A159922 Least index m such that the five numbers 2*prime(m+k) + 3^n, k=0 to 4, are five consecutive primes.
643266, 8813528, 1644953, 440421, 2826655, 1339785, 2775232, 988180, 196973, 643136, 4122122, 3477939, 182124, 6195602, 130854, 4937610, 2725523, 6118932, 231670, 478208, 2405748, 3913626, 1033788, 2945487, 22952758, 7168835, 15528738, 2753214, 2407038, 37795639
Offset: 1
Keywords
Examples
For n=15, prime(m=130854) = 1739401 starts the prime sequence 1739401, 1739411, 1739417, 1739443, 1739447 of five consecutive primes. With 3^n = 3^15 = 14348907, the five numbers 17827709 = 2*1739401+14348907, 17827729 = 2*1739411 + 14348907, 17827741 = 2*1739417 + 14348907, 17827793 = 2*1739443 + 14348907, 17827801 = 2*1739447 + 14348907 are consecutive primes, and m = 130854 is the smallest prime index of this kind, so a(n=15) = 130854.
Programs
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PARI
a(n) = {my(m=1, p=[2, 3, 5, 7, 11], q, x=3^n); while(ispseudoprime(q=(2*p[1]+x)) + sum(k=2, 5, (q=nextprime(q+1))==2*p[k]+x) < 5, m++; p=concat(p[2..5], nextprime(p[5]+1))); m; } \\ Jinyuan Wang, Mar 20 2020
Formula
Extensions
Edited by R. J. Mathar, Apr 28 2009
Replaced the wrong value 14348916 by 14348907 (3^15=14348907). - Pierre CAMI, May 09 2009
More terms from Jinyuan Wang, Mar 20 2020