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Length of record run mentioned in A048971.
+20
2
1, 2, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16
EXAMPLE
54, 55, 56, 57, 58 is a run of length 5 in A001221 all with d = 2.
EXTENSIONS
More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Sep 04 2006
Let d(n) = number of distinct primes dividing n ( A001221). Find smallest t such that d(t)=d(t+1)=...=d(t+n-1) but d(t-1) and d(t+n) are not = d(t); then a(n)=t.
+10
6
1, 14, 7, 2, 54, 91, 323, 141, 44360, 48919, 218972, 534078, 2699915, 526095, 17233173, 127890362, 29138958036, 146216247221, 118968284928
COMMENTS
a(n) > 10^13 for n from 20 to 22. a(23) = 585927201062. [From Donovan Johnson, Jul 30 2010]
EXAMPLE
a(3)=7 since 7,8,9 all have d = 1 but d(6) and d(10) != 1 and this is the first run of 3.
Smallest positive number k such that there are exactly n successive equal values of A001221 starting at k, i.e., such that A305234(k) = n.
+10
1
1, 4, 3, 2, 54, 91, 142, 141, 44360, 48919, 218972, 526097, 526096, 526095, 17233173, 127890362
EXAMPLE
For n = 5: A001221(91+k) = 2 for k = 0..5 and 91 is the smallest number x with exactly 5 successors that have the same value of A001221 as x, so a(5) = 91.
PROG
(PARI) a305234(n) = my(k=n+1, i=0); while(omega(k)==omega(n), i++; k++); i
a(n) = my(k=1); while(1, if(a305234(k)==n, return(k)); k++)
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