OFFSET
1,2
COMMENTS
For any prime p and any run of k contiguous terms divisible by p, say a(m+1), a(m+2), ..., a(m+k), we have v_p(a(m+i)) = i for i = 1..k and v_p(a(m)) = 0 and v_p(a(m+k+1)) = 0 (where v_p denotes the p-adic valuation of a number).
For any n > 0, A003557(a(n+1)) is a unitary divisor of a(n).
This sequence is likely a permutation of the natural numbers.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A297409
EXAMPLE
The run of multiples of 2 that contains 2^16, alongside its bounding terms, is:
n a(n) Prime factorization of a(n)
----- ----- ---------------------------
67528 67561 13 * 5197
67529 67562 2 * 11 * 37 * 83
67530 67556 2^2 * 16889
67531 67560 2^3 * 3 * 5 * 563
67532 67536 2^4 * 3^2 * 7 * 67
67533 67424 2^5 * 7^2 * 43
67534 65856 2^6 * 3 * 7^3
67535 67456 2^7 * 17 * 31
67536 67328 2^8 * 263
67537 67072 2^9 * 131
67538 66560 2^10 * 5 * 13
67539 63488 2^11 * 31
67540 61440 2^12 * 3 * 5
67541 57344 2^13 * 7
67542 49152 2^14 * 3
67543 32768 2^15
67544 65536 2^16
67545 67565 5 * 13513
MAPLE
N:= 200: # to get terms before the first term > N
Avail:= [$2..N]:
A[1]:= 1:
found:= true;
for n from 2 while found do
found:= false;
for j from 1 to nops(Avail) do
m:= Avail[j];
Fm:= ifactors(m)[2];
if andmap(t -> A[n-1] mod t[1]^(t[2]-1)=0 and A[n-1] mod t[1]^t[2] <> 0, Fm) then
found:= true;
A[n]:= m;
Avail:= subsop(j=NULL, Avail);
break
fi
od;
od:
seq(A[i], i=1..n-1); # Robert Israel, Jan 07 2018
PROG
(PARI) See Links section.
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Rémy Sigrist, Dec 29 2017
STATUS
approved