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

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

Author

Mokhtar Mohamed, Nov 21 2012

Keywords

• nonn
• easy

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

• 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).

Crossrefs

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

Cf. A007318.

Programs

GAP

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

Haskell

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

Magma

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

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 *)

PARI

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

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

Sage

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

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 [phone]*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