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a(1) = 1, a(n) = (k+1)*(2k+1), where k = Product_{i=1..n-1} a(i).
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%I #31 May 06 2023 06:17:59

%S 1,6,91,597871,213122969971321411,

%T 9680343693975641657052402556458789711774336036960631

%N a(1) = 1, a(n) = (k+1)*(2k+1), where k = Product_{i=1..n-1} a(i).

%C A sequence of pairwise relatively prime triangular (and also hexagonal) numbers.

%C As a clarification to the problem definition by Sierpinski, here we show that only one triangular (hexagonal) seed is needed to produce such a sequence.

%C This sequence can be used for proving the infinitude of primes.

%C In general: Let m = 2q, for any q > 0. There are infinitely many sequences of pairwise coprime m-gonal numbers, whose first term is any positive m-gonal number and whose general term is of the form a(n) = (k + 1)*((q - 1)*k + 1)), where k = Product_{i=1..n-1} a(i).

%D W. Sierpinski, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #42.

%F a(1) = 1, a(n) = (k+1)*(2k+1), where k = Product_{i=1..n-1} a(i).

%F a(n) ~ c^(3^n), where c = 1.1784502032269064445225839284451956694752084180050932315805089054871825498... - _Vaclav Kotesovec_, May 05 2023

%t a[1]=1; a[n_]:=Module[{k=Product[a[i],{i,1,n-1}]},(k+1)*(2*k+1)];

%t a/@Range[6]

%t Join[{1}, RecurrenceTable[{a[2] == 6, a[n+1] == (1 + a[n]*(Sqrt[1 + 8*a[n]] - 3)/4) * (1 + 2*a[n]*(Sqrt[1 + 8*a[n]] - 3)/4)}, a, {n, 2, 8}]] (* _Vaclav Kotesovec_, May 05 2023 *)

%Y Cf. A000217, A000384, A004019, A034792, A115590.

%K nonn

%O 1,2

%A _Ivan N. Ianakiev_, Feb 22 2023