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%I #42 Feb 15 2019 03:06:18
%S 13,2,151,5,73,17,67,7,5,17,31,47,7,37,67,83,23,19,17,11,97,19,2,61,5,
%T 37,71,229,53,139,7,59,31,83,167,197,283,97,127,89,61,149,37,17,149,
%U 29,7,41,19,67,461,181,157,97,19,89,83,79,167,139,107,409,353,7,443,491
%N a(n) is the smallest prime p such that q = 2 * p + prime(n+2) is prime AND A007953(p) = A007953(q) where A007953 is sum of digits.
%C Prime q is sum of even semiprimes 2*p (A100484) and prime(i+2) (i = 1, 2, ...) (A000040).
%C Equivalent conditions: q = 2*p + prime(i+2) <=> p = (q - prime(i+2))/2 <=> (q - prime(i+2))/p = 2.
%C Algorithm: Q = prime(i+2) + 2*N, search first N with digsum(N) = digsum(Q);
%C (a) if Q AND N prime: p = N;
%C (b) if N composite: for n(k) = N + 9*k, q(k) = prime(i+2) + 2*(N + 9*k)
%C (i) digsum(n(k)) = digsum(q(k)),
%C (ii) n(k) has to be prime, let k = 1, 2, ... (Dirichlet's theorem on arithmetic progressions),
%C (iii) q(k) has to be prime (k = 1, 2, ... , again Dirichlet's theorem on arithmetic progressions);
%C Smallest k with (i), (ii) and (iii) gives term of sequence p = p(i).
%C Each prime > 3 is so "associated" uniquely with a pair (p,q) of primes.
%C List of triples (prime(i+2), p, q) is given in link for 0 <= i <= 100.
%C Pairs of primes (p, q) classified according to their digsum:
%C 2: (2,11) (11,101) (2,101)
%C 4: (13,31) (31,103) (31,211)
%C 5: (5,23) (5,41) (23,113) (5,113) (41,311)
%C 7: (151,313) (7,43) (7,61) (61,223) (7,151) (61,313) (7,241) (7,331)
%C 8: (17,53) (17,71) (71,251) (53,233) (17,233) (107,521)
%C 10: (73,163) (37,127) (19,109) (17,107) (19,127) (37,181) (127,433) (37,271) (19,271) (181,613) (19,307) (37,433)
%C 11: (47,137) (83,227) (83,317) (29,281) (461,1163) (83,443) (353,1019) (443,1217)
%C 13: (67,157) (67,193) (229,571) (139,409) (283,733) (67,373) (157,571) (139,571) (409,1129) (157,661)
%C 14: (59,257) (167,491) (149,419) (149,509) (167,617) (491,1319)
%C 16: (97,277) (97,367) (97,457) (79,439) (97,547)
%C 17: (197,557) (89,359) (89,449)
%C Note: prime(i+2) is used in the sequence for 2 reasons: prime(i) could not be used for i=1 because q would be 2*p + prime(1) = 2*p + 2 that is not prime; prime(i+1) could not be used for i=1, because q would be 2*p + prime(2) = 2*p + 3, but numbers q with digsum(q) = digsum(p) are necessarily of form q = 9*k + 6 that is not prime.
%H G. C. Greubel, <a href="/A178548/b178548.txt">Table of n, a(n) for n = 1..10000</a>
%H G. C. Greubel, <a href="/A178548/a178548_1.txt">List of triples (prime(n+2), p=A178548(n), q=A178561(n)) for n=0..100</a>
%e i=1: 2*13 + prime(3) = 26+5 = 31 = prime(11), digsum(13) = digsum(31) = 4, 13 is first term.
%e i=2: 2*2 + prime(4) = 4+7 = 11 = prime(5), digsum(2) = digsum(11) = 2, 2 is 2nd term.
%e i=3: 2*151 + prime(5) = 302+11 = 313 = prime(65), digsum(151) = digsum(313) = 7, 151 is 3rd term.
%t f[n_]:= Block[{p = 2}, While[q = 2*p +Prime[n+2]; !PrimeQ[q] ||
%t (Total[IntegerDigits[p]] != Total[IntegerDigits[q]]), p = NextPrime[p]];
%t Array[f, 100] (* _G. C. Greubel_, Feb 08 2019 *)
%o (PARI) a(n) = {my(p=2, q=prime(n+2)+2*p); while (!(isprime(q) && (sumdigits(p) == sumdigits(q))), p = nextprime(p+1); q = prime(n+2) + 2*p); p;} \\ _Michel Marcus_, Feb 09 2019
%Y Cf. A000040, A007953 (digsum), A100484, A178561 (primes q).
%K base,nonn,less
%O 1,1
%A Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 29 2010
%E Edited by _Michel Marcus_, Feb 28 2013
%E a(37) and a(62) corrected by _Michel Marcus_, Feb 09 2019
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