login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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

%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