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A006261
a(n) = Sum_{k=0..5} binomial(n,k).
(Formerly M1126)
39
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
0,2
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
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).
LINKS
M. L. Cornelius, Variations on a geometric progression, Mathematics in School, 4 (No. 3, May 1975), p. 32.
M. L. Cornelius, Variations on a geometric progression, Mathematics in School, 4 (No. 3, May 1975), p. 32. (Annotated scanned copy)
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Reinhard Zumkeller, Enumerations of Divisors
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
EXAMPLE
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
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}]
PROG
(Sage) [binomial(n, 1)+binomial(n, 3)+binomial(n, 5) for n in range(1, 38)] # Zerinvary Lajos, May 17 2009
(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
(Haskell)
a006261 = sum . take 6 . a007318_row -- Reinhard Zumkeller, Nov 24 2012
(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
(PARI) a(n)=sum(k=0, 5, binomial(n, k)) \\ Charles R Greathouse IV, Apr 08 2016
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, based on a suggestion from S. C. Chan, Jun 10 1975
STATUS
approved