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A008860
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a(n) = Sum_{k=0..7} binomial(n,k).
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14
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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
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OFFSET
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0,2
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COMMENTS
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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
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REFERENCES
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
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LINKS
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FORMULA
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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]
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EXAMPLE
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a(8)=255 because there are 255 compositions of 9 into eight or fewer parts. - Geoffrey Critzer, Jan 23 2009
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MAPLE
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seq(sum(binomial(n, j), j=0..7), n=0..40); # G. C. Greubel, Sep 13 2019
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MATHEMATICA
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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 *)
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PROG
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(Sage) [binomial(n, 1)+binomial(n, 3)+binomial(n, 5)+binomial(n, 7) for n in range(1, 36)] # Zerinvary Lajos, May 17 2009
(Haskell)
(Magma) [&+[Binomial(n, k): k in [0..7]]: n in [0..55]]; // Vincenzo Librandi, May 20 2019
(Sage) [sum(binomial(n, k) for k in (0..7)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
(GAP) List([0..40], n-> Sum([0..7], k-> Binomial(n, k)) ); # G. C. Greubel, Sep 13 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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