OFFSET
1,1
COMMENTS
It may be possible to generate prime chains of any arbitrary length using minor variations of the procedure below.
This sequence consists of 147 primes, of which 74 are distinct and 73 are duplicates of earlier terms.
LINKS
Jinyuan Wang, Table of n, a(n) for n = 1..147
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).
FORMULA
a(4n+1) = 4*n^2 - 146*n + 1373,
a(4n+2) = 4*n^2 - 144*n + 1459,
a(4n+3) = 4*n^2 - 142*n + 1301,
a(4n+4) = 4*n^2 - 140*n + 1877.
PROG
(Pascal) { This procedure can probably be imported into Borland's latest programming software and run without any changes } procedure Ndegrees3; var a : array[0..16] of extended; ct: longint; n, nh, i, j : integer; ab1, ab2 : extended; begin for i := 0 to 16 do a[i] := 0; N := 5; a[0] := 1373{ FIRST TERM OF PRIME CHAIN}; writeln('1'); writeln(trunc(a[0])); writeln; nh := 1; a[1] := 1459 ; a[2] := 1301 ; a[3] := 1877 ; a[4] := 1231 ; a[5] := 1319 ; repeat for i := N downto nh do begin a[i] := a[i] - a[i-1] ; IF NH = 3 THEN A[I] := ABS(A[I]); {******} End; nh := nh + 1; until nh = n + 2; ct := 0; repeat ct := ct + 1; ab1 := a[n] + a[n-1]; for i := N-1 downto 1 do begin ab2 := a[i] + a[i-1] ; a[i] := ab1; ab1 := ab2; end; IF ODD(ct + 1) THEN A[5] := -A[5]; {******} A[3] := -A[3]; {******} a[0] := ab1; writeln(ct + 1); writeln(trunc(a[0])); {} readln; until 1<0; END;
CROSSREFS
KEYWORD
nonn,fini,full,uned
AUTHOR
Aldrich Stevens (aldrichstevens(AT)msn.com), Jun 04 2008
EXTENSIONS
Edited by Charles R Greathouse IV, Nov 03 2009
More terms from Jinyuan Wang, Jun 20 2021
STATUS
approved