A304780 - OEIS

%I #47 Sep 07 2018 03:48:28

%S 1,3,18,513,395523,234658258578,82596747478641253260993,

%T 10233334041075645341729789249315281196742910563,

%U 157081688394356396673208173772909833928515988895188885472258972148661958252271815996039831298

%N Consider a triangle whose first row is {1,2} and, for n > 1, has as its n-th row the integers k through 2k where k is the sum of the numbers in the (n-1)th row. Then a(n) is the first number in the n-th row.

%F a(1) = 1, a(n) = (3*a(n-1)*(a(n-1) + 1))/2 for n > 1.

%F a(n) ~ (2/3) * c^(2^n), where c = 1.515006464529590220430714781603262955960312205695360166833... - _Vaclav Kotesovec_, Jul 23 2018

%e Triangle begins:

%e   1, 2; (row sum = 3)

%e   3, 4, 5, 6; (row sum = 18)

%e   18, 19, 20, 21, ... 33, 34, 35, 36; (row sum = 513)

%e   513, 514, 515, 516, ..., 1023, 1024, 1025, 1026;

%e   ...

%t RecurrenceTable[{a[1] == 1, a[n] == (3*a[n-1]*(a[n-1] + 1))/2}, a, {n, 1, 10}] (* _Vaclav Kotesovec_, Jul 23 2018 *)

%t Nest[Append[#, Range[#, 2 #] &@ Total@ Last@ #] &, {{1, 2}}, 3] // Flatten (* _Michael De Vlieger_, Jul 26 2018 *)

%o (PARI) a(n) = if (n==1, 1, (3*a(n - 1)*(a(n - 1) + 1))/2); \\ _Michel Marcus_, May 24 2018

%Y Cf. A000217.

%K nonn

%O 1,2

%A _Nathaniel J. Strout_, May 18 2018