A232095 -id:A232095 - cp's OEIS frontend

A232096 a(n) = largest m such that m! divides 1+2+...+n; a(n) = A055881(A000217(n)).

1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 5, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 3, 2, 1, 1, 2, 4, 1, 1, 3, 3, 1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 4, 4, 1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 4, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 3, 2, 1, 1, 2, 5, 1, 1, 3, 3, 1, 1, 3

Offset: 1

Antti Karttunen, Nov 18 2013

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10080

A042963 gives the positions of ones and A014601 the positions of larger terms.

Cf. also A232097, A076733, A232094, A232095, A232098.

(define (A232096 n) (A055881 (A000217 n)))

a(n) = A055881(A000217(n)).

a(n) = A231719(A226061(n+1)). [Not a practical way to compute this sequence, but follows from the definitions]

A231717 After a(0)=0, a(n) = A231713(A219666(n),A219666(n-1)).

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 1, 6, 3, 3, 3, 2, 1, 6, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 2, 3, 10, 3, 3, 3, 2, 1, 6, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 1, 3, 5, 5, 3, 10, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 1, 2, 4, 2, 4, 5, 3, 3, 9, 3, 1, 3, 6, 2, 1, 2, 4, 2, 4, 5, 3, 2, 4, 3, 10

Offset: 0

Antti Karttunen, Nov 12 2013

nonn

For all n, a(A226061(n+1)) = A232095(n). This works because at the positions given by each x=A226061(n+1), it holds that A219666(x) = (n+1)!-1, which has a factorial base representation (A007623) of (n,n-1,n-2,...,3,2,1) whose digit sum (A034968) is the n-th triangular number, A000217(n). This in turn is always a new record as at those points, in each significant digit position so far employed, a maximal digit value (for factorial number system) is used, and thus the preceding term, A219666(x-1) cannot have any larger digits in its factorial base representation, and so the differences between their digits (in matching positions) are all nonnegative.

Antti Karttunen, Table of n, a(n) for n = 0..3149

A231718 gives the positions of ones.

Cf. also A230410, A231719, A232095.

(define (A231717 n) (if (zero? n) n (A231713bi (A219666 n) (A219666 (- n 1)))))

a(0)=0, and for n>=1, a(n) = A231713(A219666(n),A219666(n-1)).

A232094 a(n) = A060130(A000217(n)); number of nonzero digits in factorial base representation (A007623) of 0+1+2+...+n.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 3, 2, 2, 4, 3, 2, 2, 3, 4, 1, 3, 5, 4, 4, 3, 5, 3, 3, 3, 4, 5, 2, 4, 4, 5, 3, 2, 5, 4, 3, 3, 4, 4, 3, 3, 5, 6, 5, 4, 5, 3, 3, 3, 4, 5, 3, 5, 6, 5, 3, 4, 6, 5, 4, 4, 5, 6, 3, 5, 6, 4, 4, 4, 5, 5, 4, 4, 5, 5, 4, 4, 4, 6, 5, 2, 6, 5, 3, 4, 4, 5

Offset: 0

Antti Karttunen, Nov 18 2013

The next 1 after a(1), a(3) and a(15) occurs at n=224, as A000217(224) = 25200 = 5 * 7!.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A000217, A007623, A060130, A226061, A230410, A232095.

a[n_] := Module[{k = n*(n+1)/2, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Count[s, ?(# > 0 &)]]; Array[a, 100, 0] (* _Amiram Eldar, Feb 07 2024 *)

(define (A232094 n) (A060130 (A000217 n)))

a(n) = A060130(A000217(n)).

a(n) = A230410(A226061(n+1)). [Not a practical way to compute this sequence. Please see comments at A230410.]

cp's OEIS Frontend

A232096 a(n) = largest m such that m! divides 1+2+...+n; a(n) = A055881(A000217(n)).

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A231717 After a(0)=0, a(n) = A231713(A219666(n),A219666(n-1)).

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A232094 a(n) = A060130(A000217(n)); number of nonzero digits in factorial base representation (A007623) of 0+1+2+...+n.

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