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a(n) = Sum_{k=0..n} C(n,k)*3^[k*(k-1)/2].
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1, 2, 6, 40, 860, 63000, 14714728, 10562062112, 22960880409360, 150300904214651680, 2955814683617734854752, 174481716707875308905153664, 30905247968182392588500030233024
MATHEMATICA
Table[Sum[Binomial[n, k]*3^(Binomial[k, 2]), {k, 0, n}], {n, 0, 10}] (* G. C. Greubel, Nov 07 2016 *)
PROG
(PARI) {a(n)=sum(k=0, n, binomial(n, k)*3^(k*(k-1)/2))}
G.f.: sqrt( Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k) * 2^(k*(k-1)) ).
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1, 1, 3, 37, 2149, 532611, 539508291, 2202251249193, 36044200375109487, 2361471528989758715269, 618991271516919971774301613, 649043297118583276751832395970903, 2722266074808493870871954767765560237289, 45671958833739479081570180837023756023304348531, 3064991675467024774224369897734145197065069681513495767
COMMENTS
a(k) = 1 (mod 3) iff k = 9* A005836(n) + [0,1,3,4] for n>=0, with a(k) = 0 (mod 3) elsewhere, where A005836 lists numbers n whose base 3 representation contains no 2 (conjecture).
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 37*x^3 + 2149*x^4 + 532611*x^5 + 539508291*x^6 +...
where
A(x)^2 = 1 + 2*x + 7*x^2 + 80*x^3 + 4381*x^4 + 1069742*x^5 + 1080096067*x^6 +...+ A135756(n)*x^n +... such that
A135756(n) = Sum_{k=0..n} binomial(n,k) * 2^(k*(k-1)).
The residue of the terms modulo 3 begin:
[1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, ...]
where a(k) appears to be congruent to 1 (mod 3) at k = 9* A005836(n) + [0,1,3,4] for n>=0, and congruent to zeros elsewhere.
PROG
(PARI) {a(n) = polcoeff( sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)*2^(k*(k-1))) +x*O(x^n))^(1/2), n)}
for(n=0, 20, print1(a(n), ", "))
G.f. A(x) satisfies: A(x) = 1 + x * A(4*x/(1 + 3*x)) / (1 - x).
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1, 1, 5, 73, 4301, 1065361, 1079026325, 4404504773593, 72088402948928861, 4722943066827454121761, 1237982543178169058402322725, 1298086594246614900499652230482793, 5444532149619463867564918804810528611821, 91343917667481554378430257939829428893551284401
FORMULA
G.f.: Sum_{k>=0} 2^(k*(k-1)) * (x/(1 - x))^k.
a(n) = Sum_{k=0..n} binomial(n-1,k-1) * 2^(k*(k-1)).
MATHEMATICA
nmax = 13; A[_] = 0; Do[A[x_] = 1 + x A[4 x/(1 + 3 x)]/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n - 1, k - 1] 2^(k (k - 1)), {k, 0, n}], {n, 0, 13}]
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