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Sum of the first n nonzero tribonacci numbers, in ascending order, as bases, with the same, in descending order, as exponents.
18

%I #20 Jul 25 2024 08:40:06

%S 1,2,4,8,17,54,472,27216,84738887,299164114847940,

%T 311903053042108587337426568,

%U 5846720173185251353387753850814872871131756204168

%N Sum of the first n nonzero tribonacci numbers, in ascending order, as bases, with the same, in descending order, as exponents.

%F a(n) = Sum_{i=1..n} A000073(i+1)^A000073(n-i+2).

%e For the tribonacci sequence starting t(1)=t(2)=1, t(3)=2, that is, the nonzero terms of A000073:

%e a(1) = t(1)^t(1) = 1^1 = 1.

%e a(2) = t(1)^t(2) + t(2)^t(1) = 1^1 + 1^1 = 2.

%e a(3) = t(1)^t(3) + t(2)^t(2) + t(3)^t(1) = 1^2 + 1^1 + 2^1 = 4.

%e a(4) = t(1)^t(4) + t(2)^t(3) + t(3)^t(2) + t(4)^t(1) = 1^4 + 1^2 + 2^1 + 4^1 = 8.

%e a(5) = 1^7 + 1^4 + 2^2 + 4^1 + 7^1 = 17.

%e a(6) = 1^13 + 1^7 + 2^4 + 4^2 + 7^1 + 13^1 = 54.

%t a[0] = a[1] = 0 ; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Table[Sum[a[k + 2]^(a[n - k + 1]), {k, 1, n}], {n, 1, 10}] (* _G. C. Greubel_, May 18 2017 *)

%Y Cf. A000073.

%K easy,nonn

%O 1,2

%A _Jonathan Vos Post_, Jan 04 2006

%E Name clarified by _Arthur O'Dwyer_, Jul 24 2024