A048972
Length of record run mentioned in A048971.
Original entry on oeis.org
1, 2, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16
Offset: 1
54, 55, 56, 57, 58 is a run of length 5 in A001221 all with d = 2.
More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Sep 04 2006
A048932
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.
Original entry on oeis.org
1, 14, 7, 2, 54, 91, 323, 141, 44360, 48919, 218972, 534078, 2699915, 526095, 17233173, 127890362, 29138958036, 146216247221, 118968284928, 2500769994070, 3157129230489, 22498525938216, 585927201062, 313978488186061, 571560399902283, 453918847597184
Offset: 1
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.
A305235
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.
Original entry on oeis.org
1, 4, 3, 2, 54, 91, 142, 141, 44360, 48919, 218972, 526097, 526096, 526095, 17233173, 127890362, 29138958036, 118968284929, 118968284928, 585927201065, 585927201064, 585927201063, 585927201062, 313978488186061, 453918847597185, 453918847597184, 455626105596320
Offset: 0
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.
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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++)
Showing 1-3 of 3 results.
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