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A090543 a(1) = 3, then a(n) = least prime, not already present, beginning with the digit reversal of a(n-1). 2

%I #5 Dec 05 2013 19:56:38

%S 3,31,13,311,113,3119,91139,931193,39113903,30931193,391139039,

%T 930931193,3911390393,393093119329,9239113903933,3393093119329037,

%U 7309239113903933023,320339309311932903749,94730923911390393302341

%N a(1) = 3, then a(n) = least prime, not already present, beginning with the digit reversal of a(n-1).

%e a(3) = 13 because a(2) = 31 and the digit reversal of 31 is 13 which is prime.

%e a(4) = 311 because a(3) = 13, digit reversal of 13 is 31, which is already in the list. Smallest prime that starts with 31 is 311.

%p reverse := proc (nn) local n,m; m := 0; n := nn; while (n > 0) do m := m*10 + irem(n,10,'n'); od; m; end:

%p a := proc(n,m) option remember; global currSet; local currN, i, origN, j; if n = 0 then currSet := {m}; return m; end if; currN := reverse(a(n - 1,m)); if (not (evalb(currN in currSet))) then if (isprime(currN)) then currSet := currSet union {currN}; return currN; end if; end if; origN := currN; j := 1; while (true) do origN := 10 * origN; currN := origN; i := 0; while i < (10^j) do if (isprime(currN) and (not evalb(currN in currSet))) then currSet := currSet union {currN}; return currN; end if; currN := currN + 1; i := i + 1; end do; j := j + 1; end do; return currN; end proc; (Delgau)

%Y Cf. A111463, A111462.

%K base,nonn

%O 1,1

%A _Amarnath Murthy_, Dec 09 2003; corrected Aug 04 2005

%E More terms from Chris Deugau (deugaucj(AT)uvic.ca), Nov 07 2005

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Last modified March 28 17:42 EDT 2024. Contains 371254 sequences. (Running on oeis4.)