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A119892
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Prime quartet leaders: largest number of a prime quartet.
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5
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2999, 3989, 4799, 4889, 5879, 5897, 5987, 6599, 6689, 6779, 6869, 6959, 6977, 7499, 7589, 7877, 7949, 8597, 8669, 8849, 8867, 9479, 9497, 9587, 9677, 9749, 9767, 9839, 9857, 9929, 12899, 13799, 13997, 14699, 14879, 14897, 14969, 15797, 15887, 15959
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OFFSET
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1,1
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COMMENTS
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A prime quartet is a set of four different prime numbers such that the fourth number is a 1-digit number which is the sum of the digits of the third number, the third number is the sum of the digits of the second number and the second number is the sum of the digits of the first number.
Comment from Joshua Zucker, Apr 24 2007, on the difference between this sequence and A106766: The digit sum must be the largest member of a prime trio, so the first number where the sequences differ must be with digit sum 47 and thus have at least 6 digits - so until then you get all the primes with 4 or 5 digits that have digit sum 29.
The sequence of prime quintet leaders is probably too large for the OEIS; its first term is the 334-digit prime 5*10^333-10^330-10^328-1 with sum of digits a(1) = 2999. - Charles R Greathouse IV, Mar 11 2022
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LINKS
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EXAMPLE
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2999 is in the sequence because it is the largest number of the prime quartet (2999,29,11,2).
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MATHEMATICA
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pqQ[n_]:=Module[{p1=NestList[Total[IntegerDigits[#]]&, n, 3]}, AllTrue[ Take[ p1, 3], #>9&]&&AllTrue[p1, PrimeQ]]; Select[Range[16000], pqQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 02 2020 *)
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PROG
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(PARI) DigitSum(n, b=10)=local(x); x=0; while(n, x+=n%b; n\=b); x
PrimeEnsemble(n, b=10)=local(x); x=1; while(ispseudoprime(n), if(n<b, return(x)); n=DigitSum(n, b); x++); 0
forprime(p=2, 16000, if(PrimeEnsemble(p)>=4, print1(p", "))); \\ Martin Fuller
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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Luc Stevens (lms022(AT)yahoo.com), May 27 2006
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STATUS
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approved
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