A060130 -id:A060130 - cp's OEIS frontend

A275946 Number of nonzero digits that are the sole occupants of their slope in factorial base representation: a(n) = A056169(A275734(n)). (See comments for more exact definition.)

0, 1, 1, 2, 1, 0, 1, 2, 2, 3, 2, 1, 1, 2, 0, 1, 2, 1, 1, 0, 2, 1, 0, 0, 1, 2, 2, 3, 2, 1, 2, 3, 3, 4, 3, 2, 2, 3, 1, 2, 3, 2, 2, 1, 3, 2, 1, 1, 1, 2, 2, 3, 2, 1, 0, 1, 1, 2, 1, 0, 2, 3, 1, 2, 3, 2, 2, 1, 3, 2, 1, 1, 1, 2, 0, 1, 2, 1, 2, 3, 1, 2, 3, 2, 0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 0, 0, 2, 1, 3, 2, 1, 1, 2, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1

Offset: 0

Antti Karttunen, Aug 15 2016

Total number of such nonzero digits d_x in the factorial base representation (A007623) of n for which it holds that for all other nonzero digits d_y present (i_x - d_x) <> (i_y - d_y), where i_x and i_y are the indices of the digits d_x and d_y respectively.

Equally: Number of digit slopes occupied by just one nonzero digit in the factorial base representation of n. In other words, a(n) is the number of elements with multiplicity one in multiset [(i_x - d_x) | where d_x ranges over each nonzero digit present and i_x is its position from the right].

			For n=2, in factorial base "10", there is only one slope occupied by a single nonzero digit (1 is on the sub-maximal slope as 2-1 = 1), thus a(2) = 1.
For n=3, in factorial base "11", there are two occupied slopes, each having just one digit present, thus a(3) = 2.
For n=5, in factorial base "21", there is just one distinct occupied slope, but it contains two nonzero digits (2 and 1 both occupy the maximal slope as 2-2 = 1-1 = 0), thus there are no slopes with just one nonzero digit and a(5) = 0.
For n=525, in factorial base "41311", there are three occupied slopes. The maximal slope contains the nonzero digits "3.1", the sub-maximal digits "4..1.", and the sub-sub-sub-maximal just "1..." (the 1 in the position 4 from right is the sole occupier of its own slope). Thus only one of the slopes is occupied by a lonely occupant and a(525) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Indranil Ghosh, Python program for generating this sequence
Index entries for sequences related to factorial base representation

Cf. A056169, A275734.

Cf. also A060502, A060130, A225901, A275947, A275948, A275962.

(define (A275946 n) (A056169 (A275734 n)))

a(n) = A056169(A275734(n)).
Other identities. For all n >= 0.
a(n) = A275948(A225901(n)).
A060502(n) = a(n) + A275947(n).
A060130(n) = a(n) + A275962(n).

A275962 Total number of nonzero digits that occur on the multiply occupied slopes of the factorial base representation of n: a(n) = A275812(A275734(n)). (See comments for more exact definition).

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 2, 0, 2, 0, 2, 2, 3, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 2, 0, 2, 0, 2, 2, 3, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 4, 0, 0, 2, 2, 0, 2, 0, 2, 0, 2, 2, 3, 0, 0, 2, 2, 0, 2, 0, 0, 2, 2, 0, 2, 2, 2, 3, 3, 2, 4, 0, 2, 2, 4, 2, 3, 0, 2, 0, 2, 2, 3, 0, 2, 0, 2, 2, 3, 0, 2, 2, 4, 2, 3, 2, 3, 2, 3, 3, 4, 0

Offset: 0

Antti Karttunen, Aug 15 2016

a(n) gives the total number of elements (counted with multiplicity) that have multiplicity > 1 in a multiset [(i_x - d_x) | where d_x ranges over each nonzero digit present and i_x is its position from the right].

			For n=525, in factorial base "41311", there are three occupied slopes. The maximal slope contains the nonzero digits "3.1", the sub-maximal the digits "4..1.", and the sub-sub-sub-maximal just "1..." (the 1 in the position 4 from right is the sole occupier of its own slope). There are two slopes with more than one nonzero digit, each having two such digits, and thus a(525) = 2+2 = 4.
Equally, when we form a multiset of (digit-position - digit-value) differences for all nonzero digits present in "41311", we obtain a multiset [0, 0, 1, 1, 3], in which the elements that occur multiple times are [0, 0, 1, 1], thus a(525) = 4.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Indranil Ghosh, Python program for computing this sequence
Index entries for sequences related to factorial base representation

Cf. A275734, A275812.
Cf. A275804 (indices of zeros), A275805 (of nonzeros).
Cf. also A060130, A225901, A275946, A275947, A275964.

(define (A275962 n) (A275812 (A275734 n)))

a(n) = A275812(A275734(n)).
Other identities and observations. For all n >= 0.
a(n) = A275964(A225901(n)).
a(n) = A060130(n) - A275946(n).
a(n) >= A275947(n).

A277012 Factorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

Original entry on oeis.org

0, 1, 2, 5, 6, 13, 4, 9, 10, 21, 22, 45, 12, 25, 26, 53, 54, 109, 28, 57, 58, 117, 118, 237, 8, 17, 18, 37, 38, 77, 20, 41, 42, 85, 86, 173, 44, 89, 90, 181, 182, 365, 92, 185, 186, 373, 374, 749, 24, 49, 50, 101, 102, 205, 52, 105, 106, 213, 214, 429, 108, 217, 218, 437, 438, 877, 220, 441, 442, 885, 886, 1773, 56, 113, 114, 229, 230, 461, 116, 233

Offset: 0

Antti Karttunen, Sep 25 2016

			9 = "111" in factorial base (3! + 2! + 1! = 9) is converted to three 1-bits with separating zeros between, in binary as "10101" = A007088(21), thus a(9) = 21.
91 = "3301" in factorial base (91 = 3*4! + 3*3! + 1!) is converted to binary number "1110111001" = A007088(953), thus a(91) = 953. Between the rightmost 1-runs the other zero comes from the factorial base representation, while the other zero is an extra separating zero inserted after each run of 1-bits apart from the rightmost 1-run. The single zero between the two leftmost 1-runs is similarly used to separate the two "unary representations" of 3's.

Cf. A000035, A000079, A000120, A000225, A007088, A007623, A034968, A060130, A069010, A156552, A227153, A227349.
Cf. A277008 (terms sorted into ascending order).
Cf. A277011 (a left inverse).
Differs from analogous A277022 for the first time at n=24, where a(24) = 8, while A277022(24) = 60.

(define (A277012 n) (let loop ((n n) (z 0) (i 2) (j 0)) (if (zero? n) z (let ((d (remainder n i))) (loop (quotient n i) (+ z (* (A000225 d) (A000079 j))) (+ 1 i) (+ 1 j d))))))

a(n) = A156552(A276076(n)).
Other identities. For all n >= 0:
A277011(a(n)) = n.
A005940(1+a(n)) = A276076(n).
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A000120(a(n)) = A034968(n).
A069010(a(n)) = A060130(n).
A227349(a(n)) = A227153(n).

A257510 Number of nonleading zeros in factorial base representation of n (A007623).

Original entry on oeis.org

0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 3, 2, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 3, 2, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 3, 2, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 3, 2, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 1, 1, 0, 1, 0, 4

Offset: 1

Antti Karttunen, Apr 27 2015

Sequence starts from n=1, because 0 is an ambiguous case.

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

Cf. A227157 (numbers n such that a(n) = 0), A227187 (n for which a(n) > 0).
Cf. A007623, A060130, A084558, A230403.
Cf. also A257511.
Cf. also A023416, A080791 (analogous sequences for base-2), A055641 (for base-10).

factBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > i!, i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= j!, m = m - j!; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; dList]; s = Table[FromDigits[factBaseIntDs[n]], {n, 120}]; Last@ DigitCount[#] & /@ s (* Michael De Vlieger, Apr 27 2015, after Alonso del Arte at A007623 *)

(define (A257510 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (floor->exact (/ n i)) (+ 1 i) (+ s (if (zero? (modulo n i)) 1 0)))))))

a(n) = A084558(n) - A060130(n).
Other identities and observations:
For all n >= 0, a(A000142(n+1)) = n. [(n+1)! gives the position where n first appears.]
For all n, a(n) >= A230403(n).

A275727 a(0) = 0, for n >= 1, a(n) = A275736(n) + 2*a(A257684(n)).

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 4, 5, 6, 7, 6, 7, 4, 5, 6, 7, 6, 7, 4, 5, 6, 7, 6, 7, 8, 9, 10, 11, 10, 11, 12, 13, 14, 15, 14, 15, 12, 13, 14, 15, 14, 15, 12, 13, 14, 15, 14, 15, 8, 9, 10, 11, 10, 11, 12, 13, 14, 15, 14, 15, 12, 13, 14, 15, 14, 15, 12, 13, 14, 15, 14, 15, 8, 9, 10, 11, 10, 11, 12, 13, 14, 15, 14, 15, 12, 13, 14, 15

Offset: 0

Antti Karttunen, Aug 09 2016

a(n) has ones in those positions of its base-2 representation where n has nonzero digits in its factorial base representation.

			For n=19, A007623(19) = 301, thus a(19) = 5 because A007088(5) = 101.

Cf. A000120, A007088, A007623, A048675, A060130, A257684, A275733, A275736.

Cf. also A275728.

a(0) = 0, for n >= 1, a(n) = A275736(n) + 2*a(A257684(n)).
a(n) = A048675(A275733(n)).
Other identities and observations. For all n >= 0:
A000120(a(n)) = A060130(n).

A275948 Number of nonzero digits that occur only once in factorial base representation of n: a(n) = A056169(A275735(n)).

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 0, 0, 0, 2, 1, 1, 2, 2, 1, 0, 1, 1, 2, 2, 1, 2, 3, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 1, 0, 2, 1, 1, 1, 3, 2, 1, 2, 2, 1, 0, 1, 2, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 2, 3, 3, 2, 1, 2, 1, 2, 2, 1, 2, 3, 2, 1, 1, 1, 3, 2, 2, 3, 3, 2, 1, 2, 0, 1, 1, 0, 1, 2, 1, 2, 2, 1, 2, 3, 2, 1, 1, 1, 3, 2, 2, 3, 3, 2, 1, 2, 2, 3, 3, 2, 3, 4, 1

Offset: 0

Antti Karttunen, Aug 15 2016

			For n=0, with factorial base representation (A007623) also 0, there are no nonzero digits, thus a(0) = 0.
For n=2, with factorial base representation "10", there is one nonzero digit, thus a(2) = 1.
For n=3 (= "11") there is no nonzero digit which would occur just once, thus a(3) = 0.
For n=23 (= "321") there are three nonzero digits and each of those digits occurs just once, thus a(23) = 3.
For n=44 (= "1310") there are two distinct nonzero digits ("1" and "3"), but only the other (3) occurs just once, thus a(44) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation.

Cf. A056169, A275735.

Cf. also A060130, A225901, A275806, A275946, A275949, A275964.

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

from sympy import prime, factorint
from operator import mul
from functools import reduce
import collections
def a056169(n):
    f=factorint(n)
    return 0 if n==1 else sum([1 for i in f if f[i]==1])
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 20 2017

(define (A275948 n) (A056169 (A275735 n)))

a(n) = A056169(A275735(n)).
Other identities. For all n >= 0.
a(n) = A275946(A225901(n)).
A275806(n) = a(n) + A275949(n).
A060130(n) = a(n) + A275964(n).

A275964 Total number of nonzero digits with multiple occurrences in factorial base representation of n (counted with multiplicity): a(n) = A275812(A275735(n)).

Original entry on oeis.org

0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 2, 3, 3, 4, 2, 3, 0, 2, 2, 3, 2, 4, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 0, 2, 2, 3, 2, 4, 2, 2, 2, 4, 3, 3, 0, 0, 0, 2, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 2, 2, 2, 4, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 0, 0, 0, 2, 0, 0, 0

Offset: 0

Antti Karttunen, Aug 15 2016

			For n=0, with factorial base representation (A007623) also 0, there are no nonzero digits, thus a(0) = 0.
For n=2, with factorial base representation "10", there are no nonzero digits that are present multiple times, thus a(2) = 0.
For n=3 ("11") there is one nonzero digit which occurs more than once, and it occurs two times in total, thus a(3) = 2.
For n=41 ("1221") there are two distinct nonzero digits ("1" and "2"), and both occur more than once, namely twice each, thus a(41) = 2+2 = 4.
For n=44 ("1310") there are two distinct nonzero digits ("1" and "3"), but only the other (1) occurs more than once (two times), thus a(44) = 2.
For n=279 ("21211") there are two distinct nonzero digits present that occur more than once, digit 2 twice, and digit 1 for three times, thus a(279) = 2+3 = 5.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Indranil Ghosh, Python program for computing this sequence.
Index entries for sequences related to factorial base representation.

Cf. A275735, A275812.
Cf. A265349 (indices of zeros), A265350 (of terms > 0).
Cf. also A060130, A225901, A275948, A275949.

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

(define (A275964 n) (A275812 (A275735 n)))

a(n) = A275812(A275735(n)).
Other identities and observations. For all n >= 0.
a(n) = A275962(A225901(n)).
a(n) = A060130(n) - A275948(n).
a(n) >= A275949(n).

A230415 Square array T(i,j) giving the number of differing digits in the factorial base representations of i and j, for i >= 0, j >= 0, read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 1, 0, 1, 2, 2, 2, 2, 1, 1, 0, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 1, 0, 1, 2, 1, 2, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 3, 3, 2, 2, 3, 3, 3, 3, 2, 2, 3, 3, 1, 2, 2, 1, 2, 2, 0, 2, 2, 1, 2, 2, 1, 2, 2, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 0, 1, 2, 1, 2, 2, 1, 2

Offset: 0

Antti Karttunen, Nov 10 2013

This table relates to the factorial base representation (A007623) in a somewhat similar way as A101080 relates to the binary system. See A231713 for another analog.

			The top left corner of this square array begins as:
0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, ...
1, 0, 2, 1, 2, 1, 2, 1, 3, 2, 3, ...
1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 2, ...
2, 1, 1, 0, 2, 1, 3, 2, 2, 1, 3, ...
1, 2, 1, 2, 0, 1, 2, 3, 2, 3, 1, ...
2, 1, 2, 1, 1, 0, 3, 2, 3, 2, 2, ...
1, 2, 2, 3, 2, 3, 0, 1, 1, 2, 1, ...
2, 1, 3, 2, 3, 2, 1, 0, 2, 1, 2, ...
2, 3, 1, 2, 2, 3, 1, 2, 0, 1, 1, ...
3, 2, 2, 1, 3, 2, 2, 1, 1, 0, 2, ...
2, 3, 2, 3, 1, 2, 1, 2, 1, 2, 0, ...
...
For example, T(1,2) = T(2,1) = 2 as 1 has factorial base representation '...0001' and 2 has factorial base representation '...0010', and they differ by their two least significant digits.
On the other hand, T(3,5) = T(5,3) = 1, as 3 has factorial base representation '...0011' and 5 has factorial base representation '...0021', and they differ only by their second rightmost digit.
Note that as A007623(6)='100' and A007623(10)='120', we have T(6,10) = T(10,6) = 1 (instead of 2 as in A231713, cf. also its Example section), as here we count only the number of differing digit positions, but ignore the magnitudes of their differences.

Antti Karttunen, The first 121 antidiagonals of the table, flattened

The topmost row and the leftmost column: A060130.

Only the lower triangular region: A230417. Related arrays: A230419, A231713. Cf. also A101080, A084558, A230410.

nn = 14; m = 1; While[m! < nn, m++]; m; Table[Function[w, Count[Subtract @@ Map[PadLeft[#, Max@ Map[Length, w]] &, w], k_ /; k != 0]]@ Map[IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, {i - j, j}], {i, 0, nn}, {j, 0, i}] // Flatten (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

(define (A230415 n) (A230415bi (A025581 n) (A002262 n)))
(define (A230415bi x y) (let loop ((x x) (y y) (i 2) (d 0)) (cond ((and (zero? x) (zero? y)) d) (else (loop (floor->exact (/ x i)) (floor->exact (/ y i)) (+ i 1) (+ d (if (= (modulo x i) (modulo y i)) 0 1)))))))

T(n,0) = T(0,n) = A060130(n).

Each entry T(i,j) <= A231713(i,j).

A278234 Filter-sequence for factorial base (digit slopes): Least number with the same prime signature as A275734(n).

Original entry on oeis.org

1, 2, 2, 6, 2, 4, 2, 6, 6, 30, 6, 12, 2, 6, 4, 12, 6, 12, 2, 4, 6, 12, 4, 8, 2, 6, 6, 30, 6, 12, 6, 30, 30, 210, 30, 60, 6, 30, 12, 60, 30, 60, 6, 12, 30, 60, 12, 24, 2, 6, 6, 30, 6, 12, 4, 12, 12, 60, 12, 36, 6, 30, 12, 60, 30, 60, 6, 12, 30, 60, 12, 24, 2, 6, 4, 12, 6, 12, 6, 30, 12, 60, 30, 60, 4, 12, 8, 24, 12, 36, 6, 12, 12, 36, 12, 24, 2, 4, 6, 12, 4, 8, 6

Offset: 0

Antti Karttunen, Nov 16 2016

nonn

This sequence can be used for filtering certain factorial base (A007623) related sequences, because it matches only with any such sequence b that can be computed as b(n) = f(A275734(n)), where f(n) is any function that depends only on the prime signature of n (some of these are listed under the index entry for "sequences computed from exponents in ...").

Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Indranil Ghosh, Python program for computing this sequence
Index entries for sequences computed from exponents in factorization of n
Index entries for sequences related to factorial base representation

Cf. A007623, A046523, A225901, A275734.
Other filter-sequences related to factorial base: A278225, A278235, A278236.
Sequences that partition N into same or coarser equivalence classes: A060130, A060502, A275811, A275946, A275962.

(define (A278234 n) (A046523 (A275734 n)))

a(n) = A046523(A275734(n)).

a(n) = A278235(A225901(n)).

A236855 a(n) is the sum of digits in A239903(n).

Original entry on oeis.org

0, 1, 1, 2, 3, 1, 2, 2, 3, 4, 3, 4, 5, 6, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 4, 5, 6, 7, 3, 4, 4, 5, 6, 5, 6, 7, 8, 6, 7, 8, 9, 10, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 4, 5, 6, 7, 2, 3, 3, 4, 5, 3, 4, 4, 5, 6, 5, 6, 7, 8, 4, 5, 5, 6, 7, 6, 7, 8, 9, 7, 8, 9, 10, 11, 3, 4

Offset: 0

Antti Karttunen, Apr 18 2014

			As the 0th Catalan String is empty, indicated by A239903(0)=0, a(0)=0.
As the 18th Catalan String is [1,0,1,2] (A239903(18)=1012), a(18) = 1+0+1+2 = 4.
Note that although the range of validity of A239903 is inherently limited by the decimal representation employed, it doesn't matter here: We have a(58785) = 55, as the corresponding 58785th Catalan String is [1,2,3,4,5,6,7,8,9,10], even though A239903 cannot represent that unambiguously.

Antti Karttunen, Table of n, a(n) for n = 0..16796

Cf. A239903, A014420, A237449, A236859, A009766, A071155, A081291, A034968, A060130.

A236855list[m_] := With[{r = 2*Range[2, m]-1}, Reverse[Map[Total[r-#] &, Select[Subsets[Range[2, 2*m-1], {m-1}], Min[r-#] >= 0 &]]]];
A236855list[6] (* Generates C(6) terms *) (* Paolo Xausa, Feb 19 2024 *)

(define (A236855 n) (apply + (A239903raw n)))
(define (A239903raw n) (if (zero? n) (list) (let loop ((n n) (row (- (A081288 n) 1)) (col (- (A081288 n) 2)) (srow (- (A081288 n) 2)) (catstring (list 0))) (cond ((or (zero? row) (negative? col)) (reverse! (cdr catstring))) ((> (A009766tr row col) n) (loop n srow (- col 1) (- srow 1) (cons 0 catstring))) (else (loop (- n (A009766tr row col)) (+ row 1) col srow (cons (+ 1 (car catstring)) (cdr catstring))))))))
;; Alternative definition:
(define (A236855 n) (let ((x (A071155 (A081291 n)))) (- (A034968 x) (A060130 x))))

a(n) = A034968(x) - A060130(x), where x = A071155(A081291(n)).
For up to n = A000108(11)-2 = 58784, a(n) = A007953(A239903(n)).
Catalan numbers, A000108, give the positions of ones, and the n-th triangular number occurs for the first time at the position immediately before that, i.e., a(A001453(n)) = A000217(n-1).
For each n, a(n) >= A000217(A236859(n)).

cp's OEIS Frontend

A275946 Number of nonzero digits that are the sole occupants of their slope in factorial base representation: a(n) = A056169(A275734(n)). (See comments for more exact definition.)

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A275962 Total number of nonzero digits that occur on the multiply occupied slopes of the factorial base representation of n: a(n) = A275812(A275734(n)). (See comments for more exact definition).

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A277012 Factorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

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A257510 Number of nonleading zeros in factorial base representation of n (A007623).

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A275727 a(0) = 0, for n >= 1, a(n) = A275736(n) + 2*a(A257684(n)).

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A275948 Number of nonzero digits that occur only once in factorial base representation of n: a(n) = A056169(A275735(n)).

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A275964 Total number of nonzero digits with multiple occurrences in factorial base representation of n (counted with multiplicity): a(n) = A275812(A275735(n)).

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A230415 Square array T(i,j) giving the number of differing digits in the factorial base representations of i and j, for i >= 0, j >= 0, read by antidiagonals.

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