Jake Foster has authored 3 sequences.
A214343
a(n) is the smallest integer j such that the numbers of prime factors (counting multiplicity) in j, j+1, ... , j+n-1 are the full set {1,2,...,n}.
Original entry on oeis.org
2, 3, 6, 15, 77, 726, 6318, 189375, 755968, 871593371, 33714015615
Offset: 1
a(4)=15 because 15 has two prime factors, 16 has four, 17 has one and 18 has three (and 15 is the smallest number with this property).
a(5) = 77 because 77, 78, 79, 80 and 81 have 2, 3, 1, 5 and 4 prime factors.
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A214343 := proc(n)
refs := {seq(i,i=1..n)} ;
for j from 1 do
pf := {} ;
for k from 0 to n-1 do
pf := pf union {numtheory[bigomega](j+k)} ;
if nops(pf) < k+1 then
break;
end if;
end do:
if pf = refs then
return j;
end if;
end do:
end proc: # R. J. Mathar, Jul 13 2012
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f[n_] := f[n] = FactorInteger[n][[All, 2]] // Total;
n = 1;
i = 2;
While[True,
While[Union[Table[f[j], {j, i, i + n - 1}]] != Range[n],
i += 1; f[i] =.
];
Print[i]; n += 1;
];
A196223
Natural numbers n such that Sum_{k = 1..pi(n)-1} p(k) == p(pi(n)) mod n, where p(k) denotes the k-th prime and pi(n) is the number of primes strictly less than n.
Original entry on oeis.org
6, 7, 15, 27, 41, 55, 172, 561, 1334, 6571, 11490, 429705, 2173016, 4417701, 9063353, 9531624, 40411847, 64538709, 83537963, 121316228, 181504240, 222586609
Offset: 1
2+3+5+7+11==13 (mod 15) and so 15 has this property.
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Reap[Module[{c = 0}, For[n = 4, n <= 10^6, n++, If[PrimeQ[n - 1], c += NextPrime[n - 1, -1]]; If[Mod[c, n] == NextPrime[n, -1], Sow[n]]]]]
A146086
Number of n-digit numbers with each digit odd where the digits 1 and 3 occur an even number of times.
Original entry on oeis.org
3, 11, 45, 197, 903, 4271, 20625, 100937, 498123, 2470931, 12295605, 61300877, 305972943, 1528270391, 7636568985, 38168496017, 190799433363, 953868026651, 4768952712765, 23843601302357, 119214519727383, 596062138283711, 2980279310358945, 14901302408615897
Offset: 1
For n=2 the a(2)=11 numbers are 11, 33, 55, 57, 59, 75, 77, 79, 95, 97, 99.
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[(5^n+2*3^n+1)/4: n in [1..30]]; // Vincenzo Librandi, Dec 31 2013
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Table[(5^n + 2 3^n + 1)/4, {n, 1, 30}] (* Vincenzo Librandi, Dec 31 2013 *)
LinearRecurrence[{9,-23,15},{3,11,45},30] (* Harvey P. Dale, Dec 15 2014 *)
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a(n)=(5^n+2*3^n+1)/4; \\ Michel Marcus, Aug 22 2013
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