A261234 -id:A261234 - cp's OEIS frontend

A261236 a(n) = A261234(n) - A261237(n).

0, 1, 3, 7, 16, 40, 104, 279, 758, 2071, 5678, 15609, 43035, 119139, 331616, 928572, 2614743, 7396880, 20999683, 59784414, 170615755, 488073987, 1399625614, 4023315793, 11590737827, 33452982391

Offset: 0

Antti Karttunen, Aug 16 2015

a(n) = How many numbers whose base-3 representation begins with digit "1" are encountered before (3^n)-1 is reached when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k by k - (sum of digits in base-3 representation of k).

			For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), from which we subtract 4, to get 12 ("110" in base-3), from which we subtract 2 to get 10 ("101" in base-3), from which we subtract 2 to get 8 ("22" in base-3), which is the end point of iteration. Of the numbers encountered, 16, 12 and 10 have base-3 representations beginning with digit "1", thus a(2) = 3.

Cf. A261234, A261237, A261230.

/* Use the C-program given in A261234. */

Table[Length@ # - First@ FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, #] &@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &], {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

(define (A261236 n) (- (A261234 n) (A261237 n)))

a(n) = A261234(n) - A261237(n).

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016

A261235 First differences of A261234.

Original entry on oeis.org

1, 3, 7, 17, 45, 122, 334, 915, 2511, 6906, 19039, 52691, 146641, 410734, 1157321, 3276419, 9307640, 26509145, 75669108, 216483760, 620847130, 1784898316, 5143127022, 14847548759, 42923323624, 124213679008, 359730306331, 1042521363580, 3023562595587, 8776718908896, 25502164336020, 74179903571807, 216005883269160, 629644927525768, 1837129499003121, 5364782084798156

Offset: 0

Antti Karttunen, Aug 13 2015

nonn

Cf. A261234.

Cf. also A226060.

(define (A261235 n) (- (A261234 (+ 1 n)) (A261234 n)))

a(n) = A261234(n+1) - A261234(n).

Terms from a(23) onward computed from the output of Hiroaki Yamanouchi's program (given in A261234) by Antti Karttunen, Aug 16 2015

A276622 Simple self-inverse permutation of natural numbers: after a(0)=0, list each block of A261234(n) numbers in reverse order, from A261232(n) to A261233(1+n).

Original entry on oeis.org

0, 1, 3, 2, 8, 7, 6, 5, 4, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 123, 122, 121, 120, 119, 118, 117, 116, 115, 114, 113, 112, 111, 110, 109, 108, 107, 106, 105, 104, 103, 102, 101, 100, 99, 98, 97, 96

Offset: 0

Antti Karttunen, Sep 11 2016

Maps between A276623 and A276624.

Antti Karttunen, Table of n, a(n) for n = 0..6250
Index entries for sequences that are permutations of the natural numbers

Cf. A261232, A276621, A276623, A276624.

(define (A276622 n) (if (zero? n) n (+ (A261232 (+ -1 (A276621 n))) (- (A261232 (A276621 n)) n 1))))

a(0) = 0; for n >= 1, a(n) = A261232(A276621(n)-1) + A261232(A276621(n)) - n - 1.

A276621 After a(0)=0, each n occurs A261234(n-1) times.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Sep 11 2016

Auxiliary function for computing A276622 & A276623.

Antti Karttunen, Table of n, a(n) for n = 0..6250

Cf. A261233, A276622, A276623.

After a(0), a(n) differs from A111393(n+1) for the first time at n=46, where a(46)=5, while A111393(47)=6.

(define (A276621 n) (let loop ((k 0)) (if (>= (A261233 k) n) k (loop (+ 1 k)))))

A219661 Number of steps to go from (n+1)!-1 to n!-1 with map x -> x - (sum of digits in factorial base representation of x).

Original entry on oeis.org

1, 2, 5, 19, 83, 428, 2611, 18473, 150726, 1377548, 13851248, 152610108, 1835293041, 23925573979, 335859122743, 5049372125352, 80942722123544, 1378487515335424, 24858383452927384, 473228664468684846

Offset: 1

Antti Karttunen, Dec 03 2012

nonn

			(1!)-1 (0) is reached from (2!)-1 (1) with one step by subtracting A034968(1) from 1.
(2!)-1 (1) is reached from (3!)-1 (5) with two steps by first subtracting A034968(5) from 5 -> 2, and then subtracting A034968(2) from 2 -> 1.
(3!)-1 (5) is reached from (4!)-1 (23) with five steps by repeatedly subtracting the sum of digits in factorial expansion as: 23 - 6 = 17, 17 - 5 = 12, 12 - 2 = 10, 10 - 3 = 7, 7 - 2 = 5.
Thus a(1)=1, a(2)=2 and a(3)=5.

Hiroaki Yamanouchi, Fast Python-program for computing terms of this sequence

Row sums of A230420 and A230421.
Cf. A000142, A007623, A219651, A219652, A219662, A219663, A219665, A226061.
Cf. also A213709 (analogous sequence for base-2), A261234 (for base-3).

Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, 120]]] &, (n + 1)! - 1, # > n! - 1 &] - 1, {n, 0, 8}] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

(define (A219661 n) (if (zero? n) n (let loop ((i (-1+ (A000142 (1+ n)))) (steps 1)) (cond ((isA000142? (1+ (A219651 i))) steps) (else (loop (A219651 i) (1+ steps)))))))
(define (isA000142? n) (and (> n 0) (let loop ((n n) (i 2)) (cond ((= 1 n) #t) ((not (zero? (modulo n i))) #f) (else (loop (/ n i) (1+ i)))))))
;; Alternative definition:
(define (A219661 n) (- (A219652 (-1+ (A000142 (1+ n)))) (A219652 (-1+ (A000142 n)))))

a(n) = A219652((n+1)!-1) - A219652(n!-1).

a(n) = A219662(n) + A219663(n).

Terms a(16) - a(20) computed with Hiroaki Yamanouchi's Python-program by Antti Karttunen, Jun 27 2016

A261233 a(n) = number of steps to reach 0 when starting from k = (3^n)-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

Original entry on oeis.org

0, 1, 3, 8, 20, 49, 123, 319, 849, 2294, 6250, 17112, 47013, 129605, 358838, 998805, 2796093, 7869800, 22251147, 63141639, 179701239, 512744599, 1466635089, 4205423895, 12087339723, 34816804310, 100469592521, 290336059740, 839932833290, 2432050970420, 7047731703137, 20440131344750, 59334695322383, 172409162871823, 501489513690423, 1460214792034791

Offset: 0

Antti Karttunen, Aug 13 2015

nonn

One less than A261232.
Cf. A000244, A261231.
Cf. A261234 (the first differences).
Cf. also A218600.

a(0) = 0; for n >= 1, a(n) = A261234(n-1) + a(n-1).
a(n) = A261231((3^n)-1).
a(n) = A261232(n)-1.

Terms from a(24) onward added from the output of Hiroaki Yamanouchi's program by Antti Karttunen, Aug 16 2015

A261231 a(n) = number of steps to reach 0 when starting from k = n and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..19683

Cf. A000244, A053735, A054861, A261232, A261233, A261234, A261235.

Cf. also A071542.

from sympy.ntheory.factor_ import digits
def a054861(n): return (n - sum(digits(n, 3)[1:]))/2
def a(n): return 0 if n==0 else 1 + a(2*a054861(n)) # Indranil Ghosh, May 22 2017

a(0) = 0; for n >= 1, a(n) = 1 + a(2*A054861(n)). [Note that A054861(n) = (n - A053735(n))/2, where A053735(n) = sum of digits of n, when written in base 3.]

A261230 a(n) = A261236(n) - A261237(n).

Original entry on oeis.org

-1, 0, 1, 2, 3, 6, 12, 28, 71, 186, 494, 1317, 3478, 9045, 23265, 59856, 155779, 412413, 1108874, 3009228, 8188150, 22257484, 60462422, 164715758, 452011067, 1253176571

Offset: 0

Antti Karttunen, Aug 16 2015

sign

Cf. A261234, A261236, A261237.

Table[Length@ # - 2 First@ FirstPosition[#, k_ /; k != 2] + 1 &@ Map[First@ IntegerDigits[#, 3] &, #] &@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &], {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

(define (A261230 n) (- (A261236 n) (A261237 n)))

a(n) = A261236(n) - A261237(n).

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016

A261232 a(n) = number of steps to reach 0 when starting from k = 3^n and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

Original entry on oeis.org

1, 2, 4, 9, 21, 50, 124, 320, 850, 2295, 6251, 17113, 47014, 129606, 358839, 998806, 2796094, 7869801, 22251148, 63141640, 179701240, 512744600, 1466635090, 4205423896, 12087339724, 34816804311, 100469592522, 290336059741, 839932833291, 2432050970421, 7047731703138, 20440131344751, 59334695322384, 172409162871824, 501489513690424

Offset: 0

Antti Karttunen, Aug 13 2015

nonn

One more than A261233.
Cf. A000244, A261231.
Cf. also A213710.

a(0) = 1; for n >= 1, a(n) = A261234(n-1) + a(n-1).
a(n) = A261231(3^n).
a(n) = 1 + A261233(n).

Terms from a(24) onward added from the output of Hiroaki Yamanouchi's program by Antti Karttunen, Aug 16 2015

A261237 Number of steps needed when starting from (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k) to encounter the first number whose base-3 representation begins with a digit other than 2.

Original entry on oeis.org

1, 1, 2, 5, 13, 34, 92, 251, 687, 1885, 5184, 14292, 39557, 110094, 308351, 868716, 2458964, 6984467, 19890809, 56775186, 162427605, 465816503, 1339163192, 3858600035, 11138726760, 32199805820

Offset: 0

Antti Karttunen, Aug 16 2015

a(n) = How many numbers whose base-3 representation begins with digit "2" are encountered before (3^n)-1 is reached when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k by k - (sum of digits in base-3 representation of k). Note that (3^n)-1 (in base-3: "222...", with digit "2" repeated n times) is not included in the count, although the starting point (3^(n+1))-1 is.

			For n=0, we start from 3^(0+1) - 1 = 2 (also "2" in base-3), and subtract 2 to get 0, which doesn't begin with 2, thus a(0) = 1.
For n=1, we start from 3^(1+1) - 1 = 8 ("22" in base-3), and subtract 2*2 = 4 to get 4 ("11" in base-3) which doesn't begin with 2, thus a(1) = 1.
For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract first 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), so in two steps we have reached the first such number that does not begin with "2" in base-3, thus a(2) = 2.

Cf. A000244, A054861, A122586, A261230, A261234, A261236.

/* Use the C-program given in A261234. */

Flatten@ Table[FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &]] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

a(n)=my(k=3^(n+1)-1,t=2*3^n,s); while(k>=t, k-=sumdigits(k,3); s++); s \\ Charles R Greathouse IV, Aug 21 2015

(definec (A261237 n) (let loop ((k (- (A000244 (+ 1 n)) 1)) (s 0)) (if (< (A122586 k) 2) s (loop (* 2 (A054861 k)) (+ 1 s)))))

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016

cp's OEIS Frontend

A261236 a(n) = A261234(n) - A261237(n).

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A261235 First differences of A261234.

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A276622 Simple self-inverse permutation of natural numbers: after a(0)=0, list each block of A261234(n) numbers in reverse order, from A261232(n) to A261233(1+n).

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A276621 After a(0)=0, each n occurs A261234(n-1) times.

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A219661 Number of steps to go from (n+1)!-1 to n!-1 with map x -> x - (sum of digits in factorial base representation of x).

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A261233 a(n) = number of steps to reach 0 when starting from k = (3^n)-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

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A261231 a(n) = number of steps to reach 0 when starting from k = n and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

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A261230 a(n) = A261236(n) - A261237(n).

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