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A232095
Minimal number of factorials which add to 0+1+2+...+n; a(n) = A034968(A000217(n)).
4
0, 1, 2, 1, 3, 4, 5, 3, 3, 6, 4, 5, 4, 7, 7, 1, 5, 5, 5, 8, 7, 9, 5, 5, 6, 8, 10, 6, 9, 8, 10, 8, 6, 10, 12, 7, 10, 11, 6, 5, 7, 7, 8, 9, 5, 8, 5, 6, 8, 7, 10, 7, 11, 14, 8, 8, 6, 11, 7, 10, 7, 12, 10, 10, 12, 14, 7, 12, 9, 9, 11, 9, 12, 12, 12, 14, 10, 7, 11, 11
OFFSET
0,3
COMMENTS
1's occur at positions n=1, n=3 and n=15 as they are such natural numbers that A000217(n) is also one of the factorial numbers (A000142), as we have A000217(1) = 1 = 1!, A000217(3) = 1+2+3 = 6 = 3! and A000217(15) = 1 + 2 + ... + 15 = 120 = 5!
On the other hand, a(2)=2, as A000217(2) = 1+2 = 3 = 2! + 1!. Is this the only occurrence of 2?
Are some numbers guaranteed to occur an infinite number of times?
LINKS
FORMULA
a(n) = A034968(A000217(n)).
a(n) = A231717(A226061(n+1)). [Not a practical way to compute this sequence. Please see comments at A231717.]
For all n, a(n) >= A232094(n).
PROG
(Scheme)
(define (A232095 n) (A034968 (A000217 n)))
CROSSREFS
Sequence in context: A256100 A340383 A117407 * A279436 A082470 A101204
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 18 2013
STATUS
approved