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%I #15 Dec 25 2020 10:49:53
%S 1,2,4,6,10,20,26,42,52,110,130,260,286,442,484,798,850,1196,1306,
%T 3190,3320,4030,4290,9620,9906,11726,12168,19006,19490,22748,23546,
%U 42294,43144,50150,51346,72956,74262,87502,90692,226490,229810,242360,246390,318370,322660,356070,365690,856180,866086
%N a(1) = 1; a(2*n) = prime(n)*a(n), a(2*n+1) = prime(n)*a(n) + a(n+1), where prime(n) is the n-th prime.
%C Conjecture: for n > 1, 4 divides a(n) iff n == 0 (mod 3), otherwise A007814(a(n)) = 1.
%C Conjecture is true (proof by mathematical induction). - _Robert Israel_, Feb 25 2020
%C Given the generating triangle of A176528, replace k's with prime(k); then take powers of the triangle resulting in a left-shifted vector considered as a sequence.
%C First few rows of the generating triangle M for this sequence:
%C 1;
%C 2;
%C 2, 1;
%C 0, 3;
%C 0, 3, 1;
%C 0, 0, 5;
%C 0, 0, 5, 1;
%C 0, 0, 0, 7;
%C 0, 0, 0, 7, 1;
%C 0, 0, 0, 0, 11;
%C 0, 0, 0, 0, 11, 1;
%C 0, 0, 0, 0, 0, 13;
%C 0, 0, 0, 0, 0, 13, 1;
%C 0, 0, 0, 0, 0, 0, 17;
%C ...
%C Then sequence is Lim_{n->inf} M^n, the left-shifted vector considered as a sequence.
%H Robert Israel, <a href="/A176716/b176716.txt">Table of n, a(n) for n = 1..10000</a>
%e a(10) = 110 = prime(5) * a(5) = 11 * 10.
%e a(7) = 26 = prime(3) * 4 + 6 = 5 * 4 + 6.
%e 4 divides a(9) = 52 since 9 == 0 mod 3.
%e 2 divides a(13) = 286 but not 4 since 13 == 1 mod 3.
%o (PARI) a(n) = if(n==1,1, if (n % 2, a(n\2+1)+a(n\2)*prime(n\2), a(n/2)*prime(n/2))); \\ _Michel Marcus_, Dec 25 2020
%Y Cf. A007814, A176528, A000040.
%K nonn
%O 1,2
%A _Gary W. Adamson_, Apr 24 2010
%E Edited and more terms from _Robert Israel_, Feb 25 2020
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