with (combstruct):ZL:=[S, {S=Sequence(U, card=1)}, unlabeled]: seq(count(subs(r=6, ZL), size=m), m=0..41); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    f[n_] := Sum[Binomial[n, i], {i, 0, 4}]; Table[f@n, {n, 0, 40}] (* Robert G. Wilson v, Jun 29 2007 *)
    Total/@Table[Binomial[n-1,k],{n,50},{k,0,4}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,2,4,8,16},50] (* Harvey P. Dale, Aug 24 2011 *)
    Table[(n^4 - 6 n^3 + 23 n^2 - 18 n + 24) / 24, {n, 100}] (* Vincenzo Librandi, Feb 16 2015 *)
    a[ n_] := Binomial[n, 4] + Binomial[n, 2] + 1; (* Michael Somos, Dec 23 2017 *)
  • PARI
    a(n)=(n^4-6*n^3+23*n^2-18*n+24)/24 \\ Charles R Greathouse IV, Mar 22 2016
    
  • PARI
    {a(n) = binomial(n, 4) + binomial(n, 2) + 1}; /* Michael Somos, Dec 23 2017 */
    
  • Python
    def A000127(n): return n*(n*(n*(n - 6) + 23) - 18)//24 + 1 # Chai Wah Wu, Sep 18 2021

Formula

a(n) = C(n-1, 4) + C(n-1, 3) + ... + C(n-1, 0) = A055795(n) + 1 = C(n, 4) + C(n-1, 2) + n.
a(n) = Sum_{k=0..2} C(n, 2k). - Joel Sanderi (sanderi(AT)itstud.chalmers.se), Sep 08 2004
a(n) = (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24.
G.f.: (1 - 3*x + 4*x^2 - 2*x^3 + x^4)/(1-x)^5. (for offset 0) - Simon Plouffe in his 1992 dissertation
E.g.f.: (1 + x + x^2/2 + x^3/6 + x^4/24)*exp(x) (for offset 0). [Typos corrected by Juan M. Marquez, Jan 24 2011]
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Harvey P. Dale, Aug 24 2011
a(n) = A000124(A000217(n-1)) - n*A000217(n-2) - A034827(n), n > 1. - Melvin Peralta, Feb 15 2016
a(n) = A223718(-n). - Michael Somos, Dec 23 2017
For n > 2, a(n) = n + 1 + sum_{i=2..(n-2)}sum_{j=1..(n-i)}(1+(i-1)(j-1)). - Alec Jones, Nov 17 2019

Extensions

Formula corrected and additional references from torsten.sillke(AT)lhsystems.com
Additional correction from Jonas Paulson (jonasso(AT)sdf.lonestar.org), Oct 30 2003

A006261 a(n) = Sum_{k=0..5} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 63, 120, 219, 382, 638, 1024, 1586, 2380, 3473, 4944, 6885, 9402, 12616, 16664, 21700, 27896, 35443, 44552, 55455, 68406, 83682, 101584, 122438, 146596, 174437, 206368, 242825, 284274, 331212, 384168, 443704, 510416, 584935, 667928, 760099, 862190
Offset: 0

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Author

N. J. A. Sloane, based on a suggestion from S. C. Chan, Jun 10 1975

Keywords

Comments

a(n) is the sum of the first six terms of the n-th row in Pascal's triangle. - Geoffrey Critzer, Jan 19 2009
Also the interpolating polynomial for the divisors of 32: {a(k): 0 <= k < 6} = {1,2,4,8,16,32}. - Reinhard Zumkeller, Jun 17 2009
a(n) is the maximal number of regions in 5-space formed by n-1 4-dimensional hypercubes. - Carl Schildkraut, May 26 2015
a(n) is the number of binary words of length n matching the regular expression 1*0*1*0*1*0*. A000124, A000125, A000127 count binary words of the form 0*1*0*, 1*0*1*0*, and 0*1*0*1*0*, respectively. - Manfred Scheucher, Jun 22 2023

Examples

			a(7) = 120 because the first six terms in the 7th row of Pascal's triangle 1 + 7 + 21 + 35 + 35 + 21 = 120. - _Geoffrey Critzer_, Jan 19 2009
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a006261 = sum . take 6 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(n^5 - 5*n^4 + 25*n^3 + 5*n^2 + 94*n + 120)/120: n in [0..40]]; // Vincenzo Librandi, Jul 17 2011
    
  • Maple
    A006261:=(z**2-z+1)*(3*z**2-3*z+1)/(z-1)**6; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    CoefficientList[
      Series[(1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120) Exp[x], {x, 0,
        52}], x]*Table[n!, {n, 0, 52}]
  • PARI
    a(n)=sum(k=0,5,binomial(n,k)) \\ Charles R Greathouse IV, Apr 08 2016
  • Python
    A006261_list, m = [], [1, -3, 4, -2, 1, 1]
    for _ in range(10**2):
        A006261_list.append(m[-1])
        for i in range(5):
            m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016
    
  • Sage
    [binomial(n,1)+binomial(n,3)+binomial(n,5) for n in range(1, 38)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = A057703(n) + 1.
a(n) = binomial(n+1, 5) + binomial(n+1, 3) + binomial(n+1, 1). - Len Smiley, Oct 20 2001
G.f.: (1 - 4*x + 7*x^2 - 6*x^3 + 3*x^4)/(1-x)^6. - Geoffrey Critzer, Jan 19 2009
E.g.f.: (1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120)*exp(x).
a(n) = (n^5 - 5*n^4 + 25*n^3 + 5*n^2 + 94*n + 120)/120. - Reinhard Zumkeller, Jun 17 2009
a(n) = a(n-1) + A000127(n-1). - Christian Schroeder, Jan 04 2016

A008859 a(n) = Sum_{k=0..6} binomial(n,k).

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 127, 247, 466, 848, 1486, 2510, 4096, 6476, 9949, 14893, 21778, 31180, 43796, 60460, 82160, 110056, 145499, 190051, 245506, 313912, 397594, 499178, 621616, 768212, 942649, 1149017, 1391842, 1676116, 2007328
Offset: 0

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Author

Keywords

Comments

a(n) is the maximal number of regions in 6-space formed by n-1 5-dimensional hypercubes. - Christian Schroeder, Jan 04 2016
a(n) is the number of binary words of length n matching the regular expression 0*1*0*1*0*1*0*. A000124, A000125, A000127, A006261 count binary words of the form 0*1*0*, 1*0*1*0*, 0*1*0*1*0*, and 1*0*1*0*1*0*, respectively. - Manfred Scheucher, Jun 22 2023

References

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

Crossrefs

Programs

  • GAP
    List([0..40], n-> Sum([0..6], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019
  • Haskell
    a008859 = sum . take 7 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(&+[Binomial(n,k): k in [0..6]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019
    
  • Maple
    A008859 := proc(n)
        add(binomial(n,k),k=0..6) ;
    end proc: # R. J. Mathar, Oct 30 2015
  • Mathematica
    Table[Sum[Binomial[n,k],{k,0,6}],{n,0,40}] (* Harvey P. Dale, Jan 16 2012 *)
  • PARI
    a(n)=sum(k=0,6,binomial(n,k)) \\ Charles R Greathouse IV, Sep 24 2015
    
  • Sage
    [sum(binomial(n,k) for k in (0..6)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
    

Formula

a(n) = Sum_{k=0..3} binomial(n+1, 2*k). - Len Smiley, Oct 20 2001
O.g.f.: (1 - 5*x + 11*x^2 - 13*x^3 + 9*x^4 - 3*x^5 + x^6)/(1-x)^7. - R. J. Mathar, Apr 02 2008
a(n) = a(n-1) + A006261(n-1). - Christian Schroeder, Jan 04 2016
a(n) = (n^6 - 9*n^5 + 55*n^4 - 75*n^3 + 304*n^2 + 444*n + 720)/720. - Gerry Martens , May 04 2016
E.g.f.: (720 + 720*x + 360*x^2 + 120*x^3 + 30*x^4 + 6*x^5 + x^6)*exp(x)/6!. - Ilya Gutkovskiy, May 04 2016

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

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Author

Keywords

Comments

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

Examples

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

References

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

Crossrefs

Programs

  • GAP
    List([0..40], n-> Sum([0..7], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019
  • Haskell
    a008860 = sum . take 8 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [&+[Binomial(n, k): k in [0..7]]: n in [0..55]]; // Vincenzo Librandi, May 20 2019
    
  • Maple
    seq(sum(binomial(n,j), j=0..7), n=0..40); # G. C. Greubel, Sep 13 2019
  • Mathematica
    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 *)
  • PARI
    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
    
  • Sage
    [binomial(n,1)+binomial(n,3)+binomial(n,5)+binomial(n,7) for n in range(1, 36)] # Zerinvary Lajos, May 17 2009
    
  • Sage
    [sum(binomial(n,k) for k in (0..7)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
    

Formula

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

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Keywords

Comments

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

Examples

			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
		

References

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

Crossrefs

Programs

  • GAP
    List([0..40], n-> Sum([0..10], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019
  • Haskell
    a008863 = sum . take 11 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(&+[Binomial(n,k): k in [0..10]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019
    
  • Maple
    A008863:=n->add(binomial(n,k), k=0..10): seq(A008863(n), n=0..40); # Wesley Ivan Hurt, Apr 28 2017
  • Mathematica
    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 *)
  • PARI
    a(n)=sum(k=0,10,binomial(n,k)) \\ Charles R Greathouse IV, Apr 07 2016
    
  • Python
    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
    
  • Sage
    [sum(binomial(n,k) for k in (0..10)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
    

Formula

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

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Author

Keywords

Comments

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

Examples

			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
		

References

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

Crossrefs

Programs

  • GAP
    List([0..40], n-> Sum([0..8], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019
  • Haskell
    a008861 = sum . take 9 . a007318_row -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(&+[Binomial(n,k): k in [0..8]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019
    
  • Maple
    seq(sum(binomial(n,j), j=0..8), n=0..40); # G. C. Greubel, Sep 13 2019
  • Mathematica
    Sum[Binomial[Range[41]-1, j-1], {j,9}] (* G. C. Greubel, Sep 13 2019 *)
  • PARI
    vector(40, n, sum(j=0,8, binomial(n-1,j))) \\ G. C. Greubel, Sep 13 2019
    
  • Sage
    [sum(binomial(n,k) for k in (0..8)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
    

Formula

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

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Keywords

Comments

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

Examples

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

References

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

Crossrefs

Programs

  • GAP
    List([0..40], n-> Sum([0..9], k-> Binomial(n,k)) ); # G. C. Greubel, Sep 13 2019
  • Haskell
    a008862 = sum . take 10 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(&+[Binomial(n,k): k in [0..9]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019
    
  • Maple
    seq(add(binomial(n,j), j=0..9), n=0..40); # G. C. Greubel, Sep 13 2019
  • Mathematica
    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 *)
  • PARI
    vector(40, n, sum(j=0,9, binomial(n-1,j))) \\ G. C. Greubel, Sep 13 2019
    
  • Sage
    [sum(binomial(n,k) for k in (0..9)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
    

Formula

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

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

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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

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.
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