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A258780
a(n) is the least k such that k^2 + 1 is a semiprime p*q, p < q, and (q - p)/2^n is prime.
0
8, 12, 140, 64, 2236, 196, 1300, 1600, 6256, 5084, 248756, 246196, 484400, 36680, 887884, 821836, 1559116, 104120, 126072244, 9586736, 4156840, 542759984, 1017981724, 2744780140, 405793096, 148647496, 1671024916
OFFSET
2,1
COMMENTS
The corresponding primes are 2, 3, 71, 7, 1069, 7, 5, 5, 59, 2, 368471, 180463, 12421, 2, 29, 125683, 226169, 5, 369704891, 197, 5, 263, 7444559, 239621423, 594271, 2, 474359, ...
All terms are even, in order for k^2+1 to be odd. Otherwise, with k^2+1 being even, p-q would be odd and hence not a multiple of 2^n. - Michel Marcus, Apr 13 2019
EXAMPLE
a(2)=8 because 8^2+1 = 5*13 and (13-5)/2^2 = 2 is prime. The number 8 is the first term of the sequence 8, 22, 34, 46, 50, 58, ...
a(3)=12 because 12^2+1 = 5*29 and (29-5)/2^3 = 3 is prime. The number 12 is the first term of the sequence 12, 28, 44, 52, 76, 80, ...
a(4)=140 because 140^2+1 = 17*1153 and (1153-17)/2^4 = 71 is prime. The number 140 is the first term of the sequence 140, 296, 404, 604, ...
MATHEMATICA
lst={}; Do[k=2; While[!(Plus@@Last/@FactorInteger[k^2+1]==2&&PrimeQ[(FactorInteger[k^2+1][[-1, 1]]-FactorInteger[k^2+1][[1, 1]])/2^n]), k=k+2]; Print[n, " ", k], {n, 2, 19}]; lst
PROG
(PARI) isok(k, n) = my(kk=k^2+1, f=factor(kk)[, 1]~); (bigomega(kk) == 2) && (#f == 2) && (p=f[1]) && (q=f[2]) && (qq=(q-p)/2^n) && !frac(qq) && isprime(qq);
a(n) = my(k=2); while (!isok(k, n), k+=2); k; \\ Michel Marcus, Apr 13 2019
CROSSREFS
Sequence in context: A083128 A196077 A067923 * A228663 A377681 A377684
KEYWORD
nonn,more
AUTHOR
Michel Lagneau, Jun 10 2015
EXTENSIONS
Name edited by Jon E. Schoenfield, Sep 12 2017
a(20)-a(22) from Daniel Suteu, Apr 13 2019
a(23)-a(28) from Daniel Suteu, Nov 09 2019
STATUS
approved