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A219531
a(n) = Sum_{k=0..11} C(n, k).
11
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
0,2
COMMENTS
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.
LINKS
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).
FORMULA
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
MAPLE
seq(sum(binomial(n, j), j=0..11), n=0..40); # G. C. Greubel, Sep 13 2019
MATHEMATICA
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 *)
PROG
(Haskell)
a219531 = sum . take 12 . a007318_row -- Reinhard Zumkeller, Nov 24 2012
(Python)
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
(PARI) vector(40, n, sum(j=0, 11, binomial(n-1, j))) \\ G. C. Greubel, Sep 13 2019
(Magma) [(&+[Binomial(n, k): k in [0..11]]): n in [0..40]]; // G. C. Greubel, Sep 13 2019
(Sage) [sum(binomial(n, k) for k in (0..11)) for n in (0..40)] # G. C. Greubel, Sep 13 2019
(GAP) List([0..40], n-> Sum([0..11], k-> Binomial(n, k)) ); # G. C. Greubel, Sep 13 2019
KEYWORD
nonn,easy
AUTHOR
Mokhtar Mohamed, Nov 21 2012
STATUS
approved